NETS There Are a Number of Footnotes, Which Can Safely Be Ignored

Home , Net (mathematics), Subnet (mathematics), Ultrafilter

NETS

There are a number of footnotes, which can safely be ignored, especially on the ﬁrst reading.

1. Nets, convergence, continuity Nets are generalization of sequences needed to deal with convergence in general topological spaces, where convergent sequences are not enough to describe the topology. The idea is to replace the set of natural numbers, which is used to index elements of sequences, by arbitrary sets such that it makes sense to talk about “large” elements. Recall that a partial order on a set I is a binary relation ≺ such that (a) i ≺ i, (b) if i ≺ j and j ≺ i, then i = j, (c) if i ≺ j and j ≺ k, then i ≺ k. Definition 1.1. A set I with a partial order1 ≺ is called directed if for any i, j ∈ I there exists k ∈ I such that i ≺ k and j ≺ k. The following example will be our main source of directed sets. Example 1.2. Let X be a set and I be a collection of subsets of X closed under finite intersections, so if A, B ∈ I then A ∩ B ∈ I. Define a partial order on I by A ≺ B iff B ⊂ A. Then I is a directed set: for any A, B ∈ I, we have A ≺ A ∩ B and B ≺ A ∩ B.

Deﬁnition 1.3. A net in a set X is a collection (xi)i∈I of elements of X indexed by a nonempty directed set I, or in other words, it is a map I → X.

Deﬁnition 1.4. We say that a net (xi)i∈I in a topological space X converges to an element x ∈ X, and write xi → x or simply xi → x, if for every neighbourhood U of x there exists i0 ∈ I such that xi ∈ U for i 2 all i i0. More generally, we say that x is a cluster point of (xi)i∈I if for every neighbourhood U of x and every i0 ∈ I there exists i i0 such that xi ∈ U.

In a Hausdorﬀ topological space a net can converge to at most one point, which we then denote by lim xi. i Example 1.5. Assume X is a topological space, x ∈ X, and I is the collection of all neighbourhoods of x. Since I is closed under ﬁnite intersections, we can order it by the inverse inclusion as in Example 1.2. For every U ∈ I choose an element xU ∈ U. Then (xU )U∈I is a net in X converging to x, since for every neighbourhood U of x we have xV ∈ V ⊂ U for all V U. Proