Kevin Michael Drees a Dissertation Submitted to the Graduate College

Cp(X, Z)

Kevin Michael Drees

A Dissertation

Submitted to the Graduate College of Bowling Green State University in partial fulﬁllment of the requirements for the degree of

DOCTOR OF PHILOSOPHY

August 2009

Committee:

Warren Wm. McGovern, Advisor

Christopher Rump, Graduate Faculty Representative

Rieuwert J. Blok

Kit C. Chan ii ABSTRACT

Warren Wm. McGovern, Advisor

We examine the ring of continuous integer-valued continuous functions on a topological space X, denoted C(X, Z), endowed with the topology of pointwise convergence, denoted

Cp(X, Z). We first deal with the basic properties of the ring C(X, Z) and the space Cp(X, Z). We find that the concept of a zero-dimensional space plays an important role in our studies. In fact, we find that one need only assume that the domain space is zero-dimensional; this is similar to assume the space to be Tychonoff when studying C(X), where C(X) is the ring of real-valued continuous functions. We also find the space Cp(X, Z) is itself a zero-dimensional space.

Next, we consider some speciﬁc topological properties of the space Cp(X, Z) that can be

characterized by the topological properties of X. We show that if Cp(X, Z) is topologically

isomorphic to Cp(Y, Z), then the spaces X and Y are homeomorphic to each other, this is much like a the theorem by Nagata from 1949. We show that if X is a zero-dimensional space,

then there is a zero-dimensional space Y such that X is embedded in Cp(Y, Z). Thus every zero-dimensional space can be viewed as a collection of integer-valued continuous functions. We consider and prove the collection of all linear combinations of characteristic functions on

clopen (open and closed) subsets is a dense subspace of Cp(X, Z). We then consider when

the space Cp(X, Z) are Gδ- and Fσ-subsets of the collection of all functions from X to Z (a

Gδ-subset is a countable intersection of open subsets and a Fσ-subset is a countable union of closed subsets).

We make classifications for when Cp(X, Z) is a discrete space, metrizable space, Fréchet- Urysohn space, sequential space, and k-space. We end with some results on cardinal invariants and the relationships between the tightness and Lindelöfnumbers of related spaces. iii

God created the integers, all the rest is the work of Man.

- Leopold Kronecker iv ACKNOWLEDGMENTS

There are a number of people that I would like to acknowledge at the completion of this manuscript, and my work on my dissertation. I am going to list them numerically, however the order in which names appear does not reﬂect the importance of their contribution to this work.

1. I would like to thank my parents, Richard and Karen, for all of their support during the endless years of college. My parents are the ones who made it possible for me to make it this far. I would also like to thank my siblings Scott, Jerry, and Sarah for putting up with me as a little brother.

2. I thank my advisor and friend Dr. Warren Wm. McGovern for trying to answer my endless stream of questions during my dissertation research. When seeking out a dissertation adviser you never know ahead of time what your relationship a person is going to be like; I am pleased that everything turned out positively. I am thankful that Dr. McGovern took me as a student; it would have been hard (impossible?) to ﬁnd an adviser as talented as he.

3. I thank my Dissertation Committee, Dr. Rieuwert Blok, Dr. Kit Chan, and Dr. Christo- pher Rump, for putting up with my many typos and bad grammar.

4. I thank my best friend and classmate Papiya Bhattacharjee for all the help she has given me on my research and for all the time that she has listened to me talk about it.

5. I would like to thank Dr. K.T. Arasu, Dr. Qingbo Huang, and Dr. Steen Pedersen of Wright State University, and Dr. Kit Chan, Dr. Corneliu Hoﬀman, Dr. Warren Wm. McGovern, and Dr. Sergey Shpectorov of Bowling Green State University for teaching me how to be a mathematician.

6. I would also like to thank the Department of Mathematics and Statistics at Bowling Green State University for funding me during my time as a student. v

Table of Contents

CHAPTER 1: Preliminaries 1 1.1 Introduction ...... 1 1.2 General Topology ...... 4 1.3 Groups, Rings, and Modules ...... 17 1.4 Cardinality ...... 22

CHAPTER 2: Properties of C(X,Y ) and Cp(X,Y ) 25

2.1 Topological Properties of Cp(X,Y ) and Cp(X) ...... 25 2.2 Algebraic Properties of C(X,A)...... 30

2.3 Topological Properties of Cp(X, Z) ...... 40

CHAPTER 3: New Results on Cp(X, Z) 53

3.1 Evaluation Maps and Cp(Cp(X, Z), Z)...... 53

3.2 The Space Lp(X, Z)...... 65

3.3 Density and Cp(X,A), Where A is a Ring ...... 75

3.4 Gδ, Fσ, and the Baire Category Theorem ...... 80

CHAPTER 4: Classiﬁcation of Cp(X, Z) Into Diﬀerent Categories of Spaces 88

4.1 Metrizablility of Cp(X, Z) ...... 88

4.2 Fr´echet-Urysohn Spaces and Cp(X, Z)...... 94

4.3 When is Cp(X, Z) a Sequential Space? ...... 99 vi

4.4 Tightness and Lindel¨ofProperties for Cp(X, Z) ...... 107

CHAPTER 5: Closing Remarks 115 5.1 Conclusions ...... 115 5.2 Open Questions ...... 116

BIBLIOGRAPHY 117 1

CHAPTER 1

Preliminaries

1.1 Introduction

During the late 1950’s the object C(X), the set of all real valued continuous functions on the topological space X, became of interest because of its algebraic and topological properties. The original concept of the ring of continuous functions was brought to the attention of the mathematics community when Weirstauss considered C(X) to be an algebra. According to Henriksen there was a renewed interest in this object brought about by Cechˆ and Stone in 1937, who independently wrote the papers On Bicompact Spaces and Applications of the Theory of Boolean Rings to General Topology, respectively (refer to [8] and [22]). Since that time C(X) has been used as a source for examples and counter-examples of statements about rings with desired algebraic aspects. Furthermore, C(X) was investigated from an analytical point of view. First, we will discuss the algebraic aspects of C(X). The book Rings of Continuous Functions by Gillman and Jerison [13] is a milestone in the study of C(X) as a ring. Since C(X) is a collection of real-valued functions, addition and multiplication can be defined on the set by pointwise addition and multiplication of functions. Hence C(X) forms a ring. The ring theoretic structure of C(X) has a direct relationship with the topological properties 2 that the space X possesses. When studying C(X) it is usually assumed that the space X is a Tychonoff space, since given any space X there exists a Tychonoff space Y such that C(X) is ring isomorphic to C(Y ). Recall that a space X is Tychonoff if is Hausdorff and completely regular. Refer to [11] for the definitions of the above terms. Gillman and Jerison [13] gives a detailed look at the ring structure of C(X) but does not consider a topology on the set C(X). In the late 1970’s a number of people began studying the topology of pointwise convergence on C(X). The topology of pointwise convergence has as basic open subsets, sets of the form

W (F, ε, f) = {g ∈ C(X): |f(x) − g(x)| < ε, for all x ∈ F } where f ∈ C(X), ε > 0, and F is a ﬁnite subset of X. For notational convenience, we write

Cp(X) when the set C(X) is endowed with the topology of pointwise convergence. Note that the topology of pointwise convergence is the subspace topology on C(X) inherited from the Q product topology on x∈X R. The topology of pointwise convergence is considered because it is the weakest of the many natural topologies that can be placed on C(X). For example, if X is a compact space, then C(X) is a Banach space under the uniform norm and one of the natural topologies to consider would be the weak topology. However, the major reason why one would use the topology of pointwise convergence is that every Tychonoff space X can be viewed as a collection of real-valued continuous functions on a space Y with the topology of pointwise convergence (refer to page 9 of [3]). That is, for every Tychonoff space X there exists a Tychonoff space Y such that X is homeomorphic to a subspace of Cp(Y ). A. V. Arkhangel0ski˘ıwrote the book Topological Function Spaces [4] detailing the topological structure of the space Cp(X). The structure of Cp(X) is very well known at this point in time. It is know when Cp(X) is metrizable, a Fréchet-Urysohn space, a sequential space, a k-space, and countably tight; all of these are based on the choice of X. There are other properties that have been characterized, for example, when the space Cp(X) is Lindelöf, 3 σ-compact, σ-countably compact, and normal. Thus, the topological side of C(X), with the pointwise topology, is a well studied object also. Our interest in this dissertation is to given a detailed investigation of the lesser studied

C(X, Z), the set of integer-valued continuous functions on X. In the early 1960’s Alling [1] and Pierce [20] wrote the ﬁrst papers on C(X, Z). They were concerned with the ring structure. Their research was inspired by Gillman and Jerson’s book [13]. However, when making the change of R to Z as the codomain for the functions it has a drastic change on the structure of the ring. The trace of a maximal ideal in C(X) is a minimal prime ideal in

C(X, Z). It also inherits some properties from Z; C(X, Z) is always a Bézoutring, that is, every finitely generated ideal is principle. However, to have a plentiful amount of elements in C(X, Z) to make it interesting you need to assume that the space X is disconnected. In fact, it will be seen that the space X can be assumed to be a zero-dimensional space when dealing with C(X, Z). This conclusion is parallel to the assumption that the space X is a Tychonoff space when dealing with C(X). As in the case of C(X) assuming the space X to be Tychonoff, Pierce [20] demonstrated that for C(X, Z) one can assume the space is totally disconnected.

Since C(X, Z) is a subset of C(X) we can endow it with the subspace topology of pointwise convergence inherited from Cp(X). It is straightforward to check that this is the same as the Q subspace topology inherited from x∈X Z. When we want the properties of this topology we will write Cp(X, Z). A basic open set of Cp(X, Z) is a set of the form {f ∈ C(X, Z): f(xi) = ni, i = 1, . . . , k} where xi ∈ X and ni ∈ Z, and n ∈ N. The papers of Pierce [20], Alling [1], and Martinez [17] have shown that C(X, Z) behaves diﬀerently than C(X) with respect to its algebraic properties. We will show that Cp(X, Z) behaves diﬀerently with respect to topological properties. For example Cp(X, Z) is zero-dimensional for any space X while Cp(X) is never zero-dimensional. Though they do have their similarities. A number of natural properties that the topological space may have will be studied in regard to Cp(X, Z).

There will be a characterization for when Cp(X, Z) is a discrete space, is a metrizable, is a 4 Fréchet-Urysohn space, when it is a sequential and when it is a k-space. Along with these the connection between the Lindelöfnumber and tightness will be shown. We will consider other cardinal invariants such as the character and weight, and place a bound on the the density of certain spaces. Along the way we will find that there are a number of results that can be generalized to the object Cp(X,A), where A is a zero-dimensional topological ring.

1.2 General Topology

There are several standard sets that will be used, they are

(i) N, the set of natural numbers,

(ii) Z, the set of integers,

(iii) Q, the set of rational numbers, and

(iv) R, the set of real numbers.

There will be more said about these four standard sets as we proceed. We can construct new sets out of given sets. Let A and B be sets; BA is the set of all

A functions from A to B. Note that B may also be viewed as the product Πa∈AB, that is,

A we can view the element f ∈ B as the element (f(a))a∈A ∈ Πa∈AB. For a set S the power set, denoted P(S), is the collection of all subsets of S, i.e P(S) = {A : A ⊆ S}. If A and B are sets and f ∈ BA is the constant function deﬁned by f(a) = b, for all a ∈ A, we will denote f by b, that is, b ∈ BA means b(a) = b for all a ∈ A. The characteristic function for a subset S of a subset X, is the function χS : X → {0, 1} given by

  1 if x ∈ S χS(x) =  0 if x ∈ X\S. 5 Later we will deﬁne this in greater generality. If S and T are sets with a function f : S → T

and A is a subset of S, then the restriction of f to A, denoted f|A, is the function f|A : A → T such that f|A(x) = f(x) for all x ∈ A. Let S be a set, a partial order on the set S is a relation ≤ on S such that

(i) a ≤ a, for all a ∈ S, (reﬂexive)

(ii) if a, b ∈ S with a ≤ b and b ≤ a, then a = b, (anti-symmetric) and

(iii) if a, b, c ∈ S with a ≤ b and b ≤ c, then a ≤ c (transitive).

A partially ordered set (or poset for short) is a set S with a partial order relation ≤.

Example 1.2.1. Let X be a set, and let F be a family of subsets of X. The inclusion relation on F makes (F, ⊆) into a poset.

If we have a particular kind of relation on a set we can construct a topology on the set called the ordered topology. First, we need to deﬁne a linear order; a linear order on a set the X is a relation < on X which has the following properties for x, y, x ∈ X:

(i) If x < y and y < z, then x < z.

(ii) If x < y, then the relation y < x does not hold.

(iii) If x 6= y, then x < y or y < x.

The set X together with the linear order relation < is called a well-order set. We have that the power set of a set S is a collection of subsets of S, we can consider other collections of subsets S and what properties that they may have.

Deﬁnition 1.2.2. A topological space is a set X with a collection T of subsets of X which has the following properties:

(i) ∅,X ∈ T , 6

(ii) if Oα ∈ T for α ∈ I, then ∪α∈I Oα ∈ T , where I is any index set, and

n (iii) if Oi ∈ T for i = 1, . . . , n, then ∩i=1Oi ∈ T .

We denote a topological space as (X, T ), where T is called the topology on the set X.

We deﬁne a point as an element in the set for the topological space. It then follows that for every set X there always exists a topology, namely the power set. The discrete topology on set X is the topology on X which is precisely P(X). The set X is said to have the indiscreete topology if the topology is the collection {∅,X}. Note that there are many other topologies for the set X, in general.

Remark 1.2.3. When we say X is a space we mean that X is a set equipped with a topology, i.e. we suppress the topology notation. A notational note: throughout X, Y , and Z will be topological spaces, unless otherwise stated.

Given a topological space (X, T ) any set A ∈ T is called open and the complement of an open set is called closed. A set which is both open and closed is clopen. The closure of a subset S in a space X is

clX (S) = ∩{A : S ⊆ A and A is a closed subset of X}.

Observe that the closure of a subset S is the smallest closed subset containing S.

Lemma 1.2.4 (Theorem 17.5, [19]). If (X, T ) is a space and S is a subset of X, then

clX (S) = {x ∈ X : for all x ∈ O ∈ T when O ∩ S 6= ∅}

Remark 1.2.5. If X is a space with A and B subsets of X such that A ⊆ B, it is then clear

from the deﬁnition of closure that clX (A) ⊆ clX (B).

Let X be a space, and let S be a subset of X. We say that x ∈ X is a limit point of S (also known as an accumulation point or cluster point) if every open subset of X containing x also contains a point in S other than x.A sequence is a map f from N into a set S, 7

if f(n) = xn we will simply write {xn}n∈N as the sequence. We say the sequence {xn}n∈N converges to the point x ∈ X provided that for all open subsets U of X containing x there is a N ∈ N such that xn ∈ U for all n ≥ N. If the sequence {xn}n∈N converges to X we use the notation xn → x as n → ∞. We say that S is dense in X if clX (S) = X. S is called co-dense in X if X\S is dense in X and S called nowhere dense in X if clX (S) is co-dense in X.

Proposition 1.2.6 (Proposition 1.3.5, [11]). Let X be a space and S a subset of X, S is dense in X if and only if every non-empty open subset of X contains a point of S.

Let X be a set. A base is a collection, B, of subset of X such that the collection of arbitrary unions of elements of B form some topology on X. Let (X, T ) be a space. A base for the topology T is a subcollection B of T such that

(i) For all x ∈ X there is a U ∈ B such that x ∈ U.

(ii) If x ∈ U1 ∩ U2, where U1,U2 ∈ B, there is a U3 ∈ B such that U3 ⊆ U1 ∩ U2.

If (X, T ) is a space and B is a base for the space we say that B generates the topology T . We then have the following useful lemma.

Lemma 1.2.7 (Lemma 13.2, [19]). Let X be a space. If C is a collection of open sets of X such that for each open set U of X and x ∈ U, there is an element C ∈ C such that x ∈ C ⊆ U, then C is a base for X.

A subbase is a collection of sets in the topology such that the set of ﬁnite intersections of members of the collection forms a base for the topology. Let x be a point in space X, let Bx be a collection of open subsets of X containing x, if for every open set O containing x there is a U ∈ Bx such that x ∈ U ⊆ O we then say that Bx is a base at point x, i.e. a base at a point.

Deﬁnition 1.2.8. A topological space is said to be zero-dimensional if it has a base of clopen sets. 8 A few standard topological spaces, with their bases, which will be used are:

(i) R, with the base {(a, b): a < b, a, b ∈ R},

(ii) Q, with the base {(a, b) ∩ Q : a < b, a, b ∈ R},

(iii) Z, with the base {{k} : k ∈ Z}, and

(iv) N, with the base {{n} : n ∈ N}.

For the four spaces above the topology given will be called the usual topology for the re- spective spaces.

Example 1.2.9. A few examples of zero-dimensional spaces are N, Z, and Q all with the usual topology. However, R with the usual topology is not zero-dimensional. Also, it is clear that every discrete space is a zero-dimensional space.

We can make new topological spaces based on old topological spaces. If X is a topological

space with topology T and Y is a subset of X, then the set TY = {U ∩ Y : U ∈ T } is a topology on Y called the subspace topology. If X is a space and Y is a subset of X with the subspace topology we call Y a subspace. If P is a property of subsets of topological space, then a P subspace should be understood to mean a subspace that also satisﬁes P .

Let {Xi}i∈I be a collection of topological spaces, with index set I, then the product

topology is the topology on Πi∈I Xi given by the base

{Πi∈I Ui : Ui is an open subset of Xi and Ui = Xi for all but ﬁnitely many i ∈ I}.

There is a another way to create a new topology based on old ones, that is the sum of spaces. Let {Xi}i∈I be a collection of spaces such that Xi ∩ Xj = ∅ when i 6= j, the sum of

the spaces is the set X = ∪i∈I Xi where U is an open set in X if U ∩ Xi is open in Xi for all

i ∈ I. The sum of the spaces {Xi}i∈I is denoted ⊕i∈I Xi or X1 ⊕ ... ⊕ Xn if I = {1, . . . , n}. 9 If X is a set linearly ordered set we deﬁne a bounded interval in X as (x, y) = {z ∈ X : x < z < y} for x, y ∈ X. Thus, we deﬁne the order topology on the set X with the base {(x, y): x, y ∈ X, x < y}.

Take note that the usual topology for Z and N is the discrete topology and that the topology on Q, Z, and N all have the subspace topology of R. Moreover, the usual topologies on N, Z, Q, and R are all ordered topologies using the usual ordering of the real numbers.

Lemma 1.2.10. If X is zero-dimensional, then every subspace is zero-dimensional.

Proof. Since X is zero-dimensional it has a clopen base BX . Since BX is a base for X it is

clear that the collection BY = {Y ∩ O : O ∈ BX } is a base for the subspace topology on Y .

Now, we need just to verify that BY is a collection of clopen sets of Y . We already have that

Y ∩ O is open in Y for all O ∈ BX by the deﬁnition of a subspace. So we need to show that

Y ∩ O is a closed subset of Y for all O ∈ BX . Since all O ∈ BX are closed subsets of X we

have that Y ∩ O are closed subsets of Y , thus Y ∩ O is clopen in Y , for all O ∈ BX . Hence, it follows that BY is a clopen base of Y . Therefore Y is zero-dimensional.

Let (X, T ) and (Y, S) be topological spaces. A function f : X → Y is continuous at the point x in X if for every V ∈ S with f(x) ∈ V , there is a U ∈ T such that x ∈ U and f(U) ⊆ V .A continuous function is a map f : X → Y such that f −1(O) ∈ T for all O ∈ S. For example the constant functions are always continuous. A map is said to be bicontinuous if it is continuous with a continuous inverse map. An open map is a function f from the space X into the space Y such that f(O) is an open subset of Y whenever O is an open subset of X. Likewise, a closed map is a function f from the space X into the space Y such that f(V ) is a closed subset of Y whenever V is a closed subset of X. The notion C(X,Y ) denotes the set of all continuous functions from the space X into the space Y . Moreover, when dealing with real-valued continuous functions we write C(X)

instead of C(X, R). If X is a space and f ∈ C(X), f is said to be a bounded function if there exists M ∈ R such that |f(x)| ≤ M for all x ∈ X. The set of bounded real-valued 10 continuous functions on X is denoted C∗(X). Likewise, C∗(X, Z) is the set of all bounded integer-valued continuous functions.

Lemma 1.2.11. Let X, Y , and Z be spaces with continuous maps f : X → Y and g : Y → Z. The composition map g ◦ f is also a continuous map.

Proof. Let O be an open subset of Z. We want to show (g ◦ f)−1(O) is an open subset of X. By the continuity if g we have that g−1(O) is an open subset of Y . Then, by the continuity of f, f −1(g−1(O)) is an open subset of X. However, (g ◦ f0−1 = f −1(g−1(O)) giving g ◦ f be a continuous function.

Let X and Y be spaces, a function f from X onto Y is a quotient map when f −1(U) is open in X if and only if U is open in Y (refer to page 137 in [19]). A special map which is also continuous is the identity map on a space X is the map idX : X → X deﬁned by idX (x) = x, for all x ∈ X.

Remark 1.2.12. Let X be a discrete space and Y any space. Any map f : X → Y is a continuous map since f −1(U) is a subset of X, hence open in X, for any U ⊆ Y . Thus, it follows that C(X,Y ) = Y X .

A function between two topological spaces is called a homeomorphism if it is bijective and bicontinuous. If there is a homeomorphism between spaces the X and Y , then X and Y are said to be homeomorphic. Consider the following characterization.

Proposition 1.2.13 (Proposition 1.4.18, [11]). Let X and Y be spaces. For a continuous bijective map f : X → Y the following are equivalent:

(i) The map f is a homeomorphism.

(ii) The map f is an open map.

(iii) The map f is a closed map. 11 Now for some facts about restrictions of continuous functions.

Proposition 1.2.14 (Theorem 18.2, [19]). Let X and Y be spaces.

(i) If f : X → Y is continuous and A is a subspace of X, then f|A : A → Y is also continuous.

(ii) The map f : X → Y is continuous on X if X can be written as the union of open sets

{Ui}i∈I such that f|Ui is continuous for all i ∈ I.

Lemma 1.2.15 (Theorem 18.3, [19]). (Pasting Lemma) Let X and Y be spaces. If V1 and

V2 are closed subsets of X and f1 : V1 → Y and f2 : V2 → Y are continuous maps such that f1(x) = f2(x) when x ∈ V1 ∩ V2, then the map f : V1 ∪ V2 → Y deﬁned by

  f1(x), if x ∈ V1 f(x) =  f2(x), if x ∈ V2 is a continuous map.

We have the following characterization for clopen subsets.

Proposition 1.2.16. Let X be a space and U a subset of X. The set U is a clopen subset of X if and only if χU is a continuous map.

Proof. First, we consider {0, 1} equipped with the discrete topology.

(⇒) Suppose that U is a clopen subset of X. It follows that the functions f1 : U → {0, 1} given by f1(x) = 1 and f2 : X\U → {0, 1} given by f2(x) = 0 are both continuous since they are constant functions. Thus, it follows from the Pasting Lemma that χU is continuous since   f1(x), if x ∈ U χU (x) =  f2(x), if x ∈ X\U. 12

(⇐) Suppose that χU is a continuous map. We have that {0} and {1} are both open in χ−1 χ−1 {0, 1}, thus we have that U ({1}) = U and U ({0}) = X\U are both open subsets of X. Hence, we have that U is a clopen subset of X.

A useful collection of function when dealing with the product topology are the projection

functions. Let {Xi}i∈I be a collection of spaces and Πi∈I Xi the product of the spaces. The

th j projection map is πj :Πi∈I Xi → Xj such that πj((xi)i∈I ) = xj, for all j ∈ I. Notice that

πj is surjective for each j ∈ I. Also notice that πj is continuous for all j ∈ I and that it is an open mapping. We then have the following.

Proposition 1.2.17 (Theorem 19.6, [19]). Let {Xi}i∈I be a collection of spaces with index set I and Y a space with fj : Y → Xj for all j ∈ I, and f : Y → Πi∈I Xi by f(y) = (fi(y))i∈I . f is continuous if and only if fj is continuous for all j ∈ I.

A subset Y of a space X is connected if there are no disjoint open subsets U1 and U2 of

X such that Y = U1 ∪ U2, Y is disconnected if such open subsets exist.

Proposition 1.2.18 (Theorem 23.5, [19]). The continuous image of a connected set is connected.

Let X be a space, an open cover of S, a subset of X, is a collection {Ui}i∈I of open subsets of X such that S ⊆ ∪i∈I Ui. An open subcover is a subcollection of an open cover which is also an open cover. We say a space X is a compact space if for every open cover there is a finite open subcover. A space is countably compact if every countable open cover has a finite subcover. A space is called a Lindelöfspace if for every open cover there is a countable open subcover. A space X is called locally compact if for every x ∈ X there exists

an open subset U of X containing x such that clX (U) is a compact subspace of X. There are diﬀerent ways that spaces can be separated; we will list a few typical spaces that will be used and introduce additional spaces as need. A space X is a Hausdorﬀ space if 13

for distinct points x1 and x2 in X there are open subsets O1 and O2 of X such that x1 ∈ O1

and x2 ∈ O2 with O1 ∩ O2 = ∅.

Example 1.2.19. Let X = {a, b, c}, and let T = {∅, {a},X}. The topological space (X, T ) is a non-Hausdorﬀ space.

Remark 1.2.20. A few remarks about Hausdorﬀ spaces.

(i) Each of the spaces R, Q, Z, and N is a Hausdorﬀ space under the usual topology.

(ii) A ﬁnite Hausdorﬀ space is discrete.

(iii) For each x ∈ X, the subset {x} is a closed subset of X.

From this point on we assume that all spaces are Hausdorﬀ.

Another class of spaces which are used in the study of C(X) are the Tychonoff spaces. A Tychonoff space is a Hausdorff space X that has the property that for every closed set V of X and a point x ∈ X\V , there exists a f ∈ C(X) such that f(V ) = {0} and f(x) = 1. We will then notice the following propositions. The second of these tells us the relationship between zero-dimensional space and Ty- chonoff spaces.

Proposition 1.2.21. Every zero-dimensional space is a Tychonoﬀ space.

Proof. Let X be a zero-dimensional space. Take V to be a closed set of X and let x ∈ X\V . Since V is closed we have that X\V is open, thus there exists a clopen subset U of X such

that x ∈ U ⊆ X\V . So the map χX\U has the properties that χX\U (V ) = {1}, χX\V (x) = 0, and this is a continuous real-valued function.

A Gδ-set is a set which is a countable intersection of open sets. A P -space is a space where every Gδ-set is open.

Lemma 1.2.22. Let X be a space. X is a P -space if every countable union of closed sets is closed. 14

Proof. Let {On}n∈N be a countable collection of open subsets of X. Set Vn = x\On for all

n ∈ N. Thus, ∪n∈NVn is a closed subset of X. However, we have that following:

\ \ [ On = (X\Vn) = X\ Vn. n∈N n∈N n∈N

Therefore, ∩n∈NOn is an open subset of X implying that X is a P -space.

A Tychonoﬀ space X is called pseudocompact if C(X) = C∗(X). A Tychonoﬀ space X

is called Z-pseudocompact if C(X, Z) = C∗(X, Z). A space X is called locally compact if for

every x in X there exists an open subset O of X containing x such that clX (O) is a compact subspace of X.

Example 1.2.23. Here are a few examples of spaces with diﬀerent compactness properties.

(i) The interval [0, 1] with the subspace topology from R is a compact space.

(ii) R with the usual topology is a Lindel¨ofspace, which is not compact.

(iii) Let X be a countable set and let p ∈ X. Deﬁne a topology on X such that the open subsets of X are X and all the subsets not containing p. It follows that this is a countably compact space that is not compact. Refer to page 47 in [21].

Remark 1.2.24. Every compact space is a countably compact space. Also, the continuous image of a compact set is compact.

Now for a result dealing with Lindel¨ofspaces and P -spaces.

Lemma 1.2.25 (Proposition 4.4.9, [2]). If X is a Lindel¨of P -space, then so is Xn for all

n ∈ N.

Let X be a space, a partition of X is a collection {Ui}i∈I of subsets of X such that

X = ∪i∈I Ui and Ui ∩ Uj = ∅ when i, j ∈ I and i 6= j.A clopen partition of the space X is a partition {Ui}i∈I where Ui is a clopen subset of X for all i ∈ I. 15 We can construct a new topological space based on a a given one when we have an equivalence relation. Let X be a space, and let ∼ be an equivalence relation on the set X. We say that [x] = [y] if and only if x ∼ y, that is they are in the same equivalence class. We will use the notation X/ ∼ to represent the set of all equivalence classes of X under the relation ∼ and define the function q : X → X/ ∼ defined by q(x) = [x], the equivalence map of x. We define the quotient topology on X/ ∼ to be the family of all subset U of X/ ∼ such that q−1(U) is open in X. It is clear by the definition of the quotient topology that q is a continuous map. Let X be a set. We call ρ a metric if ρ is a function from X × X to [0, ∞) which satisfies the following properties for all x, y, z ∈ X:

(i) ρ(x, y) = 0 if and only if x = y,

(ii) ρ(x, y) = ρ(y, x), and

(iii) ρ(x, z) ≤ ρ(x, y) + ρ(y, z).

If X is a set and ρ is a metric on X we say that (X, ρ) is a metric space with the topology

generated by basic open subsets of the form Br(x) = {y ∈ X : ρ(x, y) < r}, where x ∈ X and r > 0. A topological space is metrizable if there is a metric which generates the topology (refer to page 47 in [5]). The class of zero-dimensional spaces are going to be used extensively through out the following sections, thus there are several properties about zero-dimensional that will be needed for later use. Firstly, the product of zero-dimensional spaces is a zero-dimensional space.

Proposition 1.2.26. If X is zero-dimensional space and I is an index set, then XI is zero-dimensional space.

Proof. Since X is zero-dimensional space, X has a base B of clopen sets. We want to show 16

that BI , where

BI = {Πi∈I Ui : Ui ∈ B for i ∈ F and Ui = X for i ∈ I\F , where F is ﬁnite},

I I forms a clopen base for X . Clearly all the members of BI are clopen in X . Now let

I I I x = (xi)i∈I ∈ X and let U be an open subset of X containing x. Since U is open in X there is a ﬁnite subset F of I such that U = Πi∈I Ui where Ui is open in X, for i ∈ F , and

Ui = X for i ∈ I\F with xi ∈ Ui, for all i ∈ I. Since each Ui is open in X, for i ∈ F , there is a Oi ∈ B such that xi ∈ Oi ⊆ Ui, for all i ∈ F . Thus, x ∈ Πi∈I Ei ⊆ U, where Ei = Oi for

I i ∈ F and Oi = X for i ∈ I\F . Therefore, X has a clopen base, namely BI .

Next we a result about a clopen subbase.

Lemma 1.2.27. If X is a space with a subbase of clopen subsets of X, then X is zero- dimensional.

Proof. Suppose that X is a space with a subbase S of clopen sets. Since S is a subbase we then have that [ B = {U1 ∩ ... ∩ Un : Ui ∈ S, i = 1, . . . , n} n∈N

is a base for X. Let U1,...,Un ∈ S for n ∈ N. Notice that U1 ∩ ... ∩ Un is a clopen subset

of X since Ui is clopen for each i = 1, . . . , n. Thus, X is zero-dimensional as it has a clopen base.

We also have the following result that will be used repeatedly without mention.

Proposition 1.2.28. If X is a zero-dimensional Hausdorﬀ space with x1, . . . , xn ∈ X are distinct, then there exists V1,...,Vn basic clopen subsets of X, such that xi ∈ Vi, for i =

1, . . . , n, with Vi ∩ Vj = ∅ whenever i 6= j.

Proof. Let x1, . . . , xn ∈ X be distinct. Since X is Hausdorﬀ there exists disjoint open subsets

0 0 Oi,j and Oi,j of X such that xi ∈ Oi,j and xj ∈ Oi,j, for i 6= j. Set Ui = ∩j6=iOi,j, for all 17

i = 1, . . . , n, we see that Ui is an open subset of X such that xi ∈ Ui, for all i, and xj ∈/ Ui,

for j 6= i. Let Vi be a basic clopen subset of X such that xi ∈ Vi ⊆ Ui, for all i. Therefore,

n we have that xj ∈/ Vi, for i 6= j, and ∩i=1Vi = ∅.

1.3 Groups, Rings, and Modules

A binary operation on a set S is a function φ : S × S → S.A group is a set G with a binary operation ∗ which satisﬁes for all a, b, c ∈ G

(i) (a ∗ b) ∗ c = a ∗ (b ∗ c) (associativity),

(ii) there is an element eG ∈ G such that eG ∗ a = a ∗ eG = a (eG is called the identity of G), and

−1 −1 −1 (iii) there exists a a ∈ G such that a ∗ a = a ∗ a = eG.

A group G with binary operation ∗ is called Abelian if it also satisﬁes a ∗ b = b ∗ a, for all a, b ∈ G. The notation for a group is (G, ∗), if the group is an Abelian group we use the notation (G, +); we will use −a to mean the inverse of a in an Abelian group and b − a to mean b + (−a). A subset of a group which is also a group is called a subgroup under the given operation. A ring is an Abelian group (A, +) with a second binary operation · which satisﬁes for all a, b, c ∈ A

(i) a · (b · c) = (a · b) · c,

(ii) a · (b + c) = a · b + a · c, and

(iii) (a + b) · c = a · c + b · c.

The notation for a ring is (A, +, ·). The identity element of the Abelian group (A, +) will

be denoted as 0A, i.e. the additive identity. If there is an a ∈ A such that a · b = b · a = b for 18

all b ∈ A, such an element if called the multiplicative identity and will be denoted by 1A.A ring A is commutative if for all a, b ∈ A, a · b = b · a. An element a of a ring A with identity

is called a unit if there exist a b ∈ A so that a · b = 1A.A subring is a subset of a ring, with

1A, which is itself a ring with the same binary operations. A ﬁeld is a commutative ring where every element is a unit.

We will assume all rings are commutative with the identity element 1A and

0A 6= 1A.

Remark 1.3.1. The set Z is a commutative ring with 1 with respect to the usual addition and multiplication. The sets Q and R are both ﬁelds with respect to the usual addition and multiplication. However, N is not a ring under usual addition and multiplication as it is not a group under addition.

Suppose A is a ring, a non-zero element a ∈ A is called a zero divisor if there is a non-zero

element b ∈ A such that a · b = 0A. A ring A is called an integral domain if it has no zero divisors. Let A be a ring. An ideal of A is a non-empty subset I of A which has the following properties:

(i) r · a ∈ I for all r ∈ A and a ∈ I, and

(ii) a − b ∈ I for all a, b ∈ I.

A maximal ideal of A is an ideal M of A such that if N is another ideal of A with M ⊆ N, then M = N or N = A.A minimal ideal is an ideal I of A such that if J is another ideal of A with J ⊆ I, then I = J or J = {0A}.A prime ideal is an ideal I such that if a · b ∈ I, then a ∈ I or b ∈ I. An A-module over a ring A is an Abelian group (M, +) with a map · : A × M → M which satisﬁes for all a, b ∈ A and m, n ∈ M

(i) a · (m + n) = a · m + a · n, 19 (ii) (a + b) · m = a · n + b · m,

(iii) (ab) · m = a · (b · m), and

(iv) 1A · m = m.

Pk Let M be an A-module. If m1, . . . , mk ∈ M and a1, . . . , ak ∈ A, then the sum i=1 aimi is an A-linear combination. A subset E of an A-module M is called linearly independent whenever 0M = a1m1 + ... + akmk, for ai ∈ A and mi ∈ E, then a1 = ... = ak = 0. A subset E of an A-module M is an A-basis if each element of M can be expressed as a linear combination of elements of E and E is linearly independent. A homomorphism is a function between groups (rings, fields, modules) which preserves the operation of the groups (rings, fields, modules), i.e. if f is a group homomorphism between (G, ·, 0G) and (H, ◦, 0H ), then f(a · b) = f(a) ◦ f(b). If there is a homomorphism between two groups (rings, fields, modules), then the groups (rings, fields, modules) are said to be homomorphic. An isomorphism is a bijective homomorphism. If there is an isomorphism between two groups (rings, fields, modules) G and H, then G is isomorphic to

H and denote thus by G ∼= H. Let f be a group (ring, ﬁeld) homomorphism from G to H. The kernel of f is the set

ker(f) = {g ∈ G : f(g) = eG} and the image of f is the set im(f) = {f(g): g ∈ G}. Take note that if f is a homomorphism on ring A, then ker(f) is an ideal of A. We can construct new rings from given ones; the ﬁrst of these constructions is called the quotient ring. Let A be a ring and I an ideal of A, then we call A/I the quotient ring with elements of the form a + I for a ∈ A, where a + I = {a + i : i ∈ I}. The quotient ring has the operations

(a + I) + (b + I) = (a + b) + I and (a + I) · (b + I) = a · b + I 20

for all a, b ∈ A. Another way to create a new ring is the direct product. Let {Ai}i∈I be a collection of rings. The direct product of rings is the set Πi∈I Ai together with componentwise addition and multiplication. The following theorem gives the relationship between the kernel and image of a ring homomorphism.

Theorem 1.3.2 (Theorem 7.3.7, [10]). (First Isomorphism Theorem for Rings) Suppose that

A and B are rings and φ is a homomorphism from A to B, then A/ker(φ) ∼= im(φ).

We can classify certain ideals based on the characteristics of their quotient ring. The ﬁrst classiﬁers a maximal ideal.

Proposition 1.3.3 (Proposition 7.4.12, [10]). Let A be a ring and M an ideal of A. M is a maximal ideal if and only if A/M is a ﬁeld.

We also have a ways of classifying a prime ideal.

Proposition 1.3.4 (Proposition 7.4.13, [10]). Let A be a ring and P an ideal of A. P is a prime ideal if and only if A/P is a integral domain.

Since groups, rings, ﬁelds, and modules are based on a set, together with some operations, a topology may be given to the sets which form these algebraic structures. A topological group is a group (G, ·) that is also a topological space such that · : G × G → G and −1 : G → G are continuous functions. Likewise, a topological ring is a ring (A, +, ·) that is also a topological space such that + : A × A → A, −1 : g → g, and · : A × A → A are continuous maps. Note that a topological ring isomorphism is a bicontinuous ring isomorphism of topological rings. A topological module is an A-module M on a topological ring A and (M, +) is a topological group such that · : A × M → M is a continuous map. A linear homeomorphism is a homomorphism between two topological modules such that it is a homeomorphism. Two topological modules are linearly homeomorphic if there is a linear homeomorphism between the topological modules. 21 Remark 1.3.5. Clearly Z, Q, and R are all topological rings under the usual addition and multiplication.

If A is a topological ring and I is an ideal of A we will give the ideal I the subspace topology induced by the topology of A; it is clear that I is also a topological ring. Moreover, if A is a topological ring and I is an ideal of A, then the ring A/I will be given the quotient topology, with the map q : A → A/I given by q(r) = r + I.

Proposition 1.3.6 (Problem 2.5(d), [19]). If A is a topological ring and I is an ideal of A, then the quotient ring A/I with the quotient topology is a topological ring.

From this point onward we will assume that ideals of topological rings have the subspace topology and that quotient rings have the induced quotient topology. We also assume that our topological rings are Hausdorﬀ. Since we now have an algebraic structure which has a topology on it we can talk about continuous functions into the algebraic structure. Let (A, +, ·) be a topological ring; we can consider C(X,A) for a space X. Since A is a ring we see that C(X,A) is also a ring with the pointwise addition and multiplication of the ring, that is

(i) (f + g)(x) = f(x) + g(x) and

(ii) (f · g)(x) = f(x) · g(x) for all f, g ∈ C(X,A) and for all x ∈ X. Thus, with this in mind we can start to talk about the algebraic structure of C(X,A). Since we are dealing with rings we can consider certain subsets of the space X. For all topological rings A and f ∈ C(X,A), we deﬁne z(f) = {x ∈ X : f(x) = 0A} to be the zero set of f.A cozero set, denoted coz(f), as that is coz(f) = X\z(f) for all f ∈ C(X,A). Notice that z(f) is a closed subset of X and coz(f) is an open subset of X for all f ∈ C(X,A) since singleton subsets are closed in a Hausdorﬀ space. 22 1.4 Cardinality

We assume the Axiom of Choice. The cardinality of a set, loosely, indicates the size of the set. We will express the cardinality of a set S as |S|. If S and T are sets with a bijective function f : S → T , we say that |S| = |T |. If there is a bijection between the set S and the set {1, . . . , n}, then |S| = n, where n ∈ N. The cardinality of N is denoted ℵ0. If a set has cardinality ℵ0 it is called countably infinite. The cardinality of the first infinite uncountable set is denoted ℵ1.

Remark 1.4.1. The sets Z and Q both have cardinality ℵ0, that is, they are both countably inﬁnite.

Let S be a set with cardinality τ. If κ ≤ τ, then we deﬁne [S]<κ = {T ∈ P(S): |T | < κ}.

Thus, it follows that [S]<ℵ0 is the collection of all finite subsets of S. It is possible to add and multiply cardinalities. Suppose that τ and γ are infinite cardinals, without loss of generality, assume τ < γ. Define

(i) γ + τ = τ + γ = γ and

(ii) γτ = τγ = γ.

Let S and T be two disjoint sets where |S| = τ and |T | = γ, the set S ∪ T has cardinality γ + τ. Also, if S and T where |S| = τ and |T | = γ, the cardinality of S × T is γτ. Refer to [15] for more on the addition and multiplication of cardinals. Deﬁne a cardinal invariant as a function F which assigns to each space X a cardinal F (X) such that F (X) = F (Y ) whenever X and Y are homeomorphic spaces. We now deﬁne several cardinal invariants. Our base references for cardinal invariants are [4] and [11].

Definition 1.4.2. The Lindelöfnumber of X, denoted l(X), is defined to be the smallest infinite cardinal τ such that any open cover of X contains a subcover with cardinality less than or equal to τ. 23

It is obvious from the deﬁnition that l(X) = ℵ0 if X and only if X is a Lindel¨ofspace. The following lemma will be useful to us later.

Lemma 1.4.3. If X is a space and Y is a closed subspace of X, then l(Y ) ≤ l(X).

Proof. Suppose that l(X) ≤ τ. Let ηY be a open cover of Y . Since Y is closed we have that

X\Y is open in X. It follows that ηX = η ∪ {X\Y } is a cover for X. Since ηX is an open

∗ ∗ ∗ cover for X there is a subcover ηX of ηX for X such that |ηX | ≤ τ. Let ηY = {U|U ∩ Y 6=

∗ ∗ ∅, where U ∈ ηX }. Clearly ηY ⊆ ηY and |ηY | ≤ τ. Thus, l(Y ) ≤ l(X).

Deﬁnition 1.4.4. The tightness of X denoted t(X) is the smallest cardinality τ such that

for any set A ⊆ X and any point x ∈ clX (A) there is a set B ⊆ A for which |B| ≤ τ and

x ∈ clX (B).

Proposition 1.4.5. If X is a space and Y is a closed subspace of X, then t(Y ) ≤ t(X).

Proof. Assume t(X) = τ. Let A be a subset of Y and y ∈ clY (A). First, notice that

A ∩ Y = A and clX (A ∩ Y ) = clY (A) with y ∈ clX (A ∩ Y ), since Y is a closed subspace of

X. Now, since t(X) = τ there is a B ⊆ A = A ∩ Y with |B| ≤ τ and y ∈ clX (B). Further

note clX (B) = clY (B) is closed subset of X and Y , as Y is a closed subspace of X. Thus,

we have that clX (B) is closed in Y . Therefore, B ⊆ A with |B| ≤ τ with y ∈ clY (B) and consequently t(Y ) ≤ τ.

Let us now define the character and weight of a space. For a space X and x in X, the character of a point x in X, denoted χ(x, X), is the least infinite cardinal number of a base of neighborhoods around x. The character of a space X, denoted χ(X), is the least infinite cardinal number τ such that χ(x, X) ≤ τ for all x ∈ X. The weight of a space X, denoted w(X), is the infinite cardinality of a minimal base for a space

Remark 1.4.6. If X is a space and Y a subspace of X, then w(Y ) ≤ w(X). 24 A space X is said to satisfy the ﬁrst axiom of countability, or is ﬁrst countable, if χ(x, X) ≤

ℵ0 for all x ∈ X. Also, a space X is said to satisfy the second axiom of countability, or is

second countable, if w(X) ≤ ℵ0.

A space X is called separable if there is a subset A with |A| = ℵ0 such that clX (A) = X. There is a more general concept of separable, the density of a space X is minimal inﬁnite cardinality of the subset A with clX (A) = X, we denote the density of a space X by d(X).

Clearly, if X is a separable space, then d(X) = ℵ0. A network in a space X is a collection S of subsets of the set X such that any point x ∈ X and any open set U containing x there is a P ∈ S such that x ∈ P ⊆ U. There is a second type of weight which is considered, this is the concept of network weight for a space X, denoted nw(X), this is the minimal cardinality of the networks of X.

Lemma 1.4.7. If X is a space, then d(X) ≤ nw(X).

Proof. Suppose that nw(X) = τ. Let S be a network of X such that |S| = τ and let S be indexed by the set I. Thus, it follows that |I| = τ. Take xi ∈ Pi for all i ∈ I where Pi ∈ S and let A = {xi}i∈I .

Suppose, by way of contradiction, clX (A) 6= X. Take x ∈ X\clX (A), notice that

X\clX (A) is an open set in X. Since x ∈ X and X\clX (A) is open there is a Pj ∈ S

such that x ∈ Pj ⊆ X\clX (A). Thus, xj ∈ X\clX (A) and xj ∈ A by deﬁnition, giv-

ing a contradiction. Hence, clX (A) = X, so d(X) ≤ |A| = τ. Therefore, it follows that d(X) ≤ nw(X).

We also have the concepts of τ-monolithic, a space X is said to be τ-monolithic if

nw(clX (A)) ≤ τ for every A ⊆ X such that |A| ≤ τ. In particular, a space X is ℵ0- monolithic if the closure of every countable subset in X is a space with a countable network weight. A space is call monolithic if it is a τ-monolithic for every cardinal τ, that is for every Y ⊆ X we have d(Y ) = nw(Y ). 25

CHAPTER 2

Properties of C(X,Y ) and Cp(X,Y )

2.1 Topological Properties of Cp(X,Y ) and Cp(X)

In this section we are going to provide a number of general properties for the collection of continuous functions with the topology of pointwise convergence. We will also state some known results for C(X) which we will prove analogous results to in our study of C(X, Z)

and Cp(X, Z).

Let X and Y be topological spaces. Deﬁne Cp(X,Y ) as the set C(X,Y ) endowed with the topology of pointwise convergence, also called the pointwise topology. A basic open subset

of Cp(X,Y ) has the form

W (x1, . . . , xn,U1,...,Un) = {f ∈ C(X,Y ): f(xi) ∈ Ui, i = 1, . . . , n},

where x1, . . . , xn ∈ X and U1,...,Un are open subsets of Y , and n ∈ N.

We will state some general properties for Cp(X,Y ), ﬁrst of which deals with the base for

Cp(X,Y ).

Proposition 2.1.1. Let X and Y be spaces and let B be a base for the space Y . The collection

{W (x1, . . . , xn,U1,...,Un): xi ∈ X,Ui ∈ B, i = 1, . . . , n, n ∈ N} 26

is a base for the space Cp(X,Y ).

Proof. Let f ∈ W (x1, . . . , xn,U1,...,Un), where where x1, . . . , xn ∈ X and U1 ...,Un are open subsets of Y . For each i = 1, . . . , n there is a basic open set Oi ∈ B such that f ∈ Oi ⊆ Ui. Consequently, f ∈ W (x1, . . . , xn,O1,...,On). Notice

W (x1, . . . , xn,O1,...,On) ⊆ W (x1, . . . , x1,U1,...,UN ).

Therefore, the collection

{W (x1, . . . , xn,U1,...,Un): xi ∈ X,Ui ∈ B, i = 1, . . . , n, n ∈ N}

is a base for the space Cp(X,Y ).

The next property relates subspaces of Y to certain subspaces of Cp(X,Y ).

Proposition 2.1.2. If X and Y are spaces and Z is a subspace of Y , then Cp(X,Z) is a

subspace of Cp(X,Y ). Furthermore, if Z is a closed subspace of Y , then Cp(X,Z) is a closed

of Cp(X,Y ).

Proof. Let f ∈ C(X,Z); it is clear that f ∈ C(X,Y ). Observe that for x1 . . . , xn ∈ X and

U1,...,Un open subsets of Z we have, with a straightforward argument, that

W (x1, . . . , xn,U1,...,Un) = Cp(X,Z) ∩ W (x1, . . . , xn,V1,...,Vn)

where Vi is an open subset of Y with Vi ∩ Z = Ui.

Now, assume that Z is a closed subset of Y . Let f ∈ Cp(X,Y )\Cp(X,Z). There is a x ∈ X such that f(x) ∈ Y \Z. Then f ∈ W (x, Y \Z). Notice there for g ∈ W (x, Y \Z) we

have g∈ / Cp(X,Z). Hence, Cp(X,Y )\Cp(X,Z) is an open subset of Cp(X,Y ), so Cp(X,Z) is a closed subspace. 27 Let X, Y , and Z be spaces and f : X → Y be a continuous map, denote the composition map by f ] : C(Y,Z) → C(X,Z) and given by f ](ϕ) = ϕ ◦ f, for all ϕ ∈ C(Y,Z). We will

] now state a useful fact about the composition map f from Cp(Y,Z) into Cp(X,Z), for a ﬁxed f ∈ C(X,Y ); more properties of the composition map will be stated later.

Proposition 2.1.3. If X, Y , and Z are spaces and f : X → Y is a continuous map, then

] f is a continuous map from Cp(Y,Z) to Cp(X,Z).

] Proof. Let ϕ ∈ Cp(Y,Z) and f (ϕ) = ψ. Consider a basic open set W (x1, . . . , xn,U1,...,Un)

of Cp(X,Z) containing ψ. Then

] ϕ(f(xi)) = (ϕ ◦ f)(xi) = f (ϕ)(xi) = ψ(xi).

It is clear that we have (ϕ ◦ f)(xi) = ψ(xi) ∈ Ui for all i = 1, . . . , n. With some rewriting

we see that ϕ(f(xi)) ∈ Ui. Thus

ϕ ∈ W (f(x1), . . . , f(xn),U1,...,Un),

where W (f(x1), . . . , f(xn),U1,...,Un) is a basic open subset of Cp(Y,Z). Let

g ∈ W (f(x1), . . . , f(xn),U1,...,Un),

] it follows f (g)(xi) = (g ◦ f)(xi) ∈ Ui for all i = 1, . . . , n. Providing

] f (g) ∈ W (x1, . . . , xn,U1,...,Un).

Hence

] f (W (f(x1), . . . , f(xn),U1,...,Un)) ⊆ W (x1, . . . , xn,U1,...,Un).

Consequently, f ] is continuous at the point φ. Since φ is arbitrary, the inverse image of an open subset is open. Hence, f ] is a continuous map. 28 We now recall two theorems involving the pointwise topology, the product space, and the topological sum.

Theorem 2.1.4 (Proposition 2.6.10, [11]). If X and Y are spaces and I is an index set,

I I then Cp X,Y is homeomorphic to Cp(X,Y ) .

Theorem 2.1.5 (Proposition 2.6.9, [11]). If X and Y are spaces and X is a topological sum, L say X = i∈I Xi, then Cp(X,Y ) is homeomorphic to Πi∈I Cp(Xi,Y ).

We will now consider the case when the codomain of the continuous functions, on a space X, is a topological ring. Let A be a topological ring, we already know that C(X,A) is a

ring, a natural consideration is whether or not Cp(X,A) is a topological ring when given the pointwise topology.

Theorem 2.1.6 (Remark 4.1.20, [6]). Let X be a space, and let A be a topological ring. The Q direct production x∈X A of A is a topological ring when endowed with the product topology.

Corollary 2.1.7. Let X be a space and A a topological ring. The space Cp(X,A) is a topological ring.

Q Proof. By Theorem 2.1.6 we have that x∈X A is a topological ring. Recall that C(X,A) is a Q subspace of x∈X A with the same operations. Thus, it follows that Cp(X,A) is a topological ring since continuous functions restricted to subspaces are continuous.

Now we will consider some facts about Cp(X). Notice that Cp(X) is a speciﬁc case of

Cp(X,Y ). Since we know that R has a base for the usual topology, we deﬁne a basic open subsets of Cp(X) by a basic open subsets of R, this follows from Proposition 2.1.1. Let f ∈ C(X), let x1, . . . , xn ∈ X, and let 0 < ε. Deﬁne a basic open subset of Cp(X) as

W (f, x1, . . . , xn, ε) = {g ∈ C(X): |f(xi) − g(xi)| < ε, i = 1, . . . , n}

and

{W (f, x1, . . . , xn, ε)|f ∈ C(X), x1, . . . , xn ∈ X, ε ∈ (0, ∞)}n∈N 29 is a base for Cp(X), refer to [4].

Let X be a topological space and let F be a subspace of RX , for every x ∈ X we deﬁne the evaluation map as ex : F → R given by ex(f) = f(x), for all f ∈ F. The following proposition states that the evaluation map is a continuous map.

Proposition 2.1.8 (Proposition 0.5.1, [4]). For a space X and subspace F ⊆ RX the map ex : F → R is continuous for all x ∈ X.

F For a space X, we also deﬁne the natural map eˆF : X → R bye ˆF (x) = ex, where

F ⊆ Cp(X). The natural map is a continuous function as well.

Proposition 2.1.9 (Proposition 0.5.2, [4]). For any space X and F subspace of Cp(X), the map eˆF : x → Cp(F) is continuous.

We know, by Proposition 2.1.8, that ex is a continuous map into R and, by Proposition

2.1.9, thate ˆF is a continuous map into Cp(F). So, in particular ex : Cp(X) → R and

Cp(X) eˆCp(X) : X → R are continuous maps. Now, since ex is a continuous map, we in fact have thate ˆCp(X) is a continuous mapping into Cp(X), thereforee ˆCp(X) is a map from X into

Cp(Cp(X)).

Proposition 2.1.10. For any space X, R is homeomorphic to a subspace of Cp(X).

Proof. Let C = {r ∈ C(X): r ∈ R} with the inherited subspace topology of Cp(X). If

W = W (f, x1, . . . , xn, ε), where x1, . . . , xn ∈ X and ε > 0, is a basic open subset of Cp(X), then C ∩ W is a basic open subset of C. We ﬁnd

C ∩ W = {r : |r(xi) − f(xi)| < ε, i = 1, . . . , n}.

Furthermore, if s ∈ C, then a basic open set of C has the form

{r ∈ C : |s(xi) − r(xi)| < ε, i = 1, . . . , n}. 30

Deﬁne ϕ : R → Cp(X) by ϕ(r) = r. It is clear that ϕ is a bijection between R and

C. We claim ϕ is bicontinuous. Let W = W (r, x1, . . . , xn, ε) be an basic open subset of C, where r ∈ C. It is clear that ϕ−1(W ) = (r − ε, r + ε). Consider the evaluation function

ex : Cp(X) → R restricted to C for a ﬁxed x ∈ X. It is clear that ex|C is an inverse function for ϕ and is continuous by Proposition 2.1.8. Thus R is homeomorphic to the subspace C.

Therefore, R is homeomorphic to a subspace of Cp(X).

With the results from this section in hand we now have a general idea of the behavior of

Cp(X,Y ) along with some standard results that we will call upon as needed.

2.2 Algebraic Properties of C(X,A)

In the study of C(X) it has been found that the ring structure of C(X) is related to the topological properties of the space X. There are several papers describing the ring structure of C(X, Z), refer to [20], [1], and [17]. In this section we are going to consider the ring properties of C(X,A), where A is a topological ring. The work done here is going to be help in our study of Cp(X,A) and Cp(X, Z) since the ring structure turns in to a use tool. We have the following theorem from Gillman and Jerison [13] which tells us what kind of spaces which need to be concerned when dealing with C(X).

Theorem 2.2.1 (Theorem 3.9, [13]). For every topological space Y , there exists a Tychonoﬀ

space X such that C(X) ∼= C(Y ).

When considering the case of C(X) there is a connection between the structure of the ring and the properties that space X have. The following proposition tells us something

about the space X in relation to a property of some functions in C(X, Z).

Proposition 2.2.2. Suppose X is a Tychonoﬀ space. X is a zero-dimensional space if and only if whenever x ∈ X and V is a closed subset of X not containing x there is some

f ∈ C(X, Z) such that f(x) = 1 and f(V ) = {0}. 31 Proof. (⇒) Assume that X is zero-dimensional. Let x ∈ X and V be a closed subset of X not containing x. Since X has a base of clopen sets there exists a basic clopen subset U of

X such that x ∈ U ⊆ X\V , that is U ∩ V = ∅. Consider the function χU , it follows that

χU (x) = 1 and χU (V ) = {0}. Also, since U is a clopen subset of X we have that χU is continuous. Hence, there exists f ∈ C(X, Z) such that f(x) = 1 and f(V ) = {0}. (⇐) Assume that whenever x ∈ X and V is a closed subset of X not containing x there is some f ∈ C(X, Z) such that f(x) = 1 and f(V ) = {0}. Let U be an open subset of X containing x. We have that x∈ / X\U and X\U is closed subset of X. By assumption there is an f ∈ C(X, Z) such that f(x) = 1 and f(X\U) = {0}. Since f is continuous and {0} is clopen in Z, we have that f −1({0}) is clopen in X. We also have that V ⊆ f −1({0}) with x∈ / f −1({0}). So, x ∈ X\f −1({0}). Hence, x is contained in a clopen set contained in U. Therefore, X has a base of clopen sets and so X is a zero-dimensional space.

We would like to note the relationship between clopen subsets and the characteristic function, in a more general case.

Lemma 2.2.3. Let X be a zero-dimensional Hausdorﬀ space, let V be a subset of X, and

let A be a zero-dimensional topological ring. V is a clopen subset of X if and only if χV ∈ C(X,A).

Proof. (⇒) Suppose V is a clopen subset of X. It is clear that X\V is also clopen in X.

Since χV is constant on V and X\V it is continuous on both V and X\V . Thus, it follows from the Pasting Lemma that χV ∈ C(X,A).

(⇐) Suppose χV is continuous from X to A. Let U be a clopen subset of A such that it χ−1 contains 1A and not 0A. It follows that V (U) is a clopen subset of X since U is a clopen χ−1 subset of A. Thus V is a open subset of X as V = V (∪α∈I Uα), where {Uα : α ∈ I} is the

collection of all clopen subsets of A which contain 1A and does not contain 0A. Likewise, X\V is an open subset of X. Therefore, V is a clopen subset of X.

Thus, it is clear that you need only consider Tychonoﬀ spaces when you are looking at 32 C(X). We have a similar result which tells us what kind of spaces you need to consider when

dealing with the C(X, Z) case, refer to [20]. The theorem stated is more general than what is needed and tells us more since we are going to apply the topology of pointwise convergence.

Theorem 2.2.4. Let A be a zero-dimensional topological ring. For every space Y , there

exist a zero-dimensional space X such that Cp(X,A) topologically isomorphic to Cp(Y,A). In particular, C(X,A) ∼= C(Y,A).

Proof. Let A be a zero-dimensional topological ring, and let BA be the collection of all clopen

subsets of A; BA is a clopen base for a topology on A.

Let Y be a space. We will create an equivalence relation on Y as follows: y1 ∼A y2 if

and only if f(y1) = f(y2) for all f ∈ C(Y,A). Let X = Y/ ∼A, that is, X the is set of

all equivalence class of Y under the relation ∼A. Let p : Y → X be the map which takes elements of Y into its equivalence class, i.e. p(y) = [y]. We presently construct a topology on X. First, deﬁne F = {g ∈ AX : g ◦ p ∈ C(Y,A)}.

−1 Let BX = {g (B): g ∈ F and B ∈ BA}. BX is an open subbase for a topology for X.

−1 We claim that BX is in fact a clopen subbase for X. Let C ∈ BX ; by deﬁnition C = g (B) for some g ∈ F and B ∈ BA. Since B ∈ BA it follows that A\B ∈ BA as well. Thus,

−1 −1 −1 g (A\B) ∈ BX . However, X\C = X\g (B) = g (A\B), thus C is a closed subset of X

−1 as g (A\B) ∈ BX . It follows that BX is in fact a clopen subbase. Therefore, the topology generated by BX is a zero-dimensional topology on X. We now claim the composition map p] : C(X,A) → C(Y,A) deﬁned be p](f) = f ◦p is in fact a ring isomorphism. We ﬁrst show that p] is a ring homomorphism. Let f, g ∈ C(X,A), we then have the following

p](f + g) = (f + g) ◦ p = (f ◦ p) + (g ◦ p) = p](f) + p](g) and p](f · g) = (f · g) ◦ p = (f ◦ p) · (g ◦ p) = p](f) · p](g). 33 Next, we show that p] is a bijection. Firstly, we show that p] is injective. Suppose f, g ∈ C(X,A) such that f 6= g. It follows that there is an x ∈ X such that f(x) 6= g(x). Since x ∈ X there is a y ∈ Y such that p(y) = x (by the construction of X). Thus, it follows that f(p(y)) 6= g(p(y)), which gives (f ◦ p)(y) 6= (g ◦ p)(y), i.e. we have that p](f) 6= p](g), and so p] is injective. Next, to show that p] is surjective, let f ∈ C(Y,A). We ﬁrst note

that for all x ∈ X, f(y1) = f(y2) for all y1, y2 ∈ [x]. Define g : X → A by g(x) = f(y) for all y ∈ [x]. By what we just stated g is well-defined. Furthermore, g ∈ C(X,A) since it is the restriction of the continuous function f to the quasi-components of Y . We see that (g ◦ p)(y) = g(p(y)) = g(x) = f(y0) = f(y) for all y0 ∈ p−1({p(y)}). Since p is clearly a surjective map it follows that (g ◦ p)(y) = f(y) for all y ∈ Y , hence g ◦ p = f. Also, from the definition of g, it is clear that g is a continuous map, so g ∈ C(X,A). Thus, we have that p](g) = f. Therefore, p] is surjective. Consequently, p] is a ring isomorphism. We claim that p] is in fact a homeomorphism. We need to first show that p is a continuous map. Take a basic open subset O of X, by construction of the topology on X, O has the

−1 0 form g (B) where g ∈ F and B ∈ BA. Then

p−1(O) = p−1(g−1(A)) = (p−1 ◦ g−1)(A) = (g ◦ p)−1(A).

However, recall that g ◦ p ∈ C(Y,A) by the deﬁnition of F, thus we have that (g ◦ p)−1(B) is open in Y . Hence p−1(O) is open in Y . Consequently, we have that p is a continuous map from Y to X. It then follows from Proposition 2.1.3 that p] is a continuous map. We now need to verify that (p])−1 is a continuous map, to show this we will show that p]

is an open map. Let W = W (x1, . . . , xk,U1,...,Uk) be a basic open set of Cp(X,A), where

x1, . . . , xk ∈ X and U1,...,Uk are basic clopen sets in A. Since p is a surjective map, there 34

is an yi ∈ Y such that p(yi) = xi for all i = 1, . . . , k. Now we have the following:

] ] p (W ) = p (W (x1, . . . , xk,U1,...,Uk))

= {f ◦ f : f ∈ W (x1, . . . , xk,U1,...,Uk)}

= {f ◦ p : f ∈ W (p(y1), . . . , p(yk),U1,...,Uk)}

= {g : g ∈ W (y1, . . . , yk,U1,...,Uk)}

= W (y1, . . . , yk,U1,...,Uk).

] ] Thus, p send basic open subsets of Cp(X,A) to basic open subsets of Cp(Y,A). Hence p is

] −1 an open map. It then follows that (p ) is a continuous map from Cp(Y,A) to Cp(X,A).

] Therefore, p is a homeomorphism between Cp(X,A) and Cp(Y,A). Moreover, it follows ∼ from Theorem 2.2.4 that Cp(X,A) = Cp(Y,A) by this map, thus Cp(X,A) is topologically

isomorphic to Cp(Y,A).

Note that the corollary also follows from Theorem 2.1 in [9].

Corollary 2.2.5. For every space Y , there exists a zero-dimensional space X such that

Cp(X, Z) topologically isomorphic to Cp(Y, Z).

Before we continue, we will make a note about the equivalence relation used in the proof of Theorem 2.2.4. We recall the definition of the connected components of a space. There are several different refinements for the connected components. Let X be a space, and let x ∈ X.A quasi-component of x is the intersection of all clopen subsets of X containing x, refer to page 356 in [11]. The collection of quasi-components is a partition of X, so we can

deﬁne a equivalences relation on X as follows: x ∼q y if and only if x and y are in the same clopen subsets of X.

Proposition 2.2.6. Let X be a space, and let A be a zero-dimensional Hausdorﬀ topological

ring. The relations ∼q and ∼A are the same. 35 Proof. Let X be a space and let A be a zero-dimensional Hausdorﬀ topological ring. We

shall demonstrate that if x1 ∼q x2, then x1 ∼A x2, and conversely.

(⇒) Let x1 and x2 be in X, with x1 6= x2, such that x1 ∼q x2. Since x1 ∼q x2 we have that x1 and x2 are contained in the same clopen subsets of X. Suppose, by way of contradiction, that x1 A x2, that is, there is a f ∈ C(X,A) such that f(x1) 6= f(x2). Since

A is zero-dimensional Hausdorﬀ there are disjoint clopen subsets V1 and V2 of A such that

−1 −1 −1 −1 f(x1) ∈ V1 and f(x2) ∈ V2. Thus x1 ∈ f (V1) and x2 ∈ f (V2) with f (V1)∩f (V2) = ∅,

−1 −1 and both f (V1) and f (V2) are clopen subsets of X. Hence, x1 and x2 do no lie in the

same clopen subset of X, this is a contradiction with x1 ∼q x2. Therefore, x1 ∼A x2.

(⇐) Let x1 and x2 be in X, with x1 6= x2, such that x1 ∼A x2. It then follows that for all f ∈ C(X,A) we have that f(x1) = f(x2). Suppose, by way of contradiction, that x1 q x2.

Since x1 q x2 there is a clopen subset U of X such that x1 ∈ U and x2 ∈/ U. Consider the

function f = χU ; we have that f ∈ C(X,A) and 1A = f(x1) 6= f(x2) = 0A. Thus we have a contradiction and therefore x1 ∼q x2.

We will note the importance of the assumption that the topological ring A be Hausdorﬀ.

Suppose X is a connected space and A = R with the indiscrete topology. Observe that A is still a zero-dimensional topological ring which is not a Hausdorﬀ space. It then follows that

X/ ∼A= X with the discrete topology and X/ ∼q= {x} with the discrete topology.

Example 2.2.7. With Corollary 2.2.5 in hand we have the following:

(i) For any connected space X, X/ ∼q= {x} equipped with the discrete topology. In ∼ ∼ particular, C(R, Z) = C({x}, Z) = Z;

∼ (ii) C(X, Z) = C({x1, . . . , xn}, Z), where X has n quasi-components and {x1, . . . , xn} has the discrete topology.

For the remainder of this section we assume that all spaces are zero-dimensional and Hausdorff, unless otherwise stated. 36 Let X be a zero-dimensional space, and let A be a zero-dimensional topological ring. We Pn χ define an A-simple function to be a function ϕ : X → A given by ϕ = i=1 ai Bi , where ai ∈ A and Bi is a clopen subset of X for all i = 1, . . . , n such that Bi ∩ Bj = ∅ for i 6= j. Note, that it follows from the fact that characteristic functions defined on clopen sets of X are continuous maps and that C(X,A) is closed under addition, that A-simple functions are in C(X,A). For B ⊆ A, the collection of all B-simple functions on X will be denoted as S(X,B). Notice that S(X,B) ⊆ S(X,A) ⊆ C(X,A); in fact S(X,A) is a subring of C(X,A).

Let X be a zero-dimensional space, and let f ∈ C(X, Z). The collection

−1 P = {f ({n})}n∈Z

is a clopen partition of X. Also, since P is a collection of clopen sets the characteristic

function χV is a continuous map from X to Z for all V ∈ P. Thus, any f ∈ C(X, Z) can be P expressed as the sum of characteristic functions, in particular f(x) = f(xn)χ −1 (x) n∈Z f ({n}) −1 for all x ∈ X, where xn is any element in f ({n}) for all n ∈ Z. We will state some more facts about simple functions deﬁne on zero-dimensional spaces.

Proposition 2.2.8. Let X be a zero-dimensional space. The set S(X,A) is a subring of C(X,A).

Proof. Let X be a zero-dimensional space and let ϕ, ψ ∈ S(X,A). By the deﬁnition of Pn χ Pk χ an A-simple function we have ϕ = i=1 ai Ai and ψ = j=1 bj Bj where ai, bj ∈ A and

Ai,Bj ∈ BA for i = 1, . . . , n and j = 1, . . . , k with Ai1 ∩ Ai2 = ∅ and Bj1 ∩ Bj2 = ∅ when

i1 6= i2 and j1 6= j2. First, we show addition and multiplication for characteristic functions. Let U and V be 37 clopen sets in X, with U ∩ V = ∅ and a, b ∈ A. We have the following

  a, if x ∈ U\V    a + b, if x ∈ U ∩ V aχU (x) + bχV (x) =  b, if x ∈ V \U    0, otherwise and   ab, if x ∈ U ∩ V aχU (x) · bχV (x) =  0, otherwise.

Thus, we have that aχU + bχV = aχU\V + abχU∩V + bχV \U and aχU · bχV = abχU∩V . Hence, aχU + bχV , aχU · bχV ∈ S(X,A). With the above in hand, we clearly have

n ! k ! n k X χ X χ X X χ ϕ · ψ = ai Ai bj Bj = aibj Ai∩Bj . i=1 j=1 i=1 j=1

Thus, ϕ · ψ ∈ S(X,A). So, S(X,A) is clearly closed under multiplication. The proof of the closure of addition is slightly, but here it is. Notice

n k X χ X χ ϕ + ψ = ai Ai + bj Bj i=1 j=1 n k n k X X X X = aiχ k + bjχB \(∪n A ) + (ai + bj)χA ∩B . Ai\(∪j=1Bj ) j i=1 i i j i=1 j=1 i=1 j=1

Thus, ϕ + ψ ∈ S(X,A). Therefore, S(X,A) is a subring of C(X,A).

Now to consider some properties on the ideals of C(X,A). We will make a note that Alling

[1] and Pierce [20] have described the ideal structure of the ring C(X, Z). We generalize some of their results. 38 Let x ∈ X. We deﬁne by

Mx = {f ∈ C(X, Z): f(x) = 0};

it is straightforward to check that Mx is an ideal of C(X,A). In particular, Mx is the kernel of the evaluation map ex : C(X,A) → A given by ex(f) = f(x). Compare our next result to Theorem 2.5 in [13].

Proposition 2.2.9. Let X be a zero-dimensional space, and let A be a topological ring.

Mx is a prime ideal of C(X,A) if and only if A is an integral domain. Moreover, Mx is a maximal ideal of C(X,A) if and only if A is a ﬁeld.

Proof. Consider the map ex : C(X,A) → A deﬁned by ex(f) = f(x) for all f ∈ C(X,A).

The map ex is a ring homomorphism as

ex(f + g) = f(x) + g(x) = ex(f) + ex(g) and

ex(fg) = (fg)(x) = f(x)g(x) = ex(f)ex(g),

for all f, g ∈ C(X,A). Furthermore, we have that ex(C(X,A)) = A as ex(a) = a(x) = a, for all a ∈ A, thus ex is surjective.

Notice that ker(ex) = {f ∈ C(X,A): f(x) = 0A} = Mx. Thus, C(X,A)/Mx isomorphic to A by the First Isomorphism Theorem for Rings. It follows from Proposition 1.3.4 that

Mx is a prime ideal of C(X,A) if and only if A is an integral domain. Therefore, Mx is a prime ideal of C(X,A). Moreover, it follows from Proposition 1.3.3 that Mx is a maximal ideal of C(X,A) if and only if A is a ﬁeld.

Recall the following lemma. 39 Lemma 2.2.10 (Lemma 1.1, [14]). Suppose A is a commutative ring and P is a prime ideal of A, P is a minimal ideal of A if and only if for each x ∈ P there exists a ∈ A\P such that

ax = 0A.

We then have the following.

Proposition 2.2.11. Let X be a zero-dimensional space, let x ∈ X, and let A be a discrete topological integral domain. The ideal Mx is minimal prime ideal of Cp(X,A).

Proof. Take f ∈ Mx, it follows that f(x) = 0A. Since {0A} is a clopen set of A we have that

−1 V = f ({0A}) is a clopen set in X, as f is continuous. Note that since V is a clopen subset

of X it follows that X\V is also clopen subset of X. We have that χV ∈ C(X,A), since V is a clopen subset of X.

Now since x ∈ V , it follows that χV (x) 6= 0A. Thus χV ∈ C(X, Z)\Mx. Lastly, we have from the choose of f and χV that χV (x) · f(x) = 1A · 0A = 0A for all x ∈ V and that

χV (x) · f(x) = 0A · f(x) = 0A for all x ∈ X\V , thus χV · f = 0. Therefore it follows from

Lemma 2.2.10 that Mx is a minimal prime ideal of C(X,A).

We can deﬁne an ideal for C(X, Z) based on an ideal of Z. This can not be done in the C(X) case as R is a ﬁeld and thus has no ideals other than R and {0}.

Remark 2.2.12. It is clear that a ﬁeld F has no ideals other than F and {0F }. Otherwise, if I is any other ideal of F , then a ∈ I is a unit. Thus, considering a−1 ∈ F we have 1 = aa−1 ∈ I. Therefore, r1 ∈ I for all r ∈ F , and so I = F .

The most important result to take from this section is Theorem 2.2.4, this is the theorem which tells use that when dealing with Cp(X, Z) zero-dimensionality is an important property. The property that X is a zero-dimensional space will be used extensively through out our studies. We have also shown that there are major diﬀerences between ideal structures of

C(X, Z) and C(X); what was a maximal ideal of C(X) is now a minimal prime idea of C(X, Z). These facts will be used repeatedly. 40 2.3 Topological Properties of Cp(X, Z)

We shall now consider some of the basic properties of C(X,A) when it is endowed with the pointwise topology. We will find similarities to results in the study of Cp(X), however most of the following results only hold for certain assumptions on A. Recall that Theorem 2.2.4 tells us that we need only consider X to be a zero-dimensional Hausdorff space when A is a zero-dimensional Hausdorff space.

We will ﬁrst consider a base for the space Cp(X, Z). We will denote the usual base for Z

by BZ given by {{n} : n ∈ Z}. Notice that it then follows from Proposition 2.1.1 that the

standard base for Cp(X, Z) is

{W (x1, . . . , xk,U1,...,Uk): xi ∈ X,Ui ∈ BZ, i = 1, . . . , k, k ∈ N}, where

W (x1, . . . , xk,U1,...,Uk) = {f ∈ C(X, Z): f(xi) ∈ Ui, i = 1, . . . , k}

for Ui ∈ BZ, and xi ∈ X with i = 1, . . . , k, for all k ∈ N. Since the basic open subsets of Z

consists of singleton sets, the standard basic open subsets of Cp(X, Z) can be expressed as subsets of the form

W (x1, . . . , xk, n1, . . . , nk) = {f ∈ C(X, Z): f(xi) = ni, i = 1, . . . , k},

where xi ∈ X and ni ∈ Z, for i = 1, . . . , k, where k ∈ N. Furthermore, the standard basic open subsets of Cp(Cp(X, Z), Z) have the form

V (f1, . . . .fk, n1, . . . , nk) = {F ∈ C(Cp(X, Z), Z): F (fi) = ni, i = 1, . . . , k}

where f1, . . . , fk ∈ C(X, Z), and ni ∈ Z for all i = 1, . . . , k, where k ∈ N. We will talk more about Cp(Cp(X, Z), Z) when the time comes. 41

Recall that if X is a zero-dimensional space and x ∈ X, we deﬁne Mx = {f ∈ C(X, Z):

f(x) = 0} and Mx is an ideal of the ring C(X, Z). We see that Mx is a clopen subset of

Cp(X, Z), for all x ∈ X, as the following result gives.

Proposition 2.3.1. Let X be a zero-dimensional space. The ideal Mx is clopen of Cp(X, Z), for each x ∈ X.

Proof. Let x ∈ X. Notice that Mx is open of Cp(X, Z) since Mx = W (x, 0). Now, let

g ∈ Cp(X, Z)\Mx, it follows that g(x) = n for some n ∈ Z\{0}. Hence, g ∈ W (x, n) which

is a basic open subset of Cp(X, Z) and notice that W (x, n) ∩ Mx = ∅. Thus, Cp(X, Z)\Mx

is open, so Mx is closed. Therefore, Mx is a clopen subset of Cp(X, Z)

We will now see that all basic open subsets of the form W (x, n) are clopen, for x ∈ X

and n ∈ Z.

Corollary 2.3.2. Let X be a zero-dimensional space. If x ∈ X and n ∈ Z, then W (x, n) is

a clopen subset of Cp(X, Z).

Proof. Let x ∈ X and n ∈ Z. We want to show that W (x, n) is clopen. First, notice that

since Cp(X, Z) is a topological ring, we have that clopen subsets of Cp(X, Z) are clopen

subsets of Cp(X, Z) under translations, because a translation is homeomorphic maps of

Cp(X, Z) to itself. Now, consider the translation of W (x, n) by n, that is

W (x, n) − n = {f − n ∈ C(X, Z): f(x) = 0} = {f ∈ C(X, Z): f(x) = 0} = Mx.

Thus, W (x, n) is a translation of Mx. Therefore W (x, n) is a clopen subset of Cp(X, Z) as

Mx is a clopen subset of Cp(X, Z).

We now notice that the intersection of a ﬁnite number of basic clopen subsets Cp(X, Z)

of the form W (x, n) generate an arbitrary basic open set of Cp(X, Z) and hence generate a

clopen subset of Cp(X, Z). 42

Lemma 2.3.3. Let X be a zero-dimensional space. If x1, . . . , xk ∈ X and n1, . . . , nk ∈ Z, then

W (x1, n1) ∩ ... ∩ W (xk, nk) = W (x1, . . . , xk, n1, . . . , nk).

Proof. Let x1, . . . , xk ∈ X and n1, . . . , nk ∈ Z. If f ∈ W (x1, n1) ∩ ... ∩ W (xk, nk), then f(xi) = ni for each i = 1, . . . , k. Thus, f ∈ W (x1, . . . , xk, n1, . . . , nk).

Conversely, if f ∈ W (x1, . . . , xk, n1, . . . , nk), then f(xi) = ni for i = 1, . . . , k. So, f ∈

W (xi, ni) for each i = 1, . . . , k, which gives us that f ∈ W (x1, n1)∩...∩W (xk, nk). Therefore

W (x1, n1) ∩ ... ∩ W (xk, nk) = W (x1, . . . , xk, n1, . . . , nk).

Since we now have Proposition 2.3.3 it can be said that Cp(X, Z) is always a zero- dimensional space.

Proposition 2.3.4. Let X be a zero-dimensional space. Cp(X, Z) is always zero-dimensional space.

Proof. Consider the collection B∗ = {W (x, n): x ∈ X, n ∈ Z}. It follows from Lemma 2.3.3

∗ that B is a subbase for Cp(X, Z). Furthermore, Lemma 2.3.3 also gives us that the elements

∗ in the collection B are clopen subsets of Cp(X, Z). Therefore, it follows from Lemma 1.2.27 that Cp(X, Z) is a zero-dimensional space.

Remark 2.3.5. We will make a note at this point that it can be see that Cp(X, Z) is a Hausdorﬀ space for any zero-dimensional space X. We can ﬁrst see this by taking f, g ∈

Cp(X, Z) such that f 6= g. It follows that there is a x ∈ X such that f(x) 6= g(x). So, by letting f(x) = nf and g(x) = ng, we have f ∈ W (x, nf ) and g ∈ W (x, ng). Thus it is clear that W (x, nf ) ∩ W (x, ng) = ∅. This provides that Cp(X, Z) is always a Hausdorﬀ space.

The fact that Cp(X, Z) is a zero-dimensional space is a direct result of Z being a zero- dimensional space. We will now see that Cp(X, Z) being a zero-dimensional space is property 43 which Cp(X) never shares. That is, Cp(X) is never be a zero-dimensional a, regardless of the choose of X.

Proposition 2.3.6. Let X be a Tychonoﬀ space, Cp(X) is not a zero-dimensional space.

Proof. Suppose, by way of contradiction, that Cp(X) is a zero-dimensional space. First, recall that R is a connected space. Let C = {r ∈ Cp(X): r ∈ R}, by Proposition 2.1.10 we have that R is homeomorphic to the subspace C of Cp(X). Since a subspace of a zero-dimensional space is also a zero-dimensional space, we have that the subspace C is a zero-dimensional space. It then follows that R is a zero-dimensional space since it is homeomorphic to C.

However, this would make R a disconnected space which is a contradiction. Hence, Cp(X) can not be a zero-dimensional space.

We can state a more general result in regards to Proposition 2.3.4.

Proposition 2.3.7. Let X and Y be a Tychonoﬀ spaces. Cp(X,Y ) is a zero-dimensional space if and only if Y is a zero-dimensional space.

Before we prove this, lets state the following lemma.

Lemma 2.3.8. Let X and Y be Tychonoﬀ spaces. There is a homeomorphic copy of Y inside of Cp(X,Y ).

Proof. Deﬁne ϕ : Y → Cp(X,Y ) be ϕ(y) = y, that is each point in Y is mapped to the constant function y. It is obvious that ϕ is an injective map, since any y1 and y2 in Y such that y1 6= y2 gives ϕ(y1) = y1 6= y2 = ϕ(y2). Furthermore, it is also clear that

ϕ(Y ) = {y ∈ Cp(X,Y ): y ∈ Y }.

Now we just need to verify that ϕ is a bicontinuous map. Let x1, . . . , x2 be in X and

U1,...,Un be open subsets of Y , it follows that W = W (x1, . . . , x2,U1,...,Un) is an basic

−1 open subset of Cp(X,Y ). We need to show that ϕ (W ) is open in Y . It is clear that

−1 ϕ (W ) is U1 ∩ ... ∩ Un which is an open subset of Y . Thus, it follows that ϕ is a continuous map. 44 Now, deﬁne ϕ← : ϕ(Y ) → Y by ϕ←(y) = y, we see that that ϕ(ϕ←(y)) = y and ϕ←(ϕ(y)) = y for all y ∈ Y . Hence ϕ and ϕ← are inverse functions of each other. Now,

← −1 let U be an open subset of Y , it is clear that (ϕ ) (U) = {y ∈ Cp(X,Y ): y ∈ U} which

← is an open subset of Cp(X,Y ). Thus, ϕ is a continuous map and consequently ϕ have a

continuous inverse map. Therefore, ϕ is a homeomorphic imbedding of Y into Cp(X,Y ).

Now for the proof of Proposition 2.3.7.

Proof of Proposition 2.3.7. (⇒) Assume that Cp(X,Y ) is a zero-dimensional space. By

Lemma 2.3.8 there is a homeomorphic copy of Y in Cp(X,Y ). It then follows from Lemma 1.2.10 that Y is a zero-dimensional space. (⇐) Assume that Y is a zero-dimensional space. Since Y is a zero-dimensional space

there exists a clopen base B for Y . Let x1, . . . , xn ∈ X and let U1,...,Un be basic clopen

subsets of Y . The set W (x1, . . . , x,U1,...,Un) is a standard basic open subset of Cp(X,Y ), moreover

W (x1,U1) ∩ ... ∩ W (xn,Un) = W (x1, . . . , x,U1,...,Un).

Hence, the sets of the form W (x, U), where x ∈ X and U is a basic clopen of Y , form a

subbase for Cp(X,Y ). We claim that W (x, U) is clopen, where is x ∈ X and U is in B. Let x ∈ X and U ∈ B,

we know that W (x, U) is an open subset of Cp(X,Y ) since it is a standard basic open subset

of Cp(X,Y ). Thus we need to show that it is also a closed subset. We wish to show that

Cp(X,Y )\W (x, U) is an open subset of Cp(X,Y ).

We claim that Cp(X,Y )\W (x, U) = W (x, Y \U). Since U is clopen, Y \U is also clopen.

Let f ∈ Cp(X,Y )\W (x, U), giving f(x) ∈/ U. Thus, f(x) ∈ Y \U, so f ∈ W (x, Y \U). Likewise, if f ∈ W (x, Y \U), then f(x) ∈ Y \U. Consequently, we have f(x) ∈/ U, giving that f ∈ Cp(X,Y )\W (x, U). So, it follows that Cp(X,Y )\W (x, U) = W (x, Y \U)

and Cp(X,Y )\W (x, U) is an open subset of Cp(X,Y ) since W (x, Y \U) is a open subset

of Cp(X,Y ). Finally, W (x, U) is a clopen subset of Cp(X,Y ). Therefore, the collection 45

of subsets of the form W (x, U) form a clopen subbase for Cp(X,Y ). It then follows from

Lemma 1.2.27 that Cp(X,Y ) is a zero-dimensional space.

Corollary 2.3.9. For every zero-dimensional space X, Cp(X, Q) is a zero-dimensional space.

We have that 0 has a base of clopen subsets of Cp(X, Z) which are in fact ideals of C(X, Z).

Proposition 2.3.10. Let X be a zero-dimensional space. The collection {Mx1 ∩ ... ∩ Mxn :

x1, . . . , xn ∈ X, n ∈ N} is a base of open sets around 0 in Cp(X, Z). Moreover, the collection consists of ideals of the ring C(X, Z).

Proof. Notice that 0 ∈ Mx for all x ∈ X. Thus, 0 ∈ Mx1 ∩ ...Mxn for all x1, . . . , xn ∈ X

for all n ∈ N. Hence the collection {Mx1 ∩ ... ∩ Mxn : x1, . . . , xn ∈ X, n ∈ N} is a base

of neighborhoods around 0. Moreover, since Mx is an ideal in C(X, Z), for each x ∈ X, it

follows that Xx1 ∩...∩Mxn is an ideal of C(X, Z). Therefore, {Mx1 ∩...∩Mxn : x1, . . . , xn ∈ X, n ∈ N} is a collection of ideals of C(X, Z).

Let A be a topological ring, a linear topology on A is a topology on that has an open

base at 0A which are ideals of the ring A. The beneﬁt of a linear topology on a topological ring is that the topology on the ring is constructed from the ideals of the ring, hence the topology comes from the ring structure. It follows from Proposition 2.3.10 that the topology of pointwise convergence on Cp(X, Z) is a linear topology on the ring C(X, Z). We can make this statement more general with that following proposition.

Proposition 2.3.11. Let X be a Tychonoﬀ space and let A be a topological ring. The space

Cp(X,A) has a linear topology if and only if A has a linear topology.

Proof. (⇒) Assume that Cp(X,A) has a linear topology. Deﬁne A = {a ∈ Cp(X,A): a ∈ A}, that is A is the set of constant functions from X to A, and let A have that subspace

topology of Cp(X,A). It is clear that A is topologically isomorphic to A as topological rings. 46

Let {Jα}α∈I be a base of open subsets of Cp(X,A) at 0 which are ideals of C(X,A), we have such a collection since Cp(X,A) has a linear topology. We now want to construct a base of open subsets of A at 0 which consists of ideals of A. Deﬁne Iα = Jα ∩ A for all

α ∈ I. We wish to show that {Iα}α∈I forms a base of open subsets of A at 0 consisting of ideals of A. First, we note that Iα is open in A, for all α ∈ I, since Jα is open in Cp(X,A) and A has the subspace topology of Cp(X,A). We now need to verify that Iα is an ideal for all α ∈ I and that {Iα}α∈I forms a base at 0.

First to show that Iα is an ideal for each α ∈ I.

(i) Let α ∈ I and let a, b ∈ Iα. We have that a, b ∈ Jα and a, b ∈ A. Thus, it follows that

a − b ∈ Jα, since it is an ideal and a − b ∈ A since it is a ring. Therefore, we have

a − b ∈ Iα.

(ii) Let α ∈ I, and let a ∈ Iα and b ∈ A. We have that a ∈ Jα and a ∈ A, by the deﬁnition

of Iα. Since b ∈ A we have that ab ∈ Jα, but a, b ∈ A so ab ∈ A. Thus, it follows that

ab ∈ Iα.

Hence, it follows that Iα is an ideal of A for all α ∈ I.

Lastly, we show that the collection {Iα}α∈I is a base of open subsets of A at 0. Let O be an open subset of A containing 0. Since A has the subspace topology of Cp(X,A) there is an open subset U of Cp(X,A) such that O = U ∩ A. It is clear that the subset U contains

0, thus there is a β ∈ I such that Jβ ⊆ U since {Jα}α∈I is a base at 0. Now, it follows that

Iβ ⊆ O, by construction. Thus, it follows that for each open subset O of A containing 0 there is an element of {Iα}α∈I containing 0 contained in O. Hence, the collection {Iα}α∈I is a base of open ideals of A at the point 0. Now, let ϕ be a topological isomorphism from A to A. It follows that {ϕ(Iα)}α∈I is a base of open ideals of A at the point 0A. Thus, A has a linear topology.

(⇐) Assume that A has a linear topology. We wish to show that Cp(X,A) has a linear topology. Let {Iα}α∈I be a base of ideals of A at the point 0A, for an index set I. In 47 particular, the collection {Iα}α∈I is a collection of open subsets of A. Consider the collection

B0 = {W (x1, . . . , xn,Iα1 ,...,Iαn ): xi ∈ X, αi ∈ I, i = 1, . . . , n, n ∈ N}

we have that B0 is a base at the point 0 in the topology on Cp(X,A) since it is a collection of basic open subsets of Cp(X,A) all containing 0.

We want to show that B0 is in fact a collection of ideals of C(X,A). Let O ∈ B0. Thus

O = W (x1, . . . , xn,Iα1 ,...,Iαn ) for some xi ∈ X, αi ∈ I, where i = 1, . . . , n and n ∈ N. Now to show that O is an ideal of C(X,A).

(i) Let f, g ∈ O. We have that f(xi), g(xi) ∈ Iαi for all i = 1, . . . , n. It follows that

f(xi) − g(xi) ∈ IαI for all i = 1, . . . , n since Iαi is an ideal of A. Hence, we have that f − g ∈ O.

(ii) Let f ∈ O and g ∈ Cp(X,A). We have that f(xi) ∈ Iαi and g(xi) ∈ A for all

i = 1, . . . , n. It follows that f(xi)g(xi) ∈ Iαi for all i = 1, . . . , n since Iαi is an ideal. Hence, we have that fg ∈ O.

Consequently, we have that O is an ideal of C(X,A). Thus, we have that B0 is a collection of ideals of C(X,A). Therefore it follows that Cp(X,A) has a linear topology.

We follow with some useful facts about the composition operator. Refer to page 14 in [4] for these properties of the composition operator in the Cp(X) case.

Proposition 2.3.12. Let X and Y be zero-dimensional spaces. For a surjective continuous map f : X → Y we have the following properties for the composition map f ]:

] (i) The map f is bijective if and only if f (C(Y, Z)) is dense in Cp(X, Z).

] (ii) The map f is a homeomorphism if and only if f (C(Y, Z)) = Cp(X, Z).

] (iii) If f is a quotient map, then f (Cp(Y, Z)) is a closed subspace of Cp(X, Z). 48 Proof. Let f : X → Y be a continuous surjective map.

(i) (⇒) Assume f is bijective. Let φ ∈ C(X, Z) and W (x1, . . . , xk, n1, . . . , nk) be a basic

clopen subset of Cp(X, Z) containing φ. Set yi = f(xi) for i = 1, . . . , k. Now, since f is

bijective there is a ψ ∈ Cp(Y, Z) such that ψ(yi) = φ(xi) for all i = 1, . . . , k. Thus, we have

] that f (ψ) ∈ W (x1, . . . , xk, n1, . . . , nk). Hence, it follows that W (x1, . . . , xk, n1, . . . , nk) ∩

f ](C(Y, Z)) = ∅. So, it follows that f ](C(Y, Z)) is dense in C(X, Z).

] (⇐) Assume that f (C(Y, Z)) is dense in Cp(X, Z). We have the assumption that the

function f is surjective, so all we need to show is that f is injective. Let x1, x2 ∈ X such that

x1 6= x2. Suppose, by way of contradiction, that f(x1) = f(x2). We have that (φ ◦ f)(x1) =

] (φ◦f)(x2) for all φ ∈ C(Y, Z), as f(x1) = f(x2). It follows that for g ∈ f (Cp(Y, Z)) we have

] g(x1) = g(x2). Take ψ ∈ Cp(X, Z) such that ψ(x1) 6= ψ(x2), it is clear that ψ∈ / f (Cp(Y, Z)).

] Consider ψ ∈ W (x1, x2, n1, n2). We claim that W (x1, x2, n1, n2) ∩ f (C(Y, Z)) = ∅. By

] way of contradiction suppose assume there is g ∈ W (x1, x2, n1, n2) ∩ f (Cp(Y, Z)). It follows

] that g(x1) = g(x2), as g ∈ f (C(Y, Z)), and that g(x1) 6= g(x2), as g ∈ W (x1, x2, n1, n2).

Hence g(x1) = n1 and g(x2) = n2, with n1 6= n2. Thus, we have that W (x1, x2, n1, n2) ∩ f ](C(Y, Z)) = ∅. However, this contradicts the assumption that f ](C(Y, Z)) is dense in

Cp(X, Z). Therefore f(x1) 6= f(x2), so f is injective. Consequently, f is bijective.

] (ii) (⇒) Assume that f is a homeomorphism. By part (i) f (C(Y, Z)) is dense in Cp(X, Z)

−1 −1 ] as f is bijective. Likewise, f is bijective and continuous, so (f ) (Cp(X, Z)) is dense in

] −1 ] Cp(Y, Z). Thus, we have that f (C(Y, Z)) = Cp(X, Z) and (f ) (C(X, Z)) = Cp(Y, Z).

] ] (⇐) Assume that f (Cp(Y, Z)) = Cp(X, Z). It is clear that f (C(Y, Z)) is dense in

Cp(X, Z), thus f is bijective by part (i). Suppose that f is not a homeomorphism. It follows that there is a closed set V in X such that f(V ) is not closed in Y . Now, take y ∈ Y \f(V ) such that y ∈ clY (f(V )) and ﬁx a function φ ∈ Cp(X, Z) such that φ(V ) = {0} and φ(x) = 1, where x ∈ f −1({y}), note that this φ exists since X is a zero-dimensional space.

] If φ ∈ Cp(X, Z) and f (ψ) = φ, then ψ(y) = 1 and ψ(f(V )) = {0}. Now, y ∈ clY (f(V ))

] gives use that ψ is not continuous. Hence φ∈ / f (Cp(Y, Z)). Thus we have a contradiction. 49 So, f is a homeomorphism.

] (iii) Assume that f is a quotient map. Take ψ ∈ clCp(X,Z)(f (C(Y, Z))), it follows that ] ψ ∈ Cp(X, Z). Now, take y ∈ Y . It is clear that every function φ ∈ f (Cp(Y, Z)) is constant on f −1({y}). Thus, the function ψ is constant on f −1({y}). Hence, there is a function

g : Y → Z such that ψ = g ◦ f, that is ψ = f ](g). Since f is a quotient map and ψ is continuous it follows that g is a continuous function. So, ψ ∈ f ](C(Y, Z)), that is the

] ] set f (C(Y, Z)) is a closed subset of Cp(X, Z). Hence, f (Cp(Y, Z)) is a closed subspace of

Cp(X, Z).

Note that the above proposition is the adaptation of the result of the composition operator

0 provided by Arkhangel ski˘ıon page 13 of [4] for the Cp(X) case. We also have a nice relationship involving the partitioning of a zero-dimensional space X. We have the following result which is a similar to Lemma 4.5.1 in [20], the original just considers the following function as a ring isomorphism.

Theorem 2.3.13. Let X be a zero-dimensional space and suppose that {Vi}i∈I is a clopen

partition of the space X. The map ϕ : Cp(X, Z) → Πi∈I Cp(Vi, Z), deﬁned by ϕ(f) = (f|Vi )i∈I ,

is a topological ring isomorphism between the topological rings Cp(X, Z) and the direct product

ring Πi∈I Cp(Vi, Z) with the product topology.

Proof. Suppose that {Vi}i∈I is a clopen partition of the space X. Deﬁne the map ϕ :

Cp(X, Z) → Πi∈I Cp(Vi, Z) by ϕ(f) = (f|Vi )i∈I . We ﬁrst show that ϕ is a ring homomorphism. This is a straight forward argument. Let

f, g ∈ Cp(X, Z), we have the following:

ϕ(f + g) = ((f + g)|Vi )i∈I

= (f|Vi + g|Vi )i∈I

= (f|Vi )i∈I + (g|Vi )i∈I

= ϕ(f) + ϕ(g) 50 and

ϕ(f · g) = ((f · g)|Vi )i∈I

= (f|Vi · g|Vi )i∈I

= (f|Vi )i∈I · (g|Vi )i∈I

= ϕ(f) · ϕ(g).

Thus, ϕ is a ring homomorphism. Next, we show that ϕ is bijective. We will ﬁrst show injectivity and secondly surjectivity.

Let f, g ∈ Cp(X, Z) such that ϕ(f) = ϕ(g). Since ϕ(f) = ϕ(g) is follows that (f|Vi )i∈I =

(g|Vi )i∈I . Thus, f|Vi = g|Vi for all i ∈ I, which gives f(x) = g(x) for all x ∈ Vi and for all i ∈ I. Thus, f(x) = g(x) for all x ∈ X since {Vi}i∈I is a partition of X. Hence, f = g, so ϕ is injective.

Now for surjectivity. Let (fi)i∈I ∈ Πi∈I Cp(Vi, Z). This gives that fi ∈ Cp(Vi, Z) for all i ∈ I. Now, deﬁne f : X → Z by f(x) = fi(x) when x ∈ Vi for all i ∈ I. It follows that this is well-deﬁned since {Vi}i∈I is a partition of X. It follows from Proposition 1.2.14 that f is continuous on X since X = ∪i∈I Vi and Vi is clopen for all i ∈ I. Thus f ∈ Cp(X, Z). Also, by construction we have that ϕ(f) = (fi)i∈I . So, we have that ϕ is a surjective map. Lastly, we show that ϕ is bicontinuous. First we show that ϕ is a continuous map.

Deﬁne the maps ϕi : Cp(X, Z) → Cp(Vi, Z) by ϕi(f) = f|Vi for all i ∈ I. We want to show that ϕi is continuous for each i ∈ I. Let j ∈ I and consider the map ϕj. Let

Wj = Wj(x1, . . . , xk, n1, . . . , nk) be a basis clopen set of Cp(Vj, Z), where x1, . . . , xk ∈ Vj and 51

n1, . . . , nk ∈ Z. It follows that

−1 ϕj (Wj) = {f ∈ Cp(X, Z): f|Vi ∈ Wj}

= {f ∈ Cp(X, Z): f|Vj (xi) = ni for all i = 1, . . . , k}

= {f ∈ Cp(X, Z): f(xi) = ni for all i = 1, . . . , k} (since x1, . . . , xk ∈ Vj)

= W (x1, . . . , xk, n1, . . . , nk).

Thus, we have that ϕj is a continuous map, so ϕi is continuous for all i ∈ I. It then follows

from Proposition 1.2.17 that ϕ, since ϕ(f) = (f|Vi )i∈I = (ϕi(f))i∈I . To ﬁnish the proof we will show that ϕ is an open map which will give that ϕ is a

homeomorphism by Proposition 1.2.13. We will start by showing that each ϕi is an open

map for all i ∈ I. Let W = W (x1, . . . , xk, n1, . . . , nk) be a basic open set of Cp(X, Z) where

x1, . . . , xk ∈ X and n1, . . . , nk ∈ Z. For convenience deﬁne the set Ji = {j ∈ {1, . . . , k} :

xj ∈ Vi}. We will have two cases.

Case 1. Suppose Ji = ∅. It follows that there are no requirements for the functions in W

to take the elements in Vi. Thus, it follows that ϕi(W ) = Cp(Vi, Z). Hence, ϕ(W ) is open

in Cp(Vi, Z). Therefore, ϕ is an open map.

Case 2. Suppose Ji 6= ∅. We have the following

ϕi(W ) = {f ∈ Cp(Vi, Z): f(xj) = nj, j ∈ Ji}

= ∩j∈Ji {f ∈ Cp(Vi, Z): f(xj) = nj}

= ∩j∈Ji Wi(xj, nj)

where Wi(xj, nj) is a basic clopen set of Cp(Vi, Z), for all j ∈ Ji. Since Wi(xj, nj) is a clopen

of Cp(Vi, Z), for all j ∈ Ji, it follows that ϕi(W ) is open in Cp(Vi, Z).

Now back to the map ϕ. We have that ϕ = (ϕi)i∈I , so it follows that ϕ(W ) = Πi∈I ϕi(W ).

From the above it is clear that ϕi(W ) = Cp(Vi, Z) if Ji = ∅ and ϕi(W ) = ∩j∈Ji Wi(xj, nj) 52

if Ji 6= ∅, for all i ∈ I. Since there are only ﬁnitely many Ji 6= ∅ it follows that ϕi(W ) =

Cp(Vi, Z) for all but ﬁnitely many i ∈ I. Thus, ϕ(W ) = Πi∈I ϕi(W ) is open in Πi∈I Cp(Vi, Z). So, it follows that ϕ is an open map, which gives that ϕ is a homeomorphism. Therefore, it is clear from the above that ϕ is a ring isomorphism that is bicontinuous.

Hence, ϕ is a topological ring isomorphism between Cp(X, Z) and Πi∈I Cp(Vi, Z).

This section has shown that the topological properties of A have an eﬀect on the topo-

logical properties of Cp(X,A). We have seen that if there is a relationship between the ideal structure of A and its topological structure there is then a corresponding relationship for

Cp(X,A), in particular the eﬀect that a linear topology has. Furthermore, we have shown

that the composition operator will be a useful tool in our analysis of Cp(X,A). 53

CHAPTER 3

New Results on Cp(X, Z)

3.1 Evaluation Maps and Cp(Cp(X, Z), Z)

We will now consider the evaluation and natural maps deﬁned for Cp(X, Z). We will see that both of these maps are useful in the embedding of spaces. One of the surprising results is that the collection of linear combination of evaluation maps is precisely the dual space for

Cp(X,A). The main result of this section is a theorem similar to that of Nagata from 1949 which tells us when X and Y are homeomorphic based on the relationship between Cp(X, Z) and Cp(Y, Z).

We ﬁrst note that Cp(X, Z) is a closed subspace of Cp(X), as Z is closed in R, this follows from Proposition 2.1.2. We see that the function ex : Cp(X, Z) → R is continuous by Proposition 2.1.8 (letting F = Cp(X, Z)). Now considere ˆCp(X,Z) : X → Cp(Cp(X, Z), Z) deﬁned bye ˆCp(X,Z)(x) = ex, for convenience lete ˆCp(X,Z) =e ˆZ.

Remark 3.1.1. The map φx deﬁned in the proof for Proposition 2.2.9 above is precisely the evaluation map, thus the evaluation map is a continuous ring homomorphism from Cp(X, Z) onto Z.

Before we continue we have to make a few defections for convenance later. A functional is a continuous map ϕ : Cp(X, Z) → Z, that is a functional is an element of Cp(Cp(X, Z), Z). A 54

linear functional is a functional ϕ such that ϕ(af +bg) = aϕ(f)+bϕ(g) for all f, g ∈ Cp(X, Z) and a, b ∈ Z. Note that a linear functional is a continuous module homomorphism from

Cp(X, Z) to Z.A multiplicative functional is a functional such that ϕ(f · g) = ϕ(f) · ϕ(g)

for all f, g ∈ Cp(X, Z). Thus, it is clear that is ϕ is a linear, multiplicative functional is a

continuous ring homomorphism from Cp(X, Z) to Z.

Does the same result hold fore ˆZ as in Proposition 2.1.9? The answer is that it does and it is given by the following proposition.

Proposition 3.1.2. Let X be a zero-dimensional space. The natural map eˆZ : X →

Cp(Cp(X, Z), Z) is continuous.

Proof. Let x ∈ X and V = V (f1, . . . , fn, n1, . . . , nk) be a basic open subset of Cp(Cp(X, Z), Z))

containing ex. Thus, we have ex(fi) = fi(x) = ni, where ni ∈ Z for i = 1, . . . , k. Since

fi ∈ Cp(X, Z), for all i = 1, . . . , k, we see that

eˆ−1(V ) = ∩{f −1({n }): i = 1, . . . , k}. Z i i

−1 Also, since {ni} is an open subset of Z, for each i = 1, . . . , n, we have that fi ({ni}) is an open subset of X. Hence, since the above intersection is a ﬁnite intersection, it follows that eˆ−1(V ) is an open of X containing x. Thereforee ˆ is a continuous map. Z Z

Note that, in general,e ˆZ is not a homeomorphism. This can be seen by considering a ﬁnite set X with the discrete topology; it follows that C(X, Z) = ZX , since all functions

X are continuous on a discrete topology. Thus C(Cp(X, Z), Z) = C(Z , Z) which contains a

subset of cardinality ℵ0, in particular the constant function {n : n ∈ Z}. Hence, there is no

bijective function between X and Cp(Cp(X, Z), Z). Therefore, there is no homeomorphism

between X and Cp(Cp(X, Z), Z).

However, we can consider whether or note ˆZ an homeomorphic embedding.

Proposition 3.1.3. Let X be a zero-dimensional space. The map eˆZ : X → Cp(Cp(X, Z), Z)

is a homeomorphism from X to the subspace eˆZ(X) of Cp(Cp(X, Z), Z). 55

Proof. We have from Proposition 3.1.2 thate ˆZ is a continuous map from X to Cp(Cp(X, Z), Z)

and it is clear thate ˆZ is surjective one ˆZ(X). Hence, we need to verify thate ˆZ is injective and thate ˆZ has a continuous inverse map.

First, to show thate ˆZ is injective. Let x, y ∈ X and suppose thate ˆZ(x) =e ˆZ(y). Thus,

by the deﬁnition ofe ˆZ we have that ex = ey. Hence, we have that ex(f) = ey(f) for all f ∈ C(X, Z), so f(x) = f(y) for all f ∈ C(X, Z). Now, suppose that x 6= y, it follows from Proposition 2.2.2 that there is a g ∈ C(X, Z) such that g({x}) = {0} and g(y) = 1. This

gives a contradiction, consequently it must be that x = y. Therefore,e ˆZ is injective, and

this provided thate ˆZ is bijective.

Secondly, to show thate ˆZ has a continuous inverse map. We will instead show thate ˆZ is

a closed map, this combined with injectivity gives thate ˆZ is a homeomorphism. Since X is a zero- dimensional space, take V as a basic clopen set of X, in particular V is a closed set. We will have the following:

eˆZ(V ) = {F ∈ Cp(Cp(X, Z), Z): F = ex, x ∈ V }

= {F ∈ Cp(Cp(X, Z), Z): F (f) = f(x), x ∈ V, f ∈ Cp(X, Z)}

= ∩x∈V {F ∈ Cp(Cp(X, Z), Z): F (f) = f(x), f ∈ Cp(X, Z)}

= ∩x∈V ∩f∈Cp(X,Z) {F ∈ Cp(Cp(X, Z), Z): F (f) = f(x)}

= ∩x∈V ∩f∈Cp(X,Z) W (f, f(x))

where W (f, f(x)) is a basic clopen set of Cp(Cp(X, Z), Z) when f ∈ Cp(X, Z) and x ∈ X.

Thus, it follows thate ˆZ is a closed map. Therefore, it follows thate ˆZ is a homeomorphism.

Now, with Proposition 3.1.3 we have following result which gives a strong statement about zero-dimensional spaces. There is a corresponding result for Tychonoﬀ spaces in the

Cp(X) case (refer to page 9 in [3]).

Theorem 3.1.4. Let X be a zero-dimensional space, there is a zero-dimensional space Y 56 such that X is homeomorphic to a subspace of Cp(Y, Z).

Proof. It follows from Proposition 3.1.3 that X is homeomorphic to the subspacee ˆZ(X) of Cp(Cp(X, Z), Z). Thus, letting Y = Cp(X, Z), which is zero-dimensional by Proposition

2.3.4, we have that X is homeomorphic to a subspace of Cp(Y, Z).

Also, with Proposition 3.1.3 we can identify every element in the zero-dimensional space

X with a continuous integer-valued function in the space Cp(Cp(X, Z), Z), namely an evaluation map. We will now show some additional properties that the evaluation maps have.

Recall, that if x ∈ X, f, g ∈ Cp(X, Z), and a, b ∈ Z, where X is a zero-dimensional space, we have the following:

ex(af + bg) = (af + bg)(x)

= af(x) + bg(x)

= aex(f) + bex(g) and

ex(f · g) = (f · g)(x)

= f(x) · g(x)

= ex(f) · ex(g).

That is, the evaluation maps are linear multiplicative functionals from Cp(X, Z) to Z. More- over, if x, y ∈ X and f ∈ Cp(X, Z), we then have

(ex + ey)(f) = ex(f) + ey(f) and (ex · ey)(f) = ex(f) · ey(f).

Now, deﬁne the set of linear combinations of evaluation maps for a zero-dimensional space 57 X as

( n ) X L(X, Z) = λiexi ∈ C(Cp(X, Z), Z): x1, . . . , xn ∈ X, λ1, . . . , λn, n ∈ N . i=1

Note that L(X, Z) is a Z-module, it is not, in general, a ring. Since L(X, Z) is a subset of

C(Cp(X, Z), Z), we will denote Lp(X, Z) as the set L(X, Z) when endowed with the subspace

0 topology of Cp(Cp(X, Z), Z). Take note that this is the same notation used by Arkhangel ski˘ı on page 17 of [4].

We will continue with some results for Lp(X, Z).

Proposition 3.1.5. Let X be a zero-dimensional space. If ϕ : Cp(X) → Z is a linear functional, then ϕ ∈ Lp(X, Z).

Proof. Suppose ϕ : Cp(X) → Z is a linear functional, that is ϕ is continuous and ϕ(af +bg) = aϕ(f) + bϕ(g) for all f, g ∈ Cp(X, Z) and a, b ∈ Z. Firstly, since ϕ is linear, it is clear that

ϕ(0) = ϕ(0 + 0) = ϕ(0) + ϕ(0), thus ϕ(0) = 0. Secondly, note that since ϕ is continuous a map, there is a basic clopen subset of Cp(X, Z) of the form W (x1, . . . , xn, 0,..., 0) such that ϕ(W (x1, . . . , xn, 0,..., 0)) ⊆ {0}, for x1, . . . , xn ∈ X and n ∈ N. However, we in fact have ϕ(W (x1, . . . , xn, 0,..., 0)) = {0} since ϕ(W (x1, . . . , xn, 0,..., 0)) is nonempty and {0} is a singleton subset of Z.

Assume, without loss of generality, that xi 6= xj if i 6= j. Take gi ∈ Cp(X, Z) such that gi(xi) = 1 and gi(xj) = 0 for i 6= j and i = 1, . . . , n, these gi’s exist by Proposition 2.2.2.

Now, set λi = ϕ(gi) for all i = 1, . . . , n.

We claim that for all g ∈ Cp(X, Z) we have that ϕ(g) = λ1ex1 (g) + ... + λnexn (g). We ﬁrst observe that

(λ1ex1 + ... + λnexn )(g) = λ1ex1 (g) + ... + λnexn (g). 58 0 0 0 Deﬁne g = g − g(x1)g1 + ... + g(xn)gn. It is clear that g ∈ Cp(X, Z) and g (xi) = 0

0 for all i = 1, . . . , n by the selection of the gi’s. Thus, it follows that ϕ(g ) = 0, since

0 g ∈ W (x1, . . . , xn, 0,..., 0). Now, since ϕ is linear, we have

n ! 0 X 0 = ϕ(g ) = ϕ(g) − ϕ g(xi)gi . i=1

Consequently,

n ! X ϕ(g) = ϕ g(xi)gi i=1 n X = g(xi)ϕ(gi) i=1 n X = λig(xi) i=1 n X = λiexi (g). i=1

Pn So, it follows that ϕ = i=1 λiexi . Therefore ϕ ∈ Lp(X, Z).

Note that this result is similar to a result provided by Arkhangel0ski˘ıon page 18 of [4].

The original result is for the Cp(X) and a similar construction of Lp(X). We will now consider classic result by Nagata from 1949 which is the following:

Theorem 3.1.6 (Theorem 0.6.1, [4]). (Nagata’s Theorem) Let X and Y be Tychonoﬀ spaces.

Cp(X) is topologically isomorphic to Cp(Y ) as topological rings if and only if X homeomorphic to Y .

We will now see how the corresponding result holds in the Cp(X, Z) case.

Theorem 3.1.7. Let X and Y be zero-dimensional spaces. The topological rings Cp(X, Z)

and Cp(Y, Z) are topologically isomorphic if and only if X is homeomorphic to Y .

Proof. (⇒) Assume that Cp(X, Z) and Cp(Y, Z) are topologically isomorphic. Let ϕ :

Cp(X, Z) → Cp(Y, Z) be a topological isomorphism, that is ϕ is a bicontinuous ring isomor- 59

phism. Deﬁne HomX be the collection of all non-zero continuous ring homomorphism from

Cp(X, Z) to Z, i.e. HomX is the collection of the non-zero linear, multiplicative functionals

of Cp(X, Z). It is clear thate ˆZ(X) ⊆ HomX since each evaluation map is a continuous ring

homomorphism. Since every element of HomX is a linear functional is follows from Proposi- tion 3.1.5 that HomX ⊆ Lp(X, Z). Thus, we have the followinge ˆZ(X) ⊆ HomX ⊆ Lp(X, Z).

Deﬁne HomY in the a similar manner, givinge ˆZ(Y ) ⊆ HomY ⊆ Lp(Y, Z).

We now wish to extend ϕ to a homeomorphism from HomY to HomX . Consider the

] composition operator ϕ : Cp(Cp(Y, Z), Z) → Cp(Cp(X, Z), Z). We have that ϕ is homeomorphism, so it follows from Proposition 2.1.3 that ϕ] and (ϕ−1)] are both continuous maps.

] ] Let ϕ |HomY : HomY → Cp(Cp(X, Z), Z) be the restriction of ϕ to the set HomY .

] ] We claim that ϕ |HomY is a bijection between HomY and HomX . Let F ∈ ϕ |HomY (HomY ),

then F = G ◦ ϕ for some G ∈ HomY . If f, g ∈ Cp(X, Z), then we have

F (f + g) = (G ◦ ϕ)(f + g)

= G(ϕ(f + g))

= G(ϕ(f) + ϕ(g))

= G(ϕ(f)) + G(ϕ(g))

= (G ◦ ϕ)(f) + (G ◦ ϕ)(g)

= F (f) + F (g) 60 and

F (f · g) = (G ◦ ϕ)(f · g)

= G(ϕ(f · g))

= G(ϕ(f) · ϕ(g))

= G(ϕ(f)) · G(ϕ(g))

= (G ◦ ϕ)(f) · (G ◦ ϕ)(g)

= F (f) · F (g).

] Thus, F ∈ HomX , which implies that ϕ |HomY (HomY ) ⊆ HomX .

] ] Now we need to verify that ϕ |HomY is actually a bijection. First we show that ϕ |HomY a

] ] injection. Let F,G ∈ HomY such that ϕ (F ) = ϕ (G). It follows that F ◦ϕ = G◦ϕ, but recall

−1 −1 that ϕ is bijective, so it follows that F ◦ϕ◦ϕ = G◦ϕ◦ϕ . Hence F ◦idCp(X,Z) = G◦idCp(X,Z). Therefore, it follows that F = G, implying ϕ] is an injection.

] −1 ] To show that ϕ |HomY is surjective we will instead show that (ϕ ) |HomX , the restriction

−1 ] ] of (ϕ ) to HomX , is injective and the inverse of ϕ |HomY . It is clear, by a similar argument,

−1 ] −1 ] that (ϕ ) |HomX (HomX ) ⊆ HomY . To show that (ϕ ) |HomX is injective, let F,G ∈ HomY such that (ϕ−1)](F ) = (ϕ−1)](G). Thus F ◦ ϕ−1 = G ◦ ϕ−1. However, ϕ is bijective, so ϕ−1

−1 −1 is bijective. Hence F ◦ ϕ ◦ ϕ = G ◦ ϕ ◦ ϕ giving F ◦ idCp(Y,Z) = G ◦ idCp(Y,Z). Therefore, it follows that F = G and (ϕ−1)] is an injection.

] −1 ] Lastly, we verify that ϕ |HomY and (ϕ ) |HomX are inverse maps of each other. If F ∈

] HomX , then ϕ |HomY (F ) ∈ HomX . Hence we have that following

−1 ] ] −1 ] (ϕ ) |HomX (ϕ |HomY (F )) = (ϕ ) |HomX (F ◦ ϕ)

= F ◦ ϕ ◦ ϕ−1

= F ◦ idCp(X,Z) = F. 61 −1 ] Similarly, if G ∈ HomY , then (ϕ ) |HomX (G) ∈ HomY . Thus, we have the following

] −1 ] ] −1 ϕ |HomY ((ϕ ) |HomX (G)) = ϕ |HomY (G ◦ ϕ )

= G ◦ ϕ−1 ◦ ϕ

= G ◦ idCp(Y,Z) = G.

] −1 ] Thus, we haveϕ |HomY and (ϕ ) |HomX are inverse maps of each other. Hence, it follows that

] ] ϕ |HomY is a bijective map. Therefore, we have ϕ |HomY is a homeomorphism from HomY to

HomX , implying HomX and HomY are homeomorphic.

We claim thate ˆZ(X) = HomX ande ˆZ(Y ) = HomY . To prove the claim we will only provee ˆZ(X) = HomX as the second is done similarly. Let F ∈ HomX , by the deﬁnition of

HomX we have that F is not the zero functional. Since HomX ⊆ Lp(X, Z) it follows that there are x1, . . . , xn ∈ X with xi 6= xj if i 6= j, and λ1, . . . , λn ∈ Z, with λi 6= 0 for all i = 1, . . . , n, where n ∈ N such that F = λ1ex1 + ... + λnexn .

Supposen > 1. Since x1 6= x2 there are disjoint basic clopen sets V1 and V2 in X such that x1 ∈ V1 and x2 ∈ V2 and x3, . . . , xn ∈/ V1,V2; V1 and V2 exist since X is a zero-dimensional χ χ χ χ Hausdorﬀ space. Then V1 , V2 ∈ CP (X, Z) with V1 · V2 = 0. Consequently, we have

χ χ χ F ( V1 ) = λ1ex1 ( V1 ) + ... + λnexn ( V1 ) χ χ = λ1 V1 (x1) + ... + λn V1 (xn)

= λ1 62 and

χ χ χ F ( V2 ) = λ1ex1 ( V2 ) + ... + λnexn ( V2 ) χ χ = λ1 V1 (x2) + ... + λn V2 (xn)

= λ2.

Therefore,

χ χ χ χ F ( V1 · V2 ) = F ( V1 )F ( V2 )

= λ1λ2

6= 0.

χ χ χ χ However, V1 · V2 = 0 and F ( V1 · V2 ) = 0. This is a contradiction. Hence, we have that n = 1.

Let F = λex for some λ ∈ Z\{0} and some x ∈ X, since F is a non-zero functional. Consider what F (1). We have

F (1) = λex(1) = λ1(x) = λ and

F (1) = F 12

= F (1)F (1)

= λex(1)λex(1)

= λ1(x)λ1(x)

= λ2. 63 2 Thus, we have that λ = λ. Since λ 6= 0 it follows that λ = 1. Therefore F = ex. Hence, it follows that F ∈ eˆZ(X), which implies thate ˆZ(X) = HomX . Similarly, we have that eˆZ(Y ) = HomY . ] From the above we have that HomX is homeomorphic to HomY by ϕ . Also, we have shown thate ˆZ(X) = HomX ande ˆZ(Y ) = HomY . Therefore it follows thate ˆZ(X) is homeomorphic toe ˆZ(Y ). We have from Proposition 3.1.3 thate ˆZ(X) is homeomorphic to X and

eˆZ(Y ) is homeomorphic to Y . Hence, it follows that X is homeomorphic to Y . (⇐) Assume that X and Y are homeomorphic, then there is a homeomorphism ϕ form X

] ] to Y . By Proposition 2.1.3 we have that ϕ : Cp(Y, Z) → Cp(X, Z) deﬁned by ϕ (f) = f ◦ ϕ is a continuous map. Since ϕ is a homeomorphism we have that ϕ−1 : Y → X is also a

−1 ] homeomorphism. Thus, by Proposition 2.1.3 we have that (ϕ ) : Cp(X, Z) → Cp(Y, Z) deﬁned by (ϕ−1)](g) = g ◦ ϕ−1 is a continuous map.

−1 We now wish to verify that ϕ and ϕ are inverse maps of each other. If f ∈ Cp(Y, Z), then

(ϕ−1)](ϕ](f))(y) = (ϕ−1)](f ◦ ϕ)(y)

= ((f ◦ ϕ) ◦ ϕ−1)(y)

= (f ◦ ϕ ◦ ϕ−1)(y)

= (f ◦ idY )(y)

= f(y) 64 −1 ] ] for all y ∈ Y . Thus, we have that (ϕ ) (ϕ (f)) = f. Likewise, if g ∈ Cp(X, Z), then

ϕ]((ϕ−1)](g))(x) = ϕ](g ◦ ϕ−1)(x)

= ((g ◦ ϕ−1) ◦ ϕ)(x)

= (g ◦ ϕ ◦ ϕ−1)(x)

= (g ◦ idX )(x)

= g(x)

for all x ∈ X. Therefore ϕ] and (ϕ−1)] are inverse maps of each other, implying ϕ] is a bijective map.

] Lastly, we verify that ϕ is a ring homomorphism. If f, g ∈ Cp(Y, Z), then

ϕ](f + g) = (f + g) ◦ ϕ

= (f ◦ ϕ) + (g ◦ ϕ)

= ϕ](f) + ϕ](g)

and

ϕ](f · g) = (f · g) ◦ ϕ

= (f ◦ ϕ) · (g ◦ ϕ)

= ϕ](f) · ϕ](g).

] Thus, ϕ is a topological isomorphism from Cp(Y, Z) to Cp(x, Z), which implies Cp(Y, Z) and

Cp(x, Z) are topologically isomorphic.

With Theorem 3.1.7 in hand we can start to classify Cp(X, Z) into diﬀerent categories based on the zero-dimensional space X.

Example 3.1.8. Now we can state a few simple examples that can be obtained by using 65 Theorem 3.1.7.

(i) If X and Y are ﬁnite spaces with the same cardinality, then Cp(X, Z) is topologically

isomorphic to Cp(X, Z).

(ii) If X and Y are discrete spaces with the same cardinality, then Cp(X, Z) is topologically

isomorphic to Cp(Y, Z).

We would like to reiterate that we now have the ability to characterize spaces based on their collections of integer-valued continuous functions. We now have that the collection of linear combinations of evaluation maps is the dual space for Cp(X, Z), which was a useful fact in the proof of Theorem 3.1.7. Thus, the evaluation maps are quite useful though they are rather simple.

3.2 The Space Lp(X, Z)

In Section 3.1 we introduced the space Lp(X, Z); in this section we will make a deeper instigation of the space Lp(X, Z). First we will need to state a few deﬁnitions. Since C(X, Z) is a Z-module we can consider all the linear maps from C(X, Z) to Z. Moreover, we will want to consider the continuous linear maps from Cp(X, Z) to Z, since they are both

0 topological Z-modules. We will denote the topological dual space of Cp(X, Z) by C(X, Z) ,

0 this is the set of all linear functionals on Cp(X, Z). That is, C(X, Z) is the collection of all

φ ∈ C(Cp(X, Z), Z) such that φ(af + bg) = aφ(f) + bφ(g) for all f, g ∈ C(X, Z) and a, b ∈ Z.

0 0 In particular Cp(X, Z) is a subset of C(Cp(X, Z), Z). Thus we will C(X, Z) endowed with

0 the subspace topology induced by Cp(Cp(X, Z), Z) by Cp(X, Z) .

0 We will ﬁrst notice that Cp(X, Z) is a Z-module and has a basis consisting of evaluation maps on Cp(X, Z).

Proposition 3.2.1. Let X be a zero-dimensional space, the collection {ex : x ∈ X} is a

0 0 Z-basis for Cp(X, Z) . Moreover, Cp(X, Z) = Lp(X, Z). 66

Proof. We will ﬁrst that {ex : x ∈ X} is a linearly independent set. Let x1, . . . , xn ∈ X, and

n let λ1, . . . , λn ∈ Z\{0}. Now, for each i = 1, . . . , n let {Vi}i=1 be a collection of disjoint basic Pn clopen subsets of X such that xi ∈ Vi. Suppose i=1 λiexi = 0Cp(X,Z). We claim that λi = 0 χ for all i = 1, . . . , n. Consider the function Vk for some k, we see that

χ 0 = 0Cp(X,Z)( Vk ) n X χ = λiexi ( Vk ) i=1 n X χ = λi Vk (xi) i=1 = λk.

Thus, we have λi = 0 for all i = 1, . . . , n. Hence, {ex : x ∈ X} is a linearly independent set. Next, we see that it follows from Proposition 3.1.5 that ever linear functional is of the form

Pn 0 i=1 λ1exi . Therefore Cp(X, Z) = Lp(X, Z).

We will now define a useful function on Lp(X, Z). If X is a zero-dimensional space ˆ ˆ and g ∈ C(X, Z), define Lf : Cp(Cp(X, Z), Z → Z by Lg(F ) = F (g). Furthermore, define ˆ ˆ Lg : Lp(X, Z) toZ by Lg = Lg|Lp(X,Z), the restriction of Lg to the subspace Lp(X, Z).

Lemma 3.2.2. Let X be a zero-dimensional space, and let f ∈ C(X, Z). The map Lf is a linear functional.

ˆ ˆ Proof. Let f ∈ Cp(X, Z). By the deﬁnition of Lf we see that Lf is actually an evaluation ˆ map. Thus it follows from Remark 3.1.1 that Lf is a continuous map. Now, since Lp(X, Z)

is a subspace of Cp(Cp(X, Z), Z) it follows that Lf is a continuous map. 67

We shall now verify that Lf is linear. If F,G ∈ Lp(X, Z) and let a, b ∈ Z, then

Lf (aF + bG) = (aF + bG)(f)

= (aF )(f) + (bG)(f)

= aF (f) + bG(f)

= aLf (F ) + bLf (G).

Hence, Lf is a linear functional from Lp(XZ) to Z.

Proposition 3.1.3 tells us that a zero-dimensional space X is homeomorphic toe ˆZ(X). It is also clearly thate ˆZ(X) is a subspace of Lp(X, Z). These two facts lead us to the natural question about the subspaces, and whether or not they are closed subspaces. We have the following lemma for the map Lg.

Proposition 3.2.3. Let X be a zero-dimensional space. Each of the following hold.

(i) eˆZ(X) is a closed subspace of Lp(X, Z).

(ii) Lp(X, Z) is a closed subspace of Cp(Cp(X, Z), Z).

(iii) eˆZ(X) is a closed subspace of Cp(Cp(X, Z), Z).

Pn Proof. (i) Let F ∈ Lp(X, Z)\eˆZ(X). Since F ∈ Lp(X, Z) we have F = i=1 λiexi for some x1, . . . , xn ∈ X and some λ1, . . . , λn ∈ Z\{0}.

Case 1: Suppose n ≥ 2. We want to ﬁnd an open subset of Lp(X, Z) containing F which

n is disjoint frome ˆZ(X). There exists a collection {Vi}i=1 of clopen subsets of X such that xi ∈ Vi for all i = 1, . . . , n. Now, take fi ∈ Cp(X, Z) such that fi(xi) = 1 and fi(X\Vi) = {0} for all i = 1, . . . , n. 68 We claim that F ∈ ∩n L−1( \{0}). First notice i=1 fi Z

n ! X Lfi (F ) = Lfi λjexj j=1 n X = λjLfi (exj ) j=1 n X = λjexj (fi) j=1 n X = λjfi(xj) j=1

= λi and recall that λ ∈ \{0}. Thus F ∈ L−1( \{0}) for each i = 1, . . . , n. Hence, it follows i Z fi Z that F ∈ ∩n L−1( \{0}). Now, since F ∈ L (X, )\eˆ (X) we have L (X, )\eˆ (X) ⊆ i=1 fi Z p Z Z p Z Z ∩n L−1( \{0}). Furthermore, if x ∈ X and i ≤ n with x ∈ V , then i=1 ji Z i

Lfi (ex) = ex(fi)

= fi(x)

= 0 ∈/ Z\{0}.

Therefore, we havee ˆ (X) ∩ L−1( \{0}) = ∅. Thus there is an nonempty open subset of Z fi Z

Lp(X, Z) containing F which is disjoint withe ˆZ(X). Hence,e ˆZ(X) is a closed subspace of

Lp(X, Z).

Case 2. Suppose n = 1. Then F = λex for some λ ∈ Z\{0}, and some x ∈ X. Note λ 6= 1 −1 since F/∈ eˆZ(X). Consider the function 1 ∈ Cp(X, Z). We claim that F ∈ L1 (Z\{0}) ⊆ 69

Lp(X, Z)\{1}. Notice

L1(F ) = L1(λex)

= λL1(ex)

= λex(1)

= λ

6= 1.

−1 Thus, F ∈ L1 (Z\{0}). Also, for x ∈ X, we have that L1(ex) = ex(1) = 1(x) = 1 with −1 1 ∈/ Z\{1}. Consequently L1 (Z\{1}) ⊆ Lp(X, Z)\eˆZ(X). Therefore there is a nonempty open subset of Lp(X, Z) containing F which is disjoint withe ˆZ(X). Hence,e ˆZ(X) is a closed subspace of Lp(X, Z)

(ii) This is clear the closure oﬀ the set of all linear functionals in Cp(Cp(X, Z), Z) is the

collection of all linear functionals on Cp(X, Z). It also follows from Proposition 3.1.5 that

Lp(X, Z) is the collection of all linear functionals on Cp(X, Z). Therefore, we have that

Lp(X, Z) is a closed subspace of Cp(Cp(X, Z), Z).

(iii) By (i) we have thate ˆZ(X) is a closed subspace of Lp(X, Z) and by (ii) we have that

Lp(X, Z) is a closed subspace of Cp(Cp(X, Z), Z), therefore it follows thate ˆZ(X) is a closed

subspace of Cp(Cp(X, Z), Z).

Now we can state an interesting result about the space Cp(X, Z) and its dual space. We will note that this result is similar to that of Theorem 6.7.5 in [24].

Theorem 3.2.4. Let X and Y be zero-dimensional spaces. Cp(X, Z) is linearly homeomor-

phic to Cp(Y, Z) if and only if Lp(X, Z) is linearly homeomorphic to Lp(Y, Z).

Proof. (⇒) Suppose Cp(X, Z) is linearly homeomorphic to Cp(Y, Z). Let ϕ : Cp(X, Z) →

] Cp(Y, Z) be a linear homeomorphism. Consider the composition map ϕ : Cp(Cp(Y, Z), Z) →

] Cp(Cp(X, Z), Z). We claim that ϕ |Lp(Y,Z) is a linear homeomorphism from Lp(Y, Z) to 70

Lp(X, Z).

] We ﬁrst show that ϕ |Lp(Y,Z) is a continuous map. Since ϕ is a linear homeomorphism it is a continuous map, thus is follows from Proposition 2.1.3 that ϕ] is a continuous map from

Cp(Cp(Y, Z), Z) to Cp(Cp(X, Z), Z). Also, since Lp(Y, Z) is a subspace of Cp(Cp(Y, Z), Z) it

] follows ϕ |Lp(Y,Z) is a continuous map. ] We now verify that ϕ |Lp(Y,Z) is linear. If F,G ∈ Lp(Y, Z) and a, b ∈ Z, then

] ϕ |Lp(Y,Z)(aF + bG) = (aF + bG) ◦ ϕ = (aF ) ◦ ϕ + (bG) ◦ ϕ

= a(F ◦ ϕ) + b(G ◦ ϕ)

] ] = aϕ |Lp(Y,Z)(F ) + bϕ |Lp(Y,Z)(G).

] Hence ϕ |Lp(Y,Z) is a linear map. ] Next we verify that for any F ∈ Lp(Y, Z) we have ϕ |Lp(Y,Z)(F ) ∈ Lp(X, Z). If F ∈

Lp(Y, Z), f, g ∈ Cp(X, Z), and a, b ∈ Z, then

] ϕ |Lp(Y,Z)(F )(af + bg) = (F ◦ ϕ)(af + bg) = F (ϕ(af + bg))

= F (aϕ(f) + bϕ(g))

= aF (ϕ(f)) + bF (ϕ(g))

] ] = aϕ |Lp(Y,Z)(F )(f) + bϕ |Lp(Y,Z)(F )(g).

] Therefore ϕ |Lp(Y,Z)(F ) is a linear functional on Cp(X, Z). Hence it follows from Proposition ] −1 ] 3.1.5 that ϕ |Lp(Y,Z)(F ) ∈ Lp(X, Z). Similarly, we have that we have that (ϕ ) |Lp(X,Z) is a −1 ] continuous linear map with (ϕ ) |Lp(X,Z)(G) ∈ L(Y, Z) for all G ∈ Lp(Y, Z). ] −1 ] Lastly, we claim that ϕ and (ϕ ) are inverse maps of each other. If G ∈ Cp(Cp(Y, Z), Z) 71 and F ∈ Cp(Cp(X, Z), Z), then

((ϕ−1)] ◦ ϕ])(G) = (ϕ−1)](ϕ](G))

= (ϕ−1)](G ◦ ϕ)

= G ◦ ϕ ◦ ϕ−1

= G ◦ idCp(Y,Z) = G and

(ϕ] ◦ (ϕ−1)])(G) = ϕ]((ϕ−1)](G))

= ϕ](F ◦ ϕ−1)

= F ◦ ϕ−1 ◦ ϕ

= F ◦ idCp(X,Z) = F.

] Hence, ϕ |Lp(Y,Z) is a linear homeomorphism from Lp(Y, Z) to Lp(X, Z). Therefore Lp(X, Z) is linearly homeomorphic to Lp(Y, Z)

(⇐) Suppose that Lp(X, Z) is linearly homeomorphic to Lp(Y, Z). Let ϕ : Lp(X, Z) →

Lp(Y, Z) be a linear homeomorphism. Deﬁne ψ : Cp(Y, Z) → Cp(X, Z) by ψ(f) = Lf ◦ϕ◦eˆZ.

It is clear that ψ is well-deﬁned since Lf , φ, ande ˆZ are. We claim that ψ is a linear homeomorphism between Cp(Y, Z) and Cp(X, Z).

First, we show that ψ is linear. Let f, g ∈ Cp(Y, Z), and let a, b ∈ Z. For any x ∈ X we 72 have the following

ψ(af + bg)(x) = (L(af+bg) ◦ ϕ ◦ eˆZ)(x)

= (L(af+bg) ◦ ϕ)(ex)

= L(af+bg)(ϕ(ex))

= ϕ(ex)(af + bg)

= aϕ(ex)(f) + bϕ(ex)(g)

= aLf (ϕ(ex)) + bLg(ϕ(ex))

= a(Lf ◦ ϕ)(ex) + b(Lg ◦ ϕ)(ex)

= a(Lf ◦ ϕ ◦ eˆZ)(x) + b(Lg ◦ ϕ ◦ eˆZ)(x).

Therefore, we have ψ is a linear map.

We now show that ψ is continuous. Let FX = {x1, . . . , xn} ⊆ X. For each x ∈ X there

(x) (x) (x) (x) Pnx (x) (x) exists y1 , . . . , ynx ∈ Y and k1 , . . . , knx ∈ Z\{0} such that ϕ(ex) = i=1 ai eyi . Let (x) FY = {yi : x ∈ Fx, i ≤ nx}, since FY is ﬁnite enumerate the FY as {y1, . . . , ym}. We will show that ψ is continuous at 0, in particular we want to show

ψ(W (y1, . . . , ym, 0,..., 0)) ⊆ W (x1, . . . , xn, 0,..., 0).

If f ∈ W (y1, . . . , ym, 0,..., 0), then f(yi) = 0 for all i = 1, . . . , m and all x ∈ FX . Thus we 73 have the following:

ψ(f)(x) = (Lf ◦ ϕ ◦ eˆZ)(x)

= (Lf ◦ ϕ)(ex)

= Lf (ϕ(ex))

nx ! X (x) = L k e(x) f i yi i=1 nx ! X (x) = k e(x) (f) i yi i=1 nx X (x) = k e(x)(f) i yi i=1 nx X (x) = ki f(yi) i=1 nx X (x) = ki · 0 i=1 = 0.

Therefore, ψ(f) ∈ W (x1, . . . , xn, 0,..., 0) implying ψ is continuous at 0.

Lastly, we verify that ψ is continuous on all of Cp(Y, Z). If f ∈ Cp(Y, Z), then

ψ(W (y1, . . . , ym, f(y1), . . . , f(ym))) = ψ(W (y1, . . . , ym, 0,..., 0) + f)

= ψ(W (y1, . . . , ym, 0,..., 0)) + ψ(f)

⊆ W (x1, . . . , xn, 0,..., 0) + ψ(f)

= W (x1, . . . , xn, ψ(f)(x1), . . . , ψ(f)(xn)).

Therefore, ψ is continuous at f and hence continuous on Cp(X, Z) since f is arbitrary in

Cp(Y, Z).

−1 Now, deﬁne µ : Cp(X, Z) → Cp(Y, Z) by µ(g) = Lg ◦ ϕ ◦ eˆZ. Using a similar argument, it is clear that µ is a continuous linear map. We claim that µ is the inverse function of ψ. If 74

f ∈ Cp(X, Z) and x ∈ X, then

−1 φ(µ(g))(x) = ψ(Lg ◦ ϕ ◦ eˆZ)(x)

= (L −1 ◦ ϕ ◦ eˆ )(x) (Lg◦ϕ ◦eˆZ) Z

= L −1 (ϕ(ˆe (x))) (Lg◦ϕ ◦eˆZ) Z

= L −1 (ϕ(ex)) (Lg◦ϕ ◦eˆZ) nx ! X (x) (x) = L −1 k e (Lg◦ϕ ◦eˆZ) i yi i=1 nx ! X (x) = k e(x) (L ◦ ϕ−1 ◦ eˆ ) i yi g Z i=1 nx X (x) = k e(x)(L ◦ ϕ−1 ◦ eˆ ) i yi g Z i=1 nx X (x) −1 = ki (Lg ◦ ϕ ◦ eˆZ)(yi) i=1 nx X (x) −1 = ki (Lg ◦ ϕ )(eyi ) i=1 nx ! −1 X (x) = (Lg ◦ ϕ ) ki eyi i=1 nx !! −1 X (x) = Lg ϕ ki eyi i=1 = Lg(ex)

= ex(g)

= g(x).

Therefore, φ(µ(g))(x) = g(x) for all x ∈ X and all g ∈ Cp(X, Z). Consequently φ(µ(g)) = g

−1 −1 for all g ∈ Cp(X, Z) and so µ = ψ . Similarly, ψ = µ . Hence, ψ and µ are inverse maps of each other. Thus, ψ is a linear homeomorphism from Cp(Y, Z) to Cp(X, Z) implying

Cp(X, Z) is linearly homeomorphic to Cp(Y, Z).

We have shown in this section that we can try to obtain a result similar to Theorem 75

3.1.7, but in this case we place a weaker condition on the relationship between Cp(X, Z) and Cp(Y, Z). In this situation we determine a relationship between Lp(X, Z) and Lp(Y, Z) rather than between X and Y .

3.3 Density and Cp(X,A), Where A is a Ring

In this section we will consider some dense subspaces in the certain function spaces. In particular, if X is a zero-dimensional space and A is a topological ring, then Cp(X,A) is a dense subspace of AX . This is a slight departure from what we have been doing; this section will be dealing with a more general theory, but we have all of the tools that are needed to proceed.

We will state the theorem in Cp(X)-theory that we want to make an analog to. For more on density in Cp(X)-theory refer to [18].

Theorem 3.3.1 (Theorem 1.1.2, [18]). Let X be a Tychonoﬀ space. The space Cp(X) is a dense subspace of RX .

X Recall that, for spaces X and Y , the product Πx∈X Y is equivalent to Y . Hence we can place the product topology on Y X . In general, if X and Y are space, then the basic open subsets of Y X , in the product topology, are subsets of the form

X {f ∈ Y : f(xi) ∈ Oi, i = 1, . . . , n}

where Oi is open in Y and xi ∈ X for all i = 1, . . . , n. When considering subsets of this form to be base for Y X it is clear that the product topology on that set corresponds to the topology of pointwise convergence. Let X be a zero-dimensional space, and let A be a zero-dimensional topological ring. Our ﬁrst result considers the density of S(X,B) in AX where B is a dense subset of A. Recall that S(X,B) is the collection of all B-simple functions on X mapping into B. 76 Theorem 3.3.2. If X is a zero-dimensional space, A is a topological ring, and B is a dense subset of A, then S(X,B) is dense in AX .

X X Proof. Let f ∈ A , and let Wf = W (x1, . . . , xn,O1,...,On) as a basic open subset of A

containing f, where O1,...,On are open subset of A.

We want to ﬁnd a ϕ ∈ S(X,B) such that ϕ ∈ Wf . Let V1,...,Vn be basic clopen subsets

n in X such that xi ∈ Vi, for i = 1, . . . , n, xi ∈/ Vj, for i 6= j, and ∩i=1Vi = ∅. Such Vi’s exists

since X is Hausdorﬀ and zero-dimensional. We have f(xi) ∈ Ui, for i = 1, . . . , n, and B is

dense in A, thus there is bi ∈ (B ∩ Ui)\{f(xi)}, for each i = 1, . . . , n. Deﬁne ϕ : X → B Pn χ by ϕ = i=1 bi Vi . It is clear that ϕ ∈ S(X,B). We need to verify that ϕ ∈ Wf . Notice

ϕ(xi) = bi ∈ Oi for every i, thus ϕ ∈ Wf since . Therefore, it follows that S(X,B) is a dense subset of AX .

We have the immediate corollary.

Corollary 3.3.3. If X is a zero-dimensional space, then S(X, Q) is a dense subset of RX .

We also have the following:

Corollary 3.3.4. If X is a zero-dimensional space, A is a topological ring, and B a dense

X subset of A, then Cp(X,B) is a dense subspace of A .

X Proof. It follows from Theorem 3.3.2 that S(X,B) is a dense subset of A , thus clAX (S(X,B)) =

X X A . However, S(X,B) ⊆ Cp(X,B) ⊆ A , hence

X clAX (S(X,B)) ⊆ clAX (Cp(X,B)) ⊆ A .

X X So it follows that clAX (Cp(X,B)) = A since clAX (S(X,B)) = A . Therefore, Cp(X,B) is a dense subset of AX .

We of course have the following corollary. 77

Corollary 3.3.5. If X is a zero-dimensional space, then Cp(X, Q) is a dense subspace of RX .

We will make a further generalization of certain results that we have. Let X be a zero- dimensional space, and let A be a topological ring. Suppose B is a clopen base for the space X and B is a subset of A. Deﬁne

( n ) X χ S(X,B, B) = bi Vi : bi ∈ B,Vi ∈ B,Vi’s are disjoint in B, n ∈ N . i=1

We have the following theorem.

Theorem 3.3.6. Let X be a zero-dimensional space, and let A be a topological ring. If B is a dense subset of A and B is a clopen base for X, then S(X,B, B) is a dense subset of

Cp(X,A).

Proof. Suppose B is a dense subset of A and B is clopen base for X. Let f ∈ Cp(X,A), and let Wf = W (x1, . . . , xn; U1 . . . , un) be a basic open subset of Cp(X,A) containing f. Take f(xi) = ai for all i. Since ai ∈ UU and B is a dense subset of A there is bi ∈ B ∩ Ui for all i.

n Now, let {Vi}i=1 be a disjoint collection of clopen subsets of X in B such that xi ∈ Vi for all Pn χ i. Deﬁne ϕ = i=1 bi Vi .

We claim that ϕ ∈ S(X,B, B) and ϕ ∈ Wf . It is clear that ϕ ∈ S(X,B, B. We will Pn χ verify that ϕ ∈ Wf . Notice ϕ(xi) = i=1 bi Vi (xi) = bi for all i. Thus, ϕ(xi) ∈ Ui for all i implying that ϕ ∈ Wf .

Lastly, we will consider a bound on the density of Cp(X,A). Recall that the density of the space is the minimal cardinality of a dense subset of the space. In the above we have seen that the collection, S(X,A), of simple functions is a dense subset of Cp(X,A) for a zero-dimensional space X and a topological ring A. The subset S(X,A) will be used to ﬁnd a bound on the density for the space Cp(X,A). We ﬁrst need the following lemma. 78 Lemma 3.3.7. Let X be a zero-dimensional space, there exists a clopen base of X with cardinality w(X).

Proof. Let w(X) = τ, and let B(X) be a base of X with |B(X)| = τ. We will construct a

clopen base of X with cardinality τ. Let I be an index set for BX , and let Uα be a clopen

subset of X contained in Oα ∈ B(X), for all α ∈ I; we can ﬁnd such a Uα since X is zero-dimensional. Deﬁne

B1 = {Uα : α ∈ I},

n B2 = {∩i=1Uαi : αi ∈ I, n ∈ N},

B3 = {X\U : U ∈ B2} ∪ B2, and

n B = {∩i=1Ui : Ui ∈ B3, n ∈ N}.

We note that B is a collections of clopen subsets of X since it is constructed by finite intersections of clopen subsets of X. We want to show that B is a base of X with cardinality τ. First show that B has cardinality τ. It is clear that |B1| = τ, since |B(X)| = τ. Thus, it follows that |B2| = τ since it is the collection of finite intersections of elements in B1. It follows that |B3| = τ since it it the union of two sets of cardinality τ. Hence, |B| = τ since it is the collection of finite interesting of elements in B3. Next we verity that B is a base for X.

(i) Let x ∈ X, there is a Uα ∈ B(X) such that x ∈ Uα for some α ∈ I. Take Oα ∈ B1, we

have x ∈ Oα or x ∈ X\Oα. However, by construction Oα and X\Oα are both in B. Hence, for each x ∈ X there is a clopen set contained in B containing x.

(ii) Let x ∈ X and let O1,O2 ∈ B such that x ∈ O1 ∩ O2. By deﬁnition we have that

n k O1 = ∩i=1Vi for some Vi ∈ B3, for i = 1, . . . , n, and O2 = ∩j=1Wj for some Wj ∈ B3,

n k for j = 1, . . . , k. It then follows that O3 = (∩i=1Vi) ∩ (∩j=1Wj) ∈ B by deﬁnition of B. 79

Therefore, x ∈ O3 ⊆ O1 ∩ O2 with O3 ∈ B.

Thus, B is a clopen base of X with cardinality equal to w(X).

We will make a remark about the relationship of density with continuous functions.

Proposition 3.3.8 (Theorem 1.4.10, [11]). Let X and Y be space. If there is a continuous surjective map from X to Y , then d(Y ) ≤ d(X).

Now for our theorem for the bound on the density on Cp(X,A).

Theorem 3.3.9. If X is a zero-dimensional space and A is a topological ring, then d(A) ≤

d(Cp(X,A)) ≤ d(A) · w(X).

Proof. We will ﬁrst show that d(A) ≤ d(Cp(X,A)). We call that the map ex : Cp(X,A) → A is continuous and subjective for all x ∈ X. Thus it follows from Proposition 3.3.8 that

d(A) ≤ d(Cp(X,A)). Suppose B be a dense subset of A and B is a clopen base for X with |B| = w(X); the base B exist by Lemma 3.3.7. We will show that the cardinality of S(X,B, B) is bounded by |B| · w(X), which gives the desired result. Consider the following string of inequalities:

∞ ( n ) [ X |S(X,B, B)| = b χ : b ∈ B,U ∈ B,U ∩ U = ∅ if i 6= i i Ui i i i1 i1 1 2 n=1 i=1 ( n ) X = b χ : b ∈ B,U ∈ B,U ∩ U = ∅ if i 6= i i Ui i i i1 i1 1 2 i=1 ≤ |B| · |B|

= |B| · w(X)

= d(A) · w(X).

Since clCp(X,A)(S(X,B, B)) = Cp(X,A), we have that d(Cp(X,A)) ≤ |S(X,B, B)|. There- fore, d(A) ≤ d(Cp(X,A)) ≤ d(A) · w(X).

We have the following corollaries to Theorem 3.3.9. 80

Corollary 3.3.10. If X is a zero-dimensional space, then ℵ0 ≤ d(Cp(X, Z)) ≤ w(X).

The second corollary is the following.

Corollary 3.3.11. If X is a second countable zero-dimensional space and A is a separable topological ring, then Cp(X,A) is separable.

We provide the following examples to demonstrate that the product of d(A) and w(X) is needed for the upper bound on the density of Cp(X,A).

Example 3.3.12. 1. Let Rd denote the set of real numbers R with the discrete topology.

It is clear that Rd is a zero-dimensional topological ring with d(Rd) = c. Thus, it follows

from Theorem 3.3.9 that c ≤ d(Cp(Z, Rd)) ≤ w(Z) · c. Therefore, d(Cp(Z, Rd)) = c

since w(Z) = ℵ0.

2. Again, let Rd be the set R with the discrete topology. We notice that w(Rd) = c and

d(Z) = ℵ0. Thus, ℵ0 ≤ d(Cp(Rd, Z)) ≤ c.

This section has shown that there is a nice connection with the density of Cp(X,A) and the density of A. We note that this result requires the use of the fact that in the case of

Cp(X, Z) we may assume that X is a zero-dimensional space.

3.4 Gδ, Fσ, and the Baire Category Theorem

X We know that if X is a discrete space, then Cp(X, Z) = Z . In this section we would like to remove the discrete assume in the above statement and determining the interaction of

X Cp(X, Z) in Z . In our discussion we are going to show when Cp(X, Z) are certain kinds of

X subsets of Z . In particular, we are going to investigate whether or not Cp(X, Z) is a Gδ or

X Fσ-subset of Z .

Let X be a space, recall that U a subset of X is called a Gδ-subset if it is the countable intersection of open subsets of X. We also have that U is called a Fσ-set if it is the countable union of closed subsets of X. 81 Before we continue we will have to deﬁne the concept of Cech-complete,ˇ in order to deﬁne

this we will have to lay down a few deﬁnitions. Let F = {Fα}α∈I be a collection of subsets

of X, we say that X has the ﬁnite intersection property if F 6= ∅ and Fα1 ∩ ...Fαk 6= ∅

for every ﬁnite subset {α1, . . . , αk} ⊆ I. We say, for a subset A of a space X, that the

diameter of A is less than a cover A = {Aα}α∈I of X provided there is an α ∈ I such that ˇ A ⊆ Aα. A space X is called Cech-complete if there exists a countable family {An}n∈N of open covers of X with the property that any family F of closed subsets of X, which has the

ﬁnite intersection property and contains sets of diameter less than An, for all n ∈ N, has non-empty intersection (refer to page 196 in [11]). We can now state the general Baire Category Theorem and properties of Cech-completeˇ spaces.

Theorem 3.4.1 (Theorem 3.9.3, [11]). (Baire Category Theorem) In a Cech-completeˇ space

X the union ∪n∈NAn, where An is nowhere dense subset of X for all n ∈ N, is a co-dense subset of X.

We have the following propositions about Cech-completeˇ space.

Proposition 3.4.2 (Page 196, [11]). Every locally compact space is a Cech-completeˇ space.

We see that Z is a Cech-completeˇ space since any ﬁnite open set containing any point is a compact subset of Z and hence is locally compact. Furthermore, we need:

Theorem 3.4.3 (Theorem 3.9.8, [11]). The product space of countably many Cech-completeˇ space is Cech-complete.ˇ

It follows that ZX is a Cech-completeˇ space for any countable space X. We are now

X ready for the heart of this section; we will see that Cp(X, Z) is not a Gδ-set in Z . First we

X have to characterize which the spaces X are need for Cp(X, Z) to contain a Gδ-subset of Z .

Lemma 3.4.4. Let X be a zero-dimensional space. If Cp(X, Z) contains a nonempty Gδ- subset of ZX , then X is the topological sum of a countable space and a discrete space. 82 The following proof is similar to the one given by van Mill for Lemma 6.31 in [24] in the

Cp(X) case.

X Proof. Suppose that S is a Gδ-subset of Z which is contained in Cp(X, Z). Since S is a

X X Gδ-subset of Z take open subsets Un of Z , for n ∈ N, such that S = ∩nNUn. If f ∈ S,

then f ∈ Un for all n ∈ N.

X Now, for every n there exists a ﬁnite subset Fn of X such that for every g ∈ Z with

f|Fn = g|Fn we have that g ∈ Un. This is clear since we have that

f ∈ W x(n), . . . , x(n), f x(n) , . . . , f x(n) ⊆ U , 1 kn 1 kn n

n o where x(n), . . . , x(n) is an enumeration of F . Now, deﬁne F = ∪ F , we see that F is 1 kn n n∈N n a countable set since it is the countable union of ﬁnite sets and for every g ∈ ZX with the

property that f|F = g|F , we ﬁnd that g ∈ ∩nNUn. Hence g ∈ S. However, since S is a subset

of Cp(X, Z), we have that g is a continuous functions. Consider the remaining part of the space X, that is consider A ⊆ X\F . We claim that X\F is a discrete subspace of the space X. We will show this by considering the function

χA. We see that χA|F = 0, thus it follows that (f + χA)|F = f|F + χA|F = f|F . Hence, by

the above, we have that f + χA is a continuous map on X. Since f is also a continuous map

on X we have that χA = (f + χA) + f is a continuous map on X. Therefore, A is a clopen subset of X\F . It follows that every subset of X\F is clopen since A is an arbitrary subset of X\F implying X\F is a discrete space. Consequently, it follows that X is a topological sum of a countable space F and a discrete space X\F .

Next, we will have a lemma in regards to the Cp(X, Z) containing a dense Gδ-subset of ZX .

Lemma 3.4.5. Let X be a countable zero-dimensional space. If Cp(X, Z) contains a dense

X Gδ-subset of Z , then X is discrete. 83 X Proof. Assume that Cp(X, Z) contains a dense Gδ-subset of Z . Suppose X is not a discrete

X X space. It follows that Z \Cp(X, Z) 6= ∅. Let g ∈ Z \Cp(X, Z).

X X Deﬁne ϕg : Z → Z by ϕg(f) = f + g. We claim that ϕg is a homeomorphism from

X X X X Z to Z . We ﬁrst show that ϕg is a bijective map. We will deﬁne ψg : Z → Z by

X ψg(f) = f − g and claim that ψg is the inverse function of ϕg. Let f ∈ Z . It following that

(ϕg ◦ ψg)(f) = ϕg(ψg(f)) = ϕg(f − g) = f − g + g = f

and

(ψg ◦ ϕg)(f) = ψg(ϕg(f)) = ψg(f + g) = f + g − g = f.

Hence, ϕg and ψg are inverse functions of each other, thus ϕg is bijective.

X X X Now, to show that ϕg is continuous. We have that + : Z × Z → Z is a continuous map. Thus it follows that the map +g : ZX → ZX deﬁned by +g(f) = f + g is a continuous

map as it is a restriction of +. However, ϕg = +g, so it follows that ϕg is a continuous

map. A similar argument gives that ψg is also continuous. Therefore, we have that ϕg is a

homeomorphism between ZX and ZX .

X Now, let S be a dense Gδ-subset of Z that is contained in Cp(X, Z). Since S is a

X Gδ-subset of Z there is a collection {On}n∈N open subsets of X such that S = ∩n∈NOn. X Since ϕg is a homeomorphism we have that ϕg(On) is open in Z , for all n ∈ N, and

X ϕg(S) = ϕg(∩n∈NOn) = ∩n∈Nϕg(On). Thus, we have that ϕg(S) is a Gδ-subset of Z . X X X Moreover, ϕg(S) ⊆ Z \Cp(X, Z) and ϕg(S) is a dense subset of Z since S is dense in Z . Recall that Z is a Cech-completeˇ space and that Theorem 3.4.3 gives use that ZX is also ˇ X X X Cech-complete space. Since S and ϕg(S) are dense in Z we have that Z \S and Z \ϕg(S) are nowhere dense in ZX . Thus, it follows from the Baire Category Theorem that

X X X Z \ Z \S ∪ Z \ϕg(S) 84 X is a dense subset of Z . However, we have that S ∩ ϕg(S) = ∅. Therefore we have

X X X X X Z \ Z \S ∪ Z \ϕg(S) = Z \ Z \(S ∩ ϕg(S))

X X = Z \ Z \∅

X X = Z \Z

= ∅.

Hence, we have a contradiction. Thus, we must have that X is a discrete space.

X Now, for the ﬁnal theorem in regards to Gδ-subsets of Z and Cp(X, Z).

X Theorem 3.4.6. Let Xbe a zero-dimensional space. If Cp(X, Z) is a Gδ-subset of Z , then X is a discrete space.

X Proof. Suppose Cp(X, Z) is a Gδ-subset of Z . We will ﬁrst recall that Cp(X, Z) is a dense subset of ZX . It follows from Lemma 3.4.4 that X is a topological sum of a countable space and a discrete space. Let C be a countable space and D a discrete space such that

X = C ⊕ D. We have that Cp(X, Z) is homeomorphic to Cp(C, Z) × Cp(D, Z). However,

D since D is discrete, we have that Cp(D, Z) = Z . Thus, Cp(X, Z) is homeomorphic to

D Cp(C, Z) × Z .

C We claim that Cp(C, Z) is a Gδ-subset of Z . We have that Cp(X, Z) is a Gδ-subset of

X X Z , thus there is a collection {On}n∈N of open subsets of Z such that Cp(X, Z) = ∩n∈NOn. ˆ ˆ ˆ Deﬁned On = {f|C : f ∈ On} for all n ∈ N. We claim that Cp(C, Z) = ∩n∈NOn and that On is an open subset of ZC for all n ∈ N. ˆ ˆ First, to show that Cp(C, Z) = ∩n∈NOn. It is clear that ∩n∈NOn ⊆ Cp(C, Z) since each ˆ function in ∩n∈NOn is a continuous function from C to Z. Now, to show that Cp(C, Z) ⊆ 85 ˆ ∩n∈NOn. For every f ∈ Cp(C, Z) and every g ∈ Cp(D, Z) deﬁne

  f(x) x ∈ C fg(x) =  g(x) x ∈ D.

It is clear that fg is a continuous function from X = C × D to Z, so fg ∈ Cp(X, Z). Thus ˆ fg ∈ ∩n∈NOn, implying fg ∈ On for all n ∈ N. However, fg|C = f, hence f ∈ On for all n ∈ N. ˆ ˆ ˆ Thus, we have that f ∈ ∩n∈NOn, so Cp(C, Z) ⊆ ∩n∈NOn. Consequently, Cp(C, Z) = ∩n∈NOn. ˆ C X Next, to show that On is open in Z for all n ∈ N. Since On is an open subset of Z we have O = ∪ W x(α), . . . , x(α), n(α), . . . , n(α) were W x(α), . . . , x(α), n(α), . . . , n(α) is n α∈I 1 kα 1 kα 1 kα 1 kα a basic clopen subset of ZX for all α ∈ I. Let α ∈ I, we have that

n o W (C) = f| : f ∈ W x(α), . . . , x(α), n(α), . . . , n(α) α C 1 kα 1 kα

(α) (α) (α) (α) = W x (C) , . . . , x (C) , n (C) , . . . , n (C) 1 kα 1 kα

n o n o where x(α) , . . . , x(α) = C ∩ x(α), . . . , x(α) . Thus W (C) is an open subset of C for all (C) (C) 1 kα α Z 1 kα ˆ (C) ˆ C α ∈ I. Moreover, On = ∪α∈I Wα for all n ∈ N. Therefore, On is an open subset of Z for all n ∈ N. ˆ C ˆ Finally, we have ∩n∈NOn is a Gδ-subset of Z and Cp(C, Z) = ∩n∈NOn. Then it follows from Lemma 3.4.5 that C is a discrete space. Therefore, since C and D are discrete we have that X = C × D is a discrete space, providing the desired result.

X We will now consider Cp(X, Z) as a Fσ-subset of Z . We will see that we have a similar result that that of Theorem 3.4.6.

X Theorem 3.4.7. Let X be a zero-dimensional space. If Cp(X, Z) is a Fσ-subset of Z , then X is a discrete space.

X Proof. Assume that Cp(X, Z) is a Fσ-subset of Z . Suppose X is not discrete. Since Cp(X, Z)

X X is an Fσ-subset of Z there exists a collection {Vn}n∈N of closed subsets in Z such that 86

Cp(X, Z) = ∪n∈NVn.

Let x0 be a non-isolated point in X, that is the subset {x0} is not an open subset of X.

∞ We will construct a sequence of clopen subsets {Un}n=0 of X containing x0 and a sequence

∞ {fn}n=0 of functions from X to {0, 1} that satisfy the following conditions for all n ∈ N:

(i) f0 ≤ fn ≤ fn+1,

(ii) U0 ⊇ Un ⊇ Un+1,

(iii) f0(x0) = fn(x0) = 1,

(iv) fn|Un\{x0} = 0,

(v) fn|Un\{x0} is continuous,

(vi) fn+1|X\Un = fn, and

X (vii) if f ∈ Z is such that f|(X\Un)∪{x0} = fn|(X\Un)∪{x0}, then f∈ / Fn+1.

We will define f0 and U0, and inductively define the remainder of the sequences. Define χ U0 = X and f0 = {x0}. Assume that we have fi and Ui for all i = 1, . . . , n. We will now construct fn+1 and Un+1. Let g ∈ Cp(X, {0, 1}) such that g(x0) = 1, g(X\Un) = {0}, and −1 −1 χ X\Un 6= g ({0}). Let Un+1 = g ({1}) and fn+1 = (X\Un+1)∪{x0}.

It is clear that Un+1 is a clopen subset of X and that fn+1 and Un+1 satisfy conditions

∞ (i)-(vi). Now, set f = limn→∞ fn and let U = ∩n=1Un. We notice that U is a closed subset of X since Un is a clopen subset of X for each n. We will also notice that f|(X\Un)∪{x0} = fn|(X\Un)∪{x0}, so it follows from (vii) that f∈ / ∪n∈NVn. Hence, f is not a continuous map from X to Z since Cp(X, Z) = ∪n∈NVn. However, we will show that f is continuous. If x ∈ X such that x∈ / U, then by (ii) we have that x∈ / Un for some n. It follows from (v) and (vi) that f is continuous at the point x. Now, if x ∈ U, then x ∈ Un for all n. Recall that Un is a clopen subset of X, so in particular Un is an open subset of X containing x. Notice f(Un) ⊆ {0, 1} by (ii) and (iii). 87 Therefore f is continuous at x by (i), (iii), (v), and (vi). Thus, f is a continuous map, giving a contradiction. So, x0 is an isolated point and thus X is a discrete space.

We have a clear understanding of what the space Cp(X, Z) looks like when X is a discrete space. This section has given us a few situations in which we can say that X is a discrete

space based on information about the space Cp(X, Z). In particular, if we know that Cp(X, Z)

X is a Gδ or Fσ-subset of Z , then X is a discrete space. 88

CHAPTER 4

Classiﬁcation of Cp(X, Z) Into Diﬀerent Categories of Spaces

4.1 Metrizablility of Cp(X, Z)

In this section we will determine which conditions on the space X and on the topological

ring A are needed to guarantee Cp(X,A) is a discrete space. Furthermore, we will determine

when Cp(X, Z) is a metrizable space. Since discreteness is a special case of a metrizable space we will consider discreteness ﬁrst and then secondly metrizablity of Cp(X, Z). We will ﬁrst talk a little bit about what happens when X is a discrete space. If X is a discrete space, then it is easy to see that it is a zero-dimensional space. We notice that if X is a discrete space and Y is any topological space, then every map f : X → Y is a continuous

X map. Thus, if X is discrete, then Cp(X, Z) = Z , so for Cp(X, Z) to be discrete X must be ﬁnite. We will show the more general case of this.

It is obvious that Cp(X) is never a discrete space. It is easy to see this by Proposition

2.3.6 which says that Cp(X) is never a zero-dimensional space. Since discrete spaces are

zero-dimensional space it follows that Cp(X) can not be discrete.

Theorem 4.1.1. Let X be a zero-dimensional space. Cp(X, Z) is a discrete space if and 89 only if X is ﬁnite.

Proof. (⇐) Assume X be ﬁnite. The result is obvious from the above.

(⇒) Assume that Cp(X, Z) is a discrete space. Let f ∈ Cp(X, Z), it follows that {f}

is an open subset of Cp(X, Z), since it is discrete. Since {f} is open there is a basic open

subset Wf = W (x1, . . . , xn, f(x1), . . . , f(xn)) of Cp(X, Z) such that f ∈ Wf ⊆ {f}, for some

x1, . . . , xn ∈ X.

Suppose, by way of contradiction, that |X| ≥ ℵ0. Since X is a Hausdorﬀ space singleton subsets are closed, thus V = {x1} ∪ ... ∪ {xn} is a closed subset of X. Let x ∈ X\V . Since X is zero-dimensional and X\V is open in X, there is a clopen subset O of X such that

x ∈ O ⊆ X\V . Consider χO, this is a continuous map from X to Z since O is a clopen

subset of X. Observe that f(xi) + χO(xi) = f(xi) for all i = 1, . . . , n since xi ∈/ O for all

i = 1, . . . , n. It follows that f + χO ∈ Wf , implying f + χO ∈ {f}. Hence, we have a

contradiction since {f} is a singleton set. Therefore, |X| < ℵ0, i.e. X is ﬁnite.

One would hope that to replace Z in Theorem 4.1.1 with a zero-dimensional topological

ring A to generalize the theorem, but that is not the case. Consider Cp(X, Q), the space Q is a zero-dimensional topological ring, but it is not discrete space. Hence, the same reasoning

used to show Cp(X) can be apply to Cp(X, Q) to show that can not be a discrete space. However, we will obtain a more general result by considering the appropriate conditions on the topological ring A.

Theorem 4.1.2. Let X be a zero-dimensional space, and let A a topological ring. Cp(X,A) is discrete if and only if A is discrete and X is ﬁnite.

Proof. (⇐) Assume that X is ﬁnite and A is a discrete space. The result is obvious.

(⇒) Assume that Cp(X,A) is discrete. We will ﬁrst show that A is discrete. Recall that

A is homeomorphic to a subspace of Cp(X,A) and that every subspace of a discrete space is discrete. Thus, it is clear that A is a discrete space. 90

Next, we show that X is ﬁnite. Assume, by way of contradiction, that |X| ≥ ℵ0. Let f ∈

Cp(X,A), it follows that {f} is an open subset of Cp(X,A), since Cp(X,A) is a discrete space.

Note there is a basis open subset W (x1, . . . , xn,U, ...,Un) of Cp(X,A) of Cp(X,A) such that f ∈ W (x1,...,Xn,U, ...,Un) ⊆ {f}. Thus, it is clear that W (x1,...,Xn,U, ...,Un) = {f} and that f(xi) ∈ Ui for all i = 1, . . . , n.

Now, subset V = {x1, . . . , xn} is a closed subset of X, since X is a Hausdorﬀ space. Thus, X\V is an open subset of X. Now, let O be a basic clopen subset of X such that O ⊆ X\V . Deﬁne g : X → A by

  0A, if x ∈ X\O g(x) =  a, if x ∈ O

for some a ∈ A\{0A}. It is clear by the Pasting Lemma that g ∈ Cp(X,A). Now, consider f + g, we see that f(xi) + g(xi) = f(xi) + 0A = f(xi) for all i = 1, . . . , n. Thus, f + g ∈

W (x1, . . . , xn,U, ...,Un), hence f + g ∈ {f}. However, f 6= f + g and {f} is a singleton set giving a contradiction. Therefore |X| < ℵ0, i.e. X is ﬁnite.

We ﬁnish this section by considering when Cp(X, Z) is it metrizable. Knowing when

Cp(X, Z) is metrizable is one of the preferable conditions we can have for a the space when there is consideration for its algebraic structure.

We will state the corresponding theorem for Cp(X).

Theorem 4.1.3 (Corollary 6.3.3, [24]). Let X be a Tychonoﬀ space. Cp(X) is metrizable if and only if X is a countable space.

We will be using a fact about topological groups to show our result. In order to determine when Cp(X, Z) is metrizable we just need to determine when it is ﬁrst-countable, as the following classic result by Birkhoﬀ shows.

Theorem 4.1.4 (Main Theorem, [7]). A Hausdorﬀ topological group is metrizable if and only if it is ﬁrst countable. 91

With Theorem 4.1.4 in hand we see that we only need consider the character of Cp(X, Z) to show metrizablity since Cp(X, Z) is a Hausdorﬀ topological group. Now, for a few lemmas and propositions needed for the main result for the metrizablity of Cp(X, Z).

X Lemma 4.1.5. For a non-ﬁnite zero-dimensional space X, w Z ≤ |X|.

Proof. Consider the standard open base subsets of ZX , i.e. the collection of subsets

B = {W (x1, . . . , xk, n1, . . . , nk): x1, . . . , xk ∈ X, n1, . . . , nk ∈ Z, k ∈ N}.

Consider the cardinality of B. We see that B can be expressed as the following:

[ [ [ B = {W (x1, . . . , xk, n1, . . . , nk)}. k∈N x1,...,xk∈X n1,...,nk∈Z

X It is easy to see that |B| ≤ |X|. Thus, w Z ≤ |X|.

Now, let us see the relationship of the character and weight of Cp(X, Z) and the cardinality of X.

Proposition 4.1.6. For a non-ﬁnite zero-dimensional space X,

|X| = χ(Cp(X, Z)) = w(Cp(X, Z)).

Proof. First note that it follows from Lemma 4.1.5 that w(ZX ) ≤ |X|. Hence, we have the following string of inequalities

X χ(Cp(X, Z)) ≤ w(Cp(X, Z)) ≤ w Z ≤ |X|.

We wish to show that |X| ≤ χ(Cp(X, Z)). Assume the contrary, that χ(Cp(X, Z)) < |X|.

Let η be a ﬁxed open base of Cp(X, Z) at 0 such that χ(Cp(X, Z)) = |η| < |X|. Without loss of generality, we may assume that the elements of η are standard basic open subsets 92

of Cp(X, Z). We may do this since there is a basic standard clopen subset of Cp(X, Z)

containing 0 contained in each element of η. For each W = W (x1, . . . , xk, 0,..., 0) ∈ η,

we shall deﬁne K(W ) = {x1, . . . , xk}, and deﬁne Y = ∪{K(W ): W ∈ η}. It is clear that Y ⊆ X, which gives |Y | ≤ |X|. Moreover, |Y | < |X|. To see this observe the following. The

function K : η → [X]<ℵ0 , given by the above, is clearly injective because [X]<ℵ0 = |X|. Also, if K is surjective, then |η| = |X| which is a contradiction with |η| < |X|. Hence, X\Y 6= ∅.

∗ ∗ Let x ∈ X\Y , and let U = W (x , 0). Consider V = W (x1, . . . , xk, 0,..., 0) ∈ η. Notice

∗ that x1, . . . , xk ∈ Y and x 6= xi for all i = 1, . . . , k. Thus, there is a function gV ∈ Cp(X, Z)

∗ such that gV (xi) = 1 for i = 1, . . . , k and gV (x ) = 0. It is clear gV ∈ U\V for all V ∈ η, hence U\V 6= ∅ for all V ∈ η. So, W (x∗, 0) does not contain any V where V ∈ η, implying

η is not a base. Therefore, |X| ≤ χ(Cp(X, Z)) and consequently the desired result.

As an immediate result of Proposition 4.1.6 we have that if the zero-dimensional space

X is countable, then Cp(X, Z) is second countable, which implies ﬁrst countable. We now have all that we need to state the condition suﬃcient and necessary on X so

that Cp(X, Z) a metrizable space.

Theorem 4.1.7. Let X be a zero-dimensional space. Cp(X, Z) is metrizable if and only if X is countable.

Proof. We have Cp(X, Z) is a topological group and that Cp(X, Z) is a Hausdorﬀ space by

Remark 2.3.5. Thus, Cp(X, Z) is metrizable if and only if it is ﬁrst countable, by Theorem

4.1.4. As a result of Proposition 4.1.6 we have that Cp(X, Z) is ﬁrst countable if and only if

X is countable. Therefore, Cp(X, Z) is metrizable if and only if |X| is countable.

Here are some examples of when Cp(X, Z) is metrizable.

Example 4.1.8. It follows from the Theorem 4.1.7 that each of the following is metrizable:

Cp(N, Z), Cp(Z, Z), and Cp(Q, Z).

We also have the following corollary to Proposition 4.1.6 and Theorem 4.1.7. 93 Corollary 4.1.9. Let X be a zero-dimensional space. The following are equivalent.

(i) Cp(X, Z) is separable and metrizable.

(ii) Cp(X, Z) is metrizable.

(iii) Cp(X, Z) is ﬁrst countable

(iv) X is countable.

Proof. (i) ⇒ (ii) Assume that Cp(X, Z) is separable and metrizable. It is clear that we have

that Cp(X, Z) is metrizable.

(ii) ⇒ (i) Assume that Cp(X, Z) is metrizable. We wish to show that Cp(X, Z) is a separable space. We have that X is countable. Thus, we have from Proposition 4.1.6 that Cp(X, Z) has weight of ℵ0. Now, let B be a base of clopen subsets for Cp(X, Z) with

cardinality ℵ0. Enumerate B as the collection {On}n∈N. Let fi ∈ Oi\ ∪i6=j Oj, and let

S = {fn}n∈N. It is clear that |S| ≤ ℵ0.

We claim that S is a dense subset of Cp(X, Z). Let f ∈ Cp(X, Z), and let O be an

open subset of Cp(X, Z) containing f. Since O is an open subset of Cp(X, Z) we have that

O = ∪i∈N Oi where N is some subset of N. Hence fi ∈ O for some i ∈ N. Thus, we have

that S is dense in Cp(X, Z). Therefore, Cp(X, Z) is a separable space implying the desired result. (ii) ⇔ (iv) The result follows directly from Theorem 4.1.7. (iii) ⇔ (iv) The result follows directly from Proposition 4.1.6.

We now know when Cp(X, Z) is a metrizable space. In the process of showing Theorem 4.1.7 we had to show some of the simple cardinal invariants, hence we ﬁnd that the cardinal

invariants play an interesting and important role in the study of Cp(X, Z). We will see later more results in regards to cardinal invariants. 94 4.2 Fr´echet-Urysohn Spaces and Cp(X, Z)

We have already determined when Cp(X, Z) will be metrizable; the suﬃcient and necessary

for Cp(X, Z) to be metrizable is for X to be countable. The next step would be to consider

when Cp(X, Z) is a Fr´echet-Urysohn space. A Fr´echet-Urysohnspace is a space X which

has the property that if A ⊆ X and x ∈ clX (A), then there is a sequence {xn}n∈N in A which converges to x. Also, deﬁne a Fr´echet-Urysohnpoint in a space X to be a x ∈ X such

there there is sequence {xn}n∈N in X which convergence to x. If X is a Fr´echet-Urysohn

space, then X is countably tight, that is t(X) = ℵ0. Also, every first countable space is a Fréchet-Urysohn space. First notice the structure of a subspace of a Fréchet-Urysohn space.

Lemma 4.2.1. A subspace of a Fr´echet-Urysohnspace is a Fr´echet-Urysohnspace.

Proof. Let X be a Fr´echet-Urysohn space, and let Y a subspace of X. Let A ⊆ Y , and let

y ∈ clY (A). Since Y is a closed subspace of X it follows that clY (A) is closed in X. Since X

is a Fr´echet-Urysohn space there is a sequence {yn}n∈N in A such that {yn}n∈N converges to y

in X. Since A ⊆ Y it follows that {yn}n∈N converges to y in Y . Thus, Y is a Fr´echet-Urysohn space.

Let S be a set, and let ζ = {Sn}n∈N a sequence of subsets of S. Deﬁne the limit inferior ∞ ∞ of the sequence ζ to be lim inf ζ = ∪k=1 ∩n=k Sn, and deﬁne limit superior of the sequence

∞ ∞ ζ to be lim sup ζ = ∩k=1 ∪n=k Sn. Also, deﬁne an ω-cover of a space X as follows, if A is a family of subsets of X, then A is a ω-cover of X if for each ﬁnite set F of X there is a

U ∈ A such that F ⊆ U (refer to page 51 of [4]). Let ζ1, and let ζ2 be ω-covers for a space,

we will use the notation ζ2 < ζ2 to mean that each U ∈ ζ1 is contained in some V ∈ ζ1. We have a few general results for ω-covers of a space.

Proposition 4.2.2. If ζ is a sequence of subsets of a space X with lim inf ζ = X, then ζ is an ω-cover of X. 95

Proof. Let ζ = {Xn}n∈N be a sequence of subsets in a space X with lim inf ζ = X. Suppose ∞ ∞ that F is a ﬁnite subset of X. We have that F ⊆ lim inf ζ, thus F ⊆ ∪k=1 ∩n=k Xn. By

∞ way of contradiction, suppose F * Xn for any n ∈ N. We would have that F * ∩n=kXn

∞ ∞ for any k ∈ N. Thus, F * ∩k=1 ∪n=k Xn which gives a contradiction to the fact that

∞ ∞ F ⊆ ∪k=1 ∩n=k Xn. Thus, F ⊆ Xn for some n ∈ N, which implies ζ is an ω-cover of X.

Another result for ω-covers is the following.

Proposition 4.2.3. If X is zero-dimensional, t(Cp(X, Z)) = τ, and η is a clopen ω-cover of X, then there is an ω-subcover η0 of η with |η0| = τ.

Proof. Assume X is zero-dimensional, t(Cp(X, Z)) = τ, and that η is a clopen ω-cover of X.

Define S = {χX\C : C ∈ η}, it is clear that S is a subset of Cp(X, Z). χ χ Notice that 0 ∈ S as 0 = ∅ = X\X , where X is a clopen set. Thus, 0 ∈ clCp(X,Z)(S). 0 0 Now, since t(Cp(X, Z)) = τ and 0 ∈ clCp(X,Z)(S) there is a S ⊆ S such that 0 ∈ clCp(X,Z)(S ) 0 0 0 and |S | ≤ τ. Now, define η = {C ∈ η : χX\C ∈ S}; it is clear that η is a subcollection of clopen subsets which are in η. We now want to show that η0 is in fact a ω-cover, show it is an ω-subcover of η with |η0| ≤ τ. It is clear that η0 is an ω-cover of X as any finite set F of X has the following property F ⊆ X ∈ S0, by definition. Therefore, there is in fact a ω-subcover η0 such that |η0| ≤ τ.

We now have the tools needed to show when Cp(X, Z) is a Fr´echet-Urysohn space.

Theorem 4.2.4. Let X be a zero-dimensional space. The following are equivalent:

(i) Cp(X, Z) is a Fr´echet-Urysohnspace.

(ii) X has the property that for any open ω-cover η of X there is a sequence ξ ⊆ η such that lim inf ζ = X.

(iii) X has the property that for any sequence {ηn}n∈N of open ω-covers of X there are

Un ∈ ηn, for all n ∈ N, such that lim inf ζ = X, where ζ = {Un}n∈N. 96

(iv) Cp(X, Z)N is a Fr´echet-Urysohnspace.

Note that conditions (ii) and (iii) are independent of Cp(X, Z), thus the proof that (ii) implies (iii) will be the same as the proof given by Arkhangel0ski˘ıon page 52 in [4]. However, the remainder of the proof will be a modiﬁcation of Arkhangel0ski˘ı’sproof so that it holds

for the Cp(X, Z) case.

Proof. Let X be a zero-dimensional space.

(i)⇒(ii) Assume that Cp(X, Z) is a Fr´echet-Urysohn space. We will show that for any open ω-cover η of X there is a sequence ζ ⊆ η such that lim inf ζ = X. Let η be an open

−1 ω-cover of X. We will deﬁne A = {f ∈ Cp(X, Z) : clX (f (Z\{0})) ⊆ U, U ∈ η}. Without

loss of generality 1 ∈ clCp(X,Z)(A), we may do this since we can translate A such that it contains 1.

Since Cp(X, Z) is a Fr´echet-Urysohn space there is a sequence {fn}n∈N in A such that fn −1 converge to 1. For each n ∈ N we will take Un ∈ η such that fn (Z\{0}) ⊆ Un. These Un’s

exist since fn ∈ A for each n ∈ N and all the elements of A satisfy the given condition. Now,

let ζ = {Un}n∈N, clearly ζ ⊆ η. Hence, we need to show that ζ has the desired condition for X.

Let x ∈ X. Since fn converges to 1, there is an n(x) ∈ N such that fn ∈ W (x, 1) for all

−1 n ≥ n(x). Then it follows that fn(x) 6= 0 for all n ≥ n(x). Thus, x ∈ fn (Z\{0}) for all

n ≥ n(x). So we have x ∈ Un for all n ≥ n(x) which implies x ∈ ∩n≥n(x)Un. It follows that

∞ ∞ x ∈ ∪k=1 ∩n=k Un. Hence, X ⊆ lim inf ζ. Therefore, for any open ω-cover η there is a ζ ⊆ η such that lim inf ζ = X. (ii)⇒(iii) We provide the following implication for completeness. Suppose for any open ω-cover η of X there is a ζ ⊆ η such that lim inf ζ = X. We wish to show that for any

sequence {ηn}n∈N of ω-covers of X there are Un ∈ ηn, for all n ∈ N, such that lim inf % = X,

where % = {Un}n∈N.

Case 1 : Suppose |X| < ℵ0 and suppose that {ηn}n∈N is a sequence of open ω-covers.

Since X is ﬁnite it is contained in some Un which is an element of ηn, for all n ∈ N. Thus, 97 take the collection % = {Un}n∈N. Since X ⊆ Un for all n ∈ N, it is clear that X ⊆ lim inf %. Thus, we have that X = lim inf %.

Case 2 : Suppose |X| ≥ ℵ0. Let S = {xn}n∈N such that xn ∈ X for all n ∈ N and xi 6= xj whenever i 6= j, and let {ηn}n∈N be a sequence of ω-covers of X. Without loss of generality ∗ we may assume that ηn < ηn+1 for all n ∈ N. Otherwise we can take the collection {ηn}n∈N, ∗ ∞ where ηn is defined as {∪k=nUk : Uk ∈ ηk, k ≥ n}, which is clearly an inscribed sequence of open ω-covers. We now wish to construct % such that it has the desired property. We shall now begin the construction of the needed %. However, we need to make an intermediary construction of an open ω-cover. Letη ˆn = {U\{xn} : U ∈ ηn} for all n ∈ N, and letη ˆ = ∪n∈Nηˆn. We claimη ˆ is actually an ω-cover as well. Let F be a finite subset of X, we want to show that F is contained in some U ∈ ηˆ. Since ηn is an ω-cover there is a Un ∈ ηn such that F ⊆ Un for each n ∈ Z. Furthermore, since F is finite there is a n0 ∈ N such that xn0 ∈/ F where xn0 ∈ S. Thus, we have F ⊆ Un0 \{xn0 }, which is inη ˆn0 . Sinceη ˆn0 ⊆ ηˆ we have Un0 ∈ ηˆ. Therefore,η ˆ is an open ω-cover. It then follows from the assumption on X that there is a %1 ⊆ ηˆ such that %1 = {Un}n∈N and lim inf %1 = X.

We will now use %1 to construct %. First, we have that Unk = Vnk \{xkn }, where Vnk ∈ ηnk for some kn ∈ N for all n ∈ N. We will consider the sequence {kn}n∈N. First notice that the sequence {kn}n∈N is unbounded. We see this by assuming that kn ≤ M for some M ∈ N, for all n ∈ N, which gives that there is no element of %1 containing the set {x1, . . . , xm}, but %1 is an ω-cover of X by Proposition 4.2.2, so {kn}n∈N can not be bounded. Thus, %2 = {Unk }n∈N gives us that lim inf %1 = lim inf %2. This implies that kn < kn+1 for all n ∈ N and so the sequence {kn}n∈N is unbounded.

Now, without loss of generality, put k1 = 1, otherwise we can shift the subsequence by k1 − 1. For a ﬁxed n ∈ N we shall take Wm ∈ ηm for m ∈ {kn, . . . , kn+1 + 1} as follows: Wkn = Vkn , Wkn−l ∈ ηkn−l such that Wkn−l+1 ⊆ Wkn−l. We deﬁne the desired % by

% = {Wn}n∈N. Lastly, we need to show that % is an open ω-cover, to do this we show that lim inf % = X, 98 it would then follow from Proposition 4.2.2 that % is an open ω-cover. Now, let x ∈ X. Take

n ∈ N such that x ∈ clX (Vkm ) for all m ≥ n. We then have x ∈ Wi for all i ≥ kn. Now,

if km < i ≤ km+1 where m ≥ n, then x ∈ Vkm+1 and hence x ∈ Wi, by the deﬁnition of %.

∞ ∞ Thus, x ∈ ∩i∈NWi, so x ∈ ∪n=1 ∩i=n Wi. Consequently, lim inf % = X, so % is a open ω-cover of X by Proposition 4.2.2. Therefore, X has the desired property.

(iii)⇒(iv) Assume that X has the property that for any sequence {ηn}n∈N of open ω- covers of X there are Un ∈ ηn, for all n ∈ N, such that lim inf ζ = X, where ζ = {Un}n∈N. First recall Theorem 2.1.4 tells us Cp(X, Z)N is homeomorphic to Cp X, ZN . Thus, it is enough to show that Cp X, ZN is a Fr´echet-Urysohn space, because of the homeomorphic equivalence.

N N Fix a constant c ∈ Z . Let B = {On}n∈N be countable base at c in Z such that

N clZN (On+1) ⊆ On, for all n ∈ N. It is enough to show that f ∈ Cp X, Z , where f(X) = {c}

is a Fr´echet-Urysohn point. Thus, we need to construct a sequence {fn}n∈N such that it converges to f.

We have that subsets of the form W (x1, . . . , xk, c, . . . , c), where k ∈ N and {x1, . . . , xk} ⊆ X, form a base at f in C X, N. Let A ⊆ C X, N such that f ∈ cl (A)\A. Deﬁne p Z p Z Cp(X,ZN) −1 ηn = {f (On): f ∈ A}. We claim that ηn is an open ω-cover of X for each n ∈ N.

We show that ηn is an open ω-cover for each n ∈ N. Let F = {x1, . . . , xk} be a ﬁnite set in X and consider η where n ∈ . Since f ∈ cl (A)\A and f ∈ W (x , . . . , x , c, . . . , c) n N Cp(X,ZN) 1 k

it follows that A ∩ W (x1, . . . , xk, c, . . . , c) 6= ∅. So there is a g ∈ A ∩ W (x1, . . . , xk, c, . . . , c).

−1 Now since c ∈ On, we have g(xi) ∈ On for all i = 1, . . . , k. Thus, {x1, . . . , xk} ⊆ g (On).

Thus, ηn is an open ω-cover of X for all n ∈ N since n is arbitrary. It follows from the

−1 assumption on X that there is a sequence {fn}n∈N in A such that % = {fn (On)}n∈N has the property that lim inf % = X.

We now need to show that the sequence {fn}n∈N converges to f. Consider the basic

open subset W (x1, . . . , xk, c, . . . , c) of Cp(X, ZN) containing f, and let F = {x1, . . . , xk}.

−1 For each x ∈ F ﬁx n(x) ∈ N such that x ∈ ∩m≥n(x)fm (Om). Let l = max{n(xi): i = 99 −1 1, . . . , k}. It follows that F ⊆ ∩m≥lfm (Om). Set p = max{l, n}, and take j ≥ p. We

−1 then have f ⊆ fj (On) which implies fj(F ) ⊆ Op ⊆ On. Finally, this gives that fi ∈

W (x1, . . . , xk, c, . . . , c) for all i ≥ p, and thus the sequence {fn}n∈N converges to f. Hence, f is a Fréchet-Urysohn point and by translation we have that Cp X, ZN is a Fréchet-Urysohn space. Since, Cp X, ZN is homeomorphic to Cp(X, Z)N, we have that Cp(X, Z)N is a Fréchet- Urysohn space.

(vi)⇒(i) Assume that Cp(X, Z)N is a Fr´echet-Urysohn space. Let A be a subset of

Cp(X, Z), and let f ∈ clCp(X,Z)(A). We want to show that there is a sequence {fn}n∈N in A ∞ such that fn convergence to f in Cp(X, Z). Fix g ∈ Cp(X, Z). It follows that A × Πi=2{g} is a subset of Cp(X, Z)N. Notice that {g} is a closed subset of Cp(X, Z) as Cp(X, Z) is a

Hausdorﬀ space, thus clCp(X,Z)({g}) = {g}. Also, notice that

cl (A × Π∞ {g}) = cl (A) × Π∞ cl ({g}) Cp(X,Z)N i=2 Cp(X,Z) i=2 Cp(X,Z)

∞ = clCp(X,Z)(A) × Πi=2{g}.

∞ Thus, there is a sequence {(fn, g, g, . . .)}n∈N in the set clCp(X,Z)(A) × Πi=2{g} such that

N {(fn, g, g, . . .)}n∈N convergence to (f, g, g, . . .) since Cp(X, Z) is a Fr´echet-Urysohn space.

Hence, {fn}n∈N convergence to f and {fn}n∈N is a sequence in clCp(X,Z)(A). Therefore

Cp(X, Z) is a Fr´echet-Urysohn space.

Finding an example of a space X which satisfies condition (ii) or (iii) in Theorem 4.2.4 is little difficult, so we will provide an example through a different characterization which will appear in a later section.

4.3 When is Cp(X, Z) a Sequential Space?

The next logical step in classifying Cp(X, Z) into diﬀerent categories of spaces is to consider when it is a sequential space; this is the goal of this section. In this section we will ﬁnd that 100

Cp(X, Z) has an interesting property that is not true in general and that is Cp(X, Z) is a Fr´echet-Urysohn space, a sequential spaces, and a k-space all at the same time. A sequential space is a space X such that each non-closed subset A ⊆ X contains a

sequence {xn}n∈N that converges to some point x ∈ X\A. Let the reader note that this section requires the use of ordinal numbers; we refer to the reader to [16] as a reference on ordinal numbers. Note the following relationship between Fr´echet-Urysohn space and sequential spaces.

Proposition 4.3.1 (Theorem 1.6.14, [11]). Every Fr´echet-Urysohnspace is a sequential space.

At the same time we will consider when Cp(X, Z) is a k-space. A space X is called a k-space if for each A ⊆ X, the set A is open in X provided that the intersection of A with any compact subspace C of X is open in C.

Proposition 4.3.2 (Theorem 3.3.20, [11]). Every sequential Hausdorﬀ space is a k-space.

Now for the main theorem of this section.

Theorem 4.3.3. Let X be a zero-dimensional space. The following are equivalent.

(i) Cp(X, Z) is a Fr´echet-Urysohnspace.

(ii) Cp(X, Z) is sequential.

(iii) Cp(X, Z) is a k-space.

We will note in general the three categories of spaces are diﬀerent, it is an interesting case that this happens for these types of function space. Before we provide the proof for Theorem 4.3.3 we need a several lemmas. We will begin by listing some properties that the space X may hold. Let X be a space, we deﬁne the following properties on X:

(α) If for any open ω-cover η of X there is a sequence ξ ⊆ η such that lim inf ξ = X. 101

(β) If for any open ω-cover η = ∪n∈Nηn of the space X, where ηn ⊂ ηn+1 for all n ∈ N,

there is a sequence ξ = {Xn}n∈N, where Xn ⊆ X for all n ∈ N, such that ηn is an open

ω-cover of Xn for each n ∈ N, and lim inf ξ = X.

(γ) If we can extract from any open ω-cover of X a countable ω-subcover.

We use use the notation that Arkhangel0ski˘ıuses to indicate that X has the property θ, namely X ` θ; we use X 0 θ to mean X does not have property θ. Also note that property α is the same condition on X that is given in part (ii) of Theorem 4.2.4.

The ﬁrst of our lemmas is independent of Cp(X) and Cp(X, Z), thus the proof will be the same as that given by Arkhangel0ski˘ıon page 53 of [4]. We provide the proof for completeness.

Lemma 4.3.4. Let X be a space. X ` α if and only if X ` β and X ` γ.

Proof. (⇒) Assume that X ` α. If follows that for any open ω-cover η of X there is a sequence ξ ⊆ η such that lim inf ξ = X. Also, it follows from Proposition 4.2.2 that ξ is a countable ω-subcover. Thus, X ` γ.

Now for the second part. We have from property α that there are Xn ∈ η such that lim inf Xn = X Notice that if Xn ∈ ηkn , then ηnk is an ω-cover for Xn, for each n ∈ N. Therefore, it follows that X ` γ. (⇐) Assume that X ` β and X ` γ. Let η be an ω-cover of X. Since X ` γ we may assume, without loss of generality, that |η| ≤ ℵ0. Furthermore, since X ` β we have that there are Xn ⊆ X such that lim inf Xn = X and Xn is ω-covered by ηn, for each n ∈ N, where

ηn is the ﬁrst N element of η for some enumeration of η. Notice that ηn is a ﬁnite ω-cover of Xn, thus there has to be some member of Yn ∈ ηn containing Xn. Hence, lim inf Yn = X since Xn ⊆ Yn ∈ ηn and lim inf Xn = X.

The next two lemmas are results for when Cp(X, Z) is as k-space. We will note that again, the proofs of the following lemmas are modeled on the one given by Arkhangel0ski˘ıon pages 54-57 of [4]. The ﬁrst lemma is for property β. 102

Lemma 4.3.5. Let X be a zero-dimensional space. If Cp(X, Z) is a k-space, then X ` β.

Proof. Assume Cp(X, Z) is a k-space. By way of contradiction, assume that X 0 β. Hence

∞ there is an open ω-cover η = {ηn}n=0 such that X does not have property β. We will deﬁne

Xf,n = {x ∈ X : f(x) < n}, where f ∈ C(X, Z) and n ∈ N. Moreover, we deﬁne

An = {f ∈ C(X, Z): Xf,n is ω-covered by the family ηn}

and A = ∪n∈NAn.

We claim that An is a closed subset of Cp(X, Z) for some n ∈ N. Let f ∈ C(X, Z)\An.

We have that Xf,n is not ω-covered by ηn. Thus, there is a ﬁnite subset F = {x1, . . . xk} of

Xf,n such that no element of ηn contains f. It then follows that f ∈ O, where

O = {g ∈ C(X, Z): g(x) < n for all x ∈ F }

= ∪0≤i1

Now notice that O is an open subset of Cp(X, Z) and O ⊆ Cp(X, Z)\An, hence An is a closed of Cp(X, Z) for some N.

Now observe that An is not a closed subset of Cp(X, Z). This observation can be seen by considering f0 = 0; we see that f0 ∈ clCp(X,Z)(A)\A implies that A is not a closed subset of

Cp(X, Z). By the assumption that Cp(X, Z) is a k-space, we choose a compact subset K of

Cp(X, Z) such that K ∩ A is not a closed subset of K. Since K is compact all integer-valued continuous maps on K are bounded in Z. That is, for all x ∈ X, there is a n(x) ∈ N such that f(x) < n(x) for all f ∈ K. Deﬁne Xn = {x ∈ X : n(x) < n}. Clearly X = ∪n∈NXn and

Xn ⊆ Xn+1 for all n ∈ N. Hence lim inf Xn = X. Since we assumed that X 0 β, there is a m ∈ N such that no ηk is an ω-cover of Xm.

However, notice that K ∩ An = ∅ whenever n > m; in fact f ∈ An whenever n > m.

Thus, Xf,n is ω-covered by the collection ηk. However Xm is not ω-covered by ηk, that is

Xm\Xf,n 6= ∅, when f(x) ≥ n > m, for some x ∈ Xm. Thus, it follows from the deﬁnition 103 ∞ of Xm that f∈ / K. However this gives that K ∩ A = ∪n=1(K ∩ An) is a closed subset of K, which is a contradiction. Therefore, X ` β.

The second of these lemmas deals with property γ.

Lemma 4.3.6. Let X be a zero-dimensional space. If Cp(X, Z) is a k-space, then X ` γ.

Proof. Assume Cp(X, Z) is a k-space. We wish to show that X ` γ. By way of contradiction assume that X 0 γ. Let η be an open ω-cover of X such that a countable ω-subcover can not be extracted. Deﬁne

A = {f ∈ C(X, Z): f(X) ⊆ {0, 1}, z(f) can be ω-cover by a countable subfamily of η}.

The set A be a not closed of Cp(X, Z) as 0 ∈ clCp(X,Z)(A)\A. Thus z(0) = X can not be ω-covered by a countable ω-subcover of η. We claim that any countable subset of A has its closure in A. Let B ⊆ A with |B| ≤ ℵ0. We have that clCp(X,Z)(B) ⊆ A. Now take g ∈ clCp(X,Z)(B). By the deﬁnition of A, it follows that there is a countable ω-subcover η0 where η0 ⊆ η which covers z(f) for all f ∈ B. We can see this by the fact that |B| ≤ ℵ0 and

0 for each f ∈ B there is a countable ω-subcover ηf which covers z(f). Thus η0 = ∪f∈Bηf is a countable ω-subcover with covers z(f) for all f ∈ B. Let F be a ﬁnite subset of X such that

F ⊆ z(g). There is then a f ∈ B such that F ⊆ z(f), since g ∈ clCp(X,Z)(B). Thus, z(g) is in fact ω-covered by ηo. This gives us that z(g) is ω-coved by η0. So z(g) is ω-covered by a countable ω-cover. Hence g ∈ A.

Now, choose C a compact set of Cp(X, Z) such that C ∩ A is not closed in C. Take g ∈ clCp(X,Z)(K ∩ A)\A. It is clear that g is a continuous map from X to {0, 1}. So z(g) is a clopen subset of X. However, z(g) is not ω-covered by a countable ω-subcover of η since g∈ / A. It is actually the case that C ∩ A is countably compact. This can bee seen by taking K ⊆ C ∩ A such that K is countable, the closure of K in C is actually contained in C ∩ A. 104

We now construct a transﬁnite recursive sequence {(fξ, ηξ,Fξ+1): ξ < ω1} which has the following properties:

(i) fξ ∈ C ∩ A, ηξ ⊆ η, Fξ+1 is a ﬁnite subset of X, and Fξ+1 ⊆ Z(g).

(ii) ηξ1 ⊆ ηξ2 , if ξ1 < ξ2 < ω1.

(iii) ηξ is an ω-cover of z(fξ), but not of Fξ+1.

(iv) If ξ1 + 1 ≤ ξ2 < ω1, then Fξ1+1 ⊆ z(fξ2 ).

Let f0 ∈ K ∩A, and let η0 be a countable subfamily of η which covers z(f0). Since there is no countable subfamily of η which ω-covers z(g), there is a ﬁnite F1 ⊆ z(g) which is contained in no member of η0. Suppose that α < ω1 and that (fξ, ηξ,Fξ+1) is constructed for all ξ < α. We wish to construct the next member of the sequence.

Case 1. Suppose α is a limit ordinal. We choose a sequence {ξn}n∈N such that ξn → α and deﬁne:

∞ (i) fα to be the limit point of {fξn }n=1 in C ∩ A,

(ii) ηα = ∪ξ<αηξ, and

(iii) Fα+1 to be a ﬁnite subset of z(g) which can not be ω-covered by ηα.

Case 2. Suppose α = β +1, i.e α has an immediate predecessor. The construction in this case is more involved. We ﬁrst note that if N is a countable subset of z(g), then there is an

∞ n f ∈ C ∩ A such that f|N = 0. We have that N = {xn}n=0 ⊆ z(g) and deﬁne Nn = {xk}k=0.

Since g ∈ clCp(X,Z)(C ∩ A), there is a fn ∈ C ∩ A such that fn|Nn−1 = 0 for all n ∈ N. We take f as any limit point of {fn}n∈N in C ∩ A. Now for the remainder of the construction of the sequence. We now construct the functions. Take fα ∈ C ∩ A such that fα|N = 0, where N = ∪ξ<βFξ+1; this function exists since

N is countable. Take a countable subfamily ηα ⊆ η such that it is an ω-cover of z(fα) and take ηβ ⊆ ηα. Lastly, take Fα+1 a ﬁnite subset of z(fα) such that it is not ω-covered by ηα. 105

The last part of the construction, we take h ∈ C as a limit point for sequence {fξ}ξ<ω1 and deﬁne T = z(h) ∩ z(g). We claim that Fα+1 ⊆ z(fβ) if and only if α + 1 ≤ β. Suppose

Fα+1 ⊆ z(fβ). It then follows that ηβ is an ω-cover of Fα+1. Thus α + 1 ≤ β. Conversely, if α + 1 ≤ β and β is an isolated ordinal, then fβ = 0 on Fα+1, i.e. Fα+1 ⊆ z(fβ). Also, if

α + 1 ≤ β, β is a limit ordinal, and α + 1 < β, then there is a n0 ∈ N such that α + 1 ≤ ξn and Fα+1 ⊆ z(fξn ) for all n > n0 since ξn → β. Therefore it follows that Fα+1 ⊆ z(fβ). Now,

since h is a limit point of {fξ}ξ<ω1 , we have that ∪ξ<ω1 Fξ+1 ⊆ T .

We now wish to show that for {Fξ+1}ξ<ω1 the following condition holds:

(†) For a ﬁnite K ⊆ X there is an open set K ⊆ G ⊆ X such that |{ξ < ω1 : Fξ+1 ⊆

G}| < ℵ0.

Notice that T a clopen subset of X, since both z(g) and z(h) are, and that ∪ξ<ω1 Fξ+1 ⊆ T . It is enough to show that (†) holds for ﬁnite K a subset of T . For each ﬁnite K ⊆ T there

is a minimal αK < ω1 such that K ⊆ z(fαK ). We claim that αK in not a limit ordinal. To

see this take K0 = K and ξ0 = βK0 where αK0 = βK0 + 1. Now the question is it possible

for αK0 = 0, i.e. is βK0 undeﬁned? Start by setting G = z(f0). Let K1 = z(fξ0 ) ∩ K0, where

ξ1 = βK1 . If βK1 is deﬁned, then αK1 = βK1 + 1. This construction gives one of the two following:

(a) either a chain K0 ) K1 ) ... ) Kn = ∅ is obtained or

(b) Kn ⊆ z(f0) for a some n ∈ N.

Now deﬁne Gk = (z(fξi+1)\z(fξi )) ∩ ∩j

If x ∈ K and (a) holds, then x ∈ Ki\Ki+1, where i < n. Thus, x ∈ Kj for all j ≤ i and

x∈ / Ki+1, that is

x ∈ (z(fξ1+1)\z(fξi )) ∩ ∩j

Now let ξ < ω1 such that ξ 6= ξi for all i < n. If ξ < ξi for all i, then Gi ⊆ X\z(fξ1 ) and 106

fξ+1 ⊆ z(fξi ) since Fξ+1 ∩ Gi = ∅ for all i < n. However, if k0 is the smallest natural number such that ξ < ξ, then ∪ G ⊆ z(f ) and F z(f ) since ξ + 1 < ξ + 1. k0 k0≤k k ξk0 +1 ξk0 +1 * ξk0 +1 k0

If x ∈ K and (b) holds, then we can reduce it to case (a) by setting z(f−1) = ∅ and

construction G from the chain K0 + K1 + ... + Kn+1 = ∅. Thus, we have property (†). Now put

ηn = {G ⊆ X : G is open and |{ξ < ω1 : Fξ ⊆ G}| ≤ n}

∞ and η = ∪n=1ηn. By property (†) we have that η is an open ω-cover of X and ηn ⊆ ηn+1 for

all n ∈ N.

Since Cp(X, Z) is a k-space we have that X ` β by Lemma 4.3.5. Also, since X ` β, there

∞ is a sequence {Xn}n=1 with lim inf Xn = X such that Xn is ω-covered by the collection ηn.

However, this is impossible since if Xn is ω-covered by the collection ηn, then Xn contains

at most n subsets Fξ+1 and lim inf Xn = X implies that Xn contains ℵ1 sets Fξ + 1. Thus, a contradiction, and therefore X ` γ.

We now have the tools need to prove Theorem 4.3.3.

Proof of Theorem 4.3.3. (i)⇒(ii) Assume Cp(X, Z) is a Fr´echet-Urysohn space, it follows

from Proposition 4.3.1 that Cp(X, Z) is sequential.

(ii)⇒(iii) Assume Cp(X, Z) is sequential, it follows from Proposition 4.3.2 that Cp(X, Z) is a k-space.

(iii)⇒(i) Assume that Cp(X, Z) is a k-space. It follows from Lemma 4.3.5 that X ` β and from Lemma 4.3.6 that X ` γ. Hence, is follows from Lemma 4.3.4 that X ` α. Therefore,

Cp(X, Z) is a Fr´echet-Urysohn space by Theorem 4.2.4.

In this section we have shown that the same conditions on the space X provide that the space Cp(X, Z) is a Fr´echet-Urysohn space, a sequential space, and a k-space. 107 4.4 Tightness and Lindel¨ofProperties for Cp(X, Z)

There are a number of results for Cp(X) relating tightness to the Lindel¨ofnumber of X. These results were shown by Asanov, Arkhangel0ski˘ı,and Pytkeev (refer to 33 and 45 pages in [4]) in 1979, 1976, and 1982, respectively. We will see how these results look when considering Cp(X, Z). We will then consider additional cardinal invariants to develop a non trivial example of a space X which make Cp(X,Z) a Fr´echet-Urysohn space.

Here is the obvious observation about the Lindel¨ofnumber and Cp(X, Z).

I Lemma 4.4.1. If X is a space, then l(Cp(X, Z)) ≤ l Cp X , Z , for any index set I.

I Proof. We show that Cp(X, Z) is homeomorphic to a closed subspace of Cp X , Z and then apply Lemma 1.4.3. The homeomorphism will be shown by using the projection map and the composition operator.

I I Consider the map πα : X → Xα, we know that πα is continuous and πα X = Xα

] ] for all α ∈ I. Also, consider the the composition operator πα given by πα(f) = f ◦ πα for

] I ] I f ∈ C(Xα, Z). Since πα : C(Xα, Z) → C X , Z , we have that πα(C(Xα, Z)) ⊆ C X , Z .

] Let Fα = πα(C(Xα, Z)), i.e. Fα = {f ◦ πα : f ∈ C(Xα, Z)}. Notice that Fα is a subspace of I Cp X , Z which is homeomorphic to Cp(Xα, Z). I Now, we need to show that Fα is a closed subspace of Cp X , Z . Since πα is an open

map it follows that πα is a quotient map. Thus, from Proposition 2.3.12, Fα is a closed

I I subspace of Cp X , Z . Therefore l(Cp(X, Z)) = l(Fα) ≤ l Cp X , Z by Lemma 1.4.3 I and Remark ??. Hence, we have the desired result of l(Cp(X, Z)) ≤ l Cp X , Z .

In 1979 Asanov proved the analog for the following theorem in the case of Cp(X) (refer to page 33 in [4]).

Theorem 4.4.2. For every zero-dimensional space X, t(X) ≤ l(Cp(X, Z)).

Proof. Suppose l(Cp(X, Z)) = τ. We will show t(X) ≤ τ. Let A be a subset of X, and let

x ∈ clX (A) ⊆ X. Let U be an open subset of X containing x. We may assume A ⊆ U, 108 otherwise take U to be the union of the open subsets of X that intersect A, which would still give x ∈ clX (A ∩ U).

Consider Mx = {f ∈ Cp(X, Z): f(x) = 0}. Clearly l(Mx) ≤ τ since Mx is a closed subset of Cp(X, Z). Consider My = {f ∈ Cp(X, Z): f(y) = 0}, which is an open subset of Cp(X, Z), for y ∈ A. Take an arbitrary f ∈ Mx. Since x ∈ clX (A), f is continuous, and f(x) = 0 there is a y ∈ A such that f(y) = 0. This is true since x ∈ f −1({0}), where f −1({0}) is an open

−1 subset of X. So y ∈ f ({0}) ∩ A 6= ∅ implying f(y) = 0. Therefore ∪{My : y ∈ A} ⊇ Mx.

Since l(Mx) ≤ τ, there is a B ⊆ A such that |B| ≤ τ and Mx ⊆ ∪{My : y ∈ B}. We claim x ∈ clX (B). By way of contradiction, suppose x∈ / clX (B). There exists, by Lemma

2.2.2, an f0 ∈ Cp(X, Z) such that f0(x) = {0} and f0(clX (B)) = {1}. Since f0(x) = 0, we have f0 ∈ Mx. Thus f0(y) = 0, since f0 ∈ My for some y ∈ B. Hence a contradiction.

Consequently, x ∈ clX (B). Therefore t(X) ≤ τ implying t(X) ≤ l(Cp(X, Z)).

Here is an immediate corollary to Theorem 4.4.2 that follows from facts about zero- dimensional spaces.

n n Corollary 4.4.3. For every zero-dimensional space X, t(X ) ≤ l(Cp(X , Z)), for all n ∈ N.

Proof. It follows from Proposition 1.2.26 that Xn is zero-dimensional for any n ∈ N. Thus,

n n it follows from Theorem 4.4.2 that t(X ) ≤ l(Cp(X , Z)).

There is no converse for Theorem 4.4.2 in the case of Cp(X, Z). The best we can do is

n to show that the tightness of Cp(X, Z) and the Lindel¨ofnumber of X are bounded by the same quantities.

Theorem 4.4.4. Let X be a zero-dimensional space. l(Xn) ≤ τ for all n ∈ N if and only if t(Cp(X, Z)) ≤ τ.

Before we prove the theorem we need the deﬁnition of small relative to η. Let X be a space, let n ∈ N, and let η be an open cover of Xn. A ﬁnite system µ of open subsets of a space X is called small relative to η (η-small) if for any V1,...,Vn ∈ µ there is a G ∈ η 109

such that V1 × ... × Vn ⊆ G (refer to page 45 in [4]). Note that the forward direction is very similar to the proof given by Arkhangel0ski˘ıfor Theorem II.1.1 in [4] and the reverse direction works the same as the proof given by Pytkeev.

n Proof of Theorem 4.4.4. (⇒) Let l(X ) ≤ τ for all n ∈ N, and let A ⊆ Cp(X, Z). Without

loss of generality we may assume that 0 ∈ clCp(X,Z)(A), as we may translate A so that clCp(X,Z)(A) contains 0. We want to show that there is a B ⊆ A such that 0 ∈ clCp(X,Z)(B) and |B| ≤ τ.

n Let ξ = (x1, . . . , xn) ∈ X . Since 0 ∈ Cp(X, Z) the basic clopen subset

W = W (x1, . . . , xn, 0,..., 0)

of Cp(X, Z) contains 0. We have that W ∩ A 6= ∅ since 0 ∈ W and 0 ∈ clCp(X,Z)(A). Thus

there exists a gξ ∈ A such that gξ(xi) = 0 for i = 1, . . . , n. By continuity of gξ there is an

open subset Oxi of X such that gξ(x) = 0 for all x ∈ Oxi for all i = 1, . . . , n.

n Let Uξ = Ox1 × ... × Oxn . It follows that Uξ is an open subset of X containing ξ. Now,

n n deﬁne ηn = {Uξ : ξ ∈ X }. It is clear that η an open cover of X as each Uξ is an open

n ∗ n ∗ subset of X . Since l(X) ≤ τ, there is an open subcover ηn ⊆ ηn of X such that |ηn| ≤ τ.

∗ Let Bn = {gξ : Uξ ∈ ηn}, and let B = ∪{Bn : n ∈ N}. By deﬁnition it is clear that B ⊆ A, and |B| ≤ τ.

n Lastly, we want to show that 0 ∈ clCp(X,Z)(B). Let y = (y1, . . . , yn) ∈ X . It follows n ∗ that there is a ξ ∈ X such that y ∈ Uξ with Uξ ∈ ηn. Clearly gξ ∈ B and gξ(yi) = 0 for i = 1, . . . , n. Hence 0 ∈ B by the construction of Uξ. Therefore f ∈ clCp(X,Z)(B) implying t(C(X, Z)) ≤ τ.

n (⇐) Let t(Cp(X, Z)) ≤ τ, ﬁx n ∈ N, and let ηn be an open cover of X . Let En denote the family of all (ﬁnite) ηn-small families of open subsets of X. For each µ ∈ En let

Aµ = {f ∈ Cp(X, Z): f(X\ ∪ µ) = {0}}, and let A = ∪{Aµ : µ ∈ En}. We want to show clCp(X,Z)(A) = Cp(X, Z). 110

Let f ∈ Cp(X, Z), and let K ⊆ X. We construct a ﬁnite family θK of open subsets of

n X such that for any collection (y1, . . . , yn) ∈ K there is V1,...,Vn ∈ θK and a G ∈ ηn for which yi ∈ Vi and V1 × ... × Vn ⊆ G. Clearly, K ⊆ ∪θK . For each x ∈ K we deﬁne

Wx = ∩{V ∈ θK : x ∈ V } and consider the family µK = {Wx : x ∈ K}. It is clear that

K ⊆ ∪µK .

n We claim the family µK is ηn-small. Indeed, consider a subset of X of the form Wx1 ×

... × Wxn . There are V1,...,Vn ∈ θK such that xi ∈ Vi and V1 × ... × Vn ⊆ G ∈ ηn. Since

Wxi ⊆ Vi for i = 1, . . . , n, it follows that Wx1 × ... × Wxn ⊆ G.

Consider a function g ∈ Cp(X, Z) such that g|K = f|K and G(X\ ∪ µK ) = {0}. Such a function exists since K ⊆ ∪µK . Clearly, g ∈ AµK ⊆ A and g lies in the same standard basic clopen subsets of f given by the ﬁnite subset K. Hence, f ∈ clCp(X,Z)(A).

In particular, take f to be the function f1 ∈ Cp(X, Z) such that f1 = 1. By the above, f1 ∈ clCp(X,Z)(A) and since t(Cp(X, Z)) ≤ τ there is a B ⊆ A such that |B| ≤ τ and f1 ∈ clCp(X,Z)(B). Thus there is a subfamily E0 ⊆ En for which B ⊆ ∪{Aµ : µ ∈ E0} and

|E0| ≤ τ.

n Let µ ∈ E0. For each ξ = (V1,...,Vn) ∈ µ we ﬁx Gξ ∈ ηn such that V1 × ... × Vn ⊆ Gξ.

n Put ηµ = {Gξ : ξ ∈ µ }. The family ηµ is ﬁnite, hence the cardinality of the family

η˜ = ∪{ηµ : µ ∈ G0} does not exceed τ.

n n We now show thatγ ˜ covers X . Let (x1, . . . , xn) ∈ X , and let U = {f ∈ Cp(X, Z): f(xi) > 0, i = 1, . . . , n}. The subset U is an open of Cp(X, Z) containing f1. Since f1 ∈

clCp(X,Z)(B) and B ⊆ ∪{Aµ : µ ∈ E0}, there is a µ0 ∈ E0 such that U ∩ Aµ0 6= ∅. For

g ∈ Aµ0 we have g(xi) > 0 for i = 1, . . . , n and g(x) = 0 for x∈ / ∪µ0. Thus, xi ∈ ∪µ0 for all i = 1, . . . , n. Take Vi ∈ µ0, i = 1, . . . , n such that xi ∈ Vi. We ﬁnally have the following

n (x1, . . . , xn) ∈ V1 × ... × Vn ⊆ Gξ ∈ γµ0 ⊆ γ˜. Therefore l(X ) ≤ τ for all n ∈ N.

We now about ready to ﬁnd the example of a space which makes Cp(X, Z) a Fr´echet- Urysohn space. The method we will use requires some theorems about τ-monolithic and τ-simplicity. 111 We wish to combine idea of τ-monolithic with strongly τ-monolithic. A space X is said to be strongly τ-monolithic if for every A ⊆ X, with |A| ≤ τ, has the property w(clX (A)) ≤ τ, where clX (A) has the subspace topology (refer to page 83 in [4]). A space X is said to be strongly monolithic if it is strongly τ-monolithic for all τ ≥ ℵ0 (refer to page 83 in [4]). Let X and Y be spaces. The space X is said to be τ-simple if for every continuous map f for X to Y , where w(Y ) ≤ τ, the cardinality of f(X) does not exceed τ (refer to page 84 in [4]). A space is called just simple if it is τ-simple for all τ ≥ ℵ0 (refer to page 84 in [4]). Before we begin using τ-simple and strongly τ-monolithic let us state a need result about the netweight.

Lemma 4.4.5. Let X be a zero-dimensional space, nw(X) = nw(Cp(X, Z))

Proof. We will ﬁrst show that nw(Cp(X, Z)) ≤ nw(X). Let S be a network for X and recall that BZ is the usual base for Z. Fix k ∈ N, let S1,...,Sk ∈ S, and let {n1},..., {nk} ∈ BZ. Deﬁne

V (S1,...,Sk, {n1},..., {nk}) = {f ∈ C(X, Z): f(Si) = {ni} for i = 1, . . . , k},

and deﬁne η = {V (S1,...,Sk, {n1},..., {nk}): S1,...,Sk ∈ S and {n1},..., {nk} ∈ BZ}.

We claim that η is a network of Cp(X, Z). Let f ∈ Cp(X, Z). The function f is in some basic open set W (x1, . . . , xk, m1, . . . , mk). Since S is a network of X there are S1,...,Sk ∈ S such that xi ∈ Si for all i = 1, . . . , k. We need to show

V (S1,...,Sk, {m1},..., {mk}) ⊆ W (x1, . . . , xk, m1, . . . , mk).

Let g ∈ V (S1,...,Sk, {m1},..., {mk}). Now since g(Si) = {mi} and xi ∈ Si for all i =

1, . . . , k we have that g(xi) = mi for all i = 1, . . . , k. Hence g ∈ W (x1, . . . , xk, m1, . . . , mk).

Thus, η is a network for Cp(X, Z) and by the construction of η is follows that |η| ≤ |S|.

Therefore we have that nw(Cp(X, Z)) ≤ nw(X). 112

Now to show nw(X) ≤ nw(Cp(X, Z). First, recall that map eZ : X → Cp(Cp(X, Z), Z) is the map deﬁned ase ˆZ(x) = ex. It follows that nw(X) ≤ nw(Cp(Cp(X, Z), Z). However, the above gives us nw(Cp(Cp(X, Z), Z)) ≤ nw(Cp(X, Z). Thus, nw(X) ≤ nw(Cp(X, Z).

Therefore, we can conclude nw(X) = nw(Cp(X, Z)).

Now on to the results that we need. Our ﬁrst is a proposition about Lindel¨of P -spaces and ℵ0-simplicity.

Proposition 4.4.6 (Theorem II.7.6, [4]). Every Lindel¨of P -space is a ℵ0-simple.

Next we have a theorem relating ℵ0-simple to strongly ℵ0-monolithic. This is an adaptation of the result given be Arkhangel0ski˘ıon page 85 of [4].

Theorem 4.4.7. A zero-dimensional space X is ℵ0-simple if and only if Cp(X, Z) is strongly

ℵ0-monolithic.

Proof. (⇒) Assume X is ℵ0-simple. Let F ⊆ Cp(X, Z) such that |F| ≤ ℵ0. Deﬁne ϕ : X →

F Z by ϕ(x) = (xf )f∈F where xf = f(x) for f ∈ F. Let Y = ϕ(X). Since X is ℵ0-simple, ϕ is continuous, and w(Y ) ≤ ℵ0, we have that |Y | ≤ ℵ0.

Y Now, let P = g ◦ f : g ∈ Z . We have that F ⊆ P . Note that P is a closed subspace

X Y of Z which is homeomorphic to Z . It follows that clCp(X,Z)(F) ⊆ P . Since P is a closed subset of Cp(X, Z), we have from Remark 1.4.6 and Lemma 4.1.5 that

Y w(clCp(X,Z)(F)) ≤ w(P ) ≤ w Z ≤ |Y | ≤ ℵ0.

Thus, it follows that Cp(X, Z) is strongly ℵ0-monolithic.

(⇐) Assume that Cp(X, Z) is strongly ℵ0-monolithic. Let ϕ : X → Y be a continuous map with w(Y ) ≤ ℵ0. It follows from Lemma 4.4.5 and Lemma 1.4.7 that

d(Cp(X, Z)) ≤ nw(Cp(X, Z)) ≤ nw(Y ) ≤ ℵ0. 113

Thus, we have that the space Cp(Y, Z) is homeomorphic to some subspace P of Cp(X, Z).

Hence, w(Cp(X, Z)) ≤ ℵ0 since Cp(Y, Z) is strongly ℵ0-monolithic and d(Cp(Y, Z)) ≤ ℵ0.

Therefore |Y | ≤ ℵ0, implying that X is ℵ0-simple.

We have the following corollary to Theorem 4.4.7.

Corollary 4.4.8. Let X be a zero-dimensional space. Cp(X, Z) is a strongly ℵ0-monolithic

n Fr´echet-Urysohnspace if and only if the space X is ℵ0-simple and l(X ) ≤ ℵ0 for all n ∈ N.

Proof. (⇒) Assume that Cp(X, Z) is a strongly ℵ0-monolithic Fréchet-Urysohn space. It is clear that t(Cp(X, Z)) ≤ ℵ0 by the definition of a Fréchet-Urysohn space. Thus, it follows

n from Theorem 4.4.4 that l(X ) ≤ ℵ0 for all n ∈ N. Hence, it follows from the previous

n statement and Theorem 4.4.7 that X is a ℵ0-simple space with the property that l(X ) ≤ ℵ0 for all n ∈ N.

n (⇐) Assume X is ℵ0-simple and l(X ) ≤ ℵ0 for all n ∈ N. By Theorem 4.4.7 we have that

Cp(X, Z) is strongly ℵ0-monolithic and from Theorem 4.4.4 we have that t(Cp(X, Z)) ≤ ℵ0.

Now we need to verify that Cp(X, Z) is in fact a Fréchet-Urysohn space. Let f ∈ Cp(X, Z) and A ⊆ Cp(X, Z) such that f ∈ clCp(X,Z)(A). Since t(Cp(X, Z)) ≤ ℵ0 there is a countable subset B ⊆ A such that f ∈ clCp(X,Z)(B). Notice that clCp(X,Z)(B) as a subspace has a ∞ countable base since B is countable. It is then clear that there is a sequence {fn}n=1 in B which convergence to f. Thus, Cp(X, Z) is a Fréchet-Urysohn space. Therefore Cp(X, Z) is a strongly ℵ0-monolithic Fréchet-Urysohn space.

Finally, the theorem we will use for our example.

Theorem 4.4.9. If X is a Lindel¨of P -space, then Cp(X, Z) is a strongly ℵ0- monolithic Fr´echet-Urysohnspace.

Proof. It follows from Lemma 1.2.25 that Xn is a Lindel¨of P -space for all n ∈ N. Also, it follows from Proposition 4.4.6 that X is a ℵ0-simple space. Therefore, Cp(X, Z) is a strongly

ℵ0-monolithic Fr´echet-Urysohn space by Corollary 4.4.8. 114

Now for an interesting example of when Cp(X, Z) is a Fr´echet-Urysohn space.

Example 4.4.10. Let D be an uncountable discrete space, let λ∈ / D, and let X = {λ} ∪ D. Deﬁne a topology on X to be the collection of open subsets of D and the open subset, O,

containing λ satisfy the condition that |D\O| ≤ ℵ0. We claim that X is a Lindel¨of P -space.

We show ﬁrst that X is a Lindel¨ofspace. Let {Ui}i∈I be an open cover of X with the

index set I. Since {Ui}i∈I is a cover there is a Uiλ such that λ ∈ Uiλ with |X\Uiλ | ≤ ℵ0.

Take an enumeration of {xn}n∈N of X\Uiλ , since {Ui}i∈I is a cover of X there is an Uin such

that xn ∈ Uin for all n ∈ Z. Thus, {Uin }n∈N ∪ {Uiλ } is a countable subcover of X. Thus, we have that X is a Lindel¨ofspace.

Next, we show that X is a P -space. Let {Un}n∈N be a collection of open subsets of X. We consider the two cases which arise.

Case 1. Suppose λ∈ / Ui for some i ∈ N. Since λ∈ / Ui it follows that λ∈ / ∩n∈NUn, thus we have ∩n∈NUn ⊆ D. It follows that ∩n∈NUn is an open subset of X.

Case 2. Suppose λ ∈ Un for all n ∈ N. It follows that λ ∈ ∩n∈NUn. We have that

Hence, X is a Lindel¨of P -space. Thus, it follows from Theorem 4.4.9 that Cp(X, Z) is a Fr´echet-Urysohn space.

Thus, we now have a not trivial example of a space X which makes Cp(X, Z) a Fr´echet- Urysohn space. We would like to make a note that the method by which we obtained the example provided us with more information about the varies cardinal invariants on topological spaces, hence we have provided with a richer picture. 115

CHAPTER 5

Closing Remarks

5.1 Conclusions

We have found that the object Cp(X, Z) and Cp(X,A) are interesting in terms of their ring structure and their structure as a topological space. The source of motivation has been the results for Cp(X), we wanted to see what would make a translation from the classic study to this new case. We have found that both Cp(X, Z) and Cp(X,A) are similar to Cp(X) in many ways, but are more different than they are alike. We have discovered that the concept of zero-dimensional space plays an important role in the study of Cp(X, Z) and in the more general version of Cp(X,A), where A is a zero- dimensional topological ring. Do to the fact that we are working with zero-dimensional spaces the construction of functions is easier than when working with Tychonoff spaces. Rather than just know the exists of a particular functions we are able to explicitly define the functions we are dealing with, this is a fact that we took advantage of repeatedly. We examined a long list of cardinal invariant during the course of this dissertation. We found that the cardinal invariants for Cp(X, Z) are a the same as Cp(X) most of the time, for example weight and character. This should not come as a major surprise since the weight and character of both Z and R are ℵ0. 116

We have shown that the space Cp(X, Z has the ability to be a discrete space, which

Cp(X) does not. We have also shown that the same conditions can be placed on the space

X to make Cp(X, Z) and Cp(X) metrizable. We have shown that Cp(X, Z) is a Fr´echet-

Urysohn space, a sequential space, and k-space all at the same time, just like Cp(X). Thus, it would seem that Cp(X, Z) has many of the same topological properties that Z possesses and inherits many of the properties Cp(X) as a subspace. One could aregue that Cp(X, Z) is a mixture of both Cp(X) and Z.

5.2 Open Questions

We provide a short list of interesting open problems that can lead to future study.

1. Can Theorem 3.1.7 be generalized to a general ring?

2. If we assume the ring A is a Tychonoﬀ space can we assume that X is Tychonoﬀ in

the study of Cp(X,A)?

3. Can we obtain a similar result to Theorem 3.3.9 without the use of X being a zero- dimensional space? Is there a corresponding result using the weak weight of the space X and some cardinal invariant of A?

4. What are the possible results of compactness that can be obtained for Cp(X,A)? What are the diﬀerent types of compactness on A that have an eﬀect on the compactness of

Cp(X,A)?

We will end our list here, though there are many diﬀerent possible questions that can be asked. The strategy in the study of Cp(X,A) is to look at Cp(X) and consider what characteristics of R are used in the results for Cp(X). 117

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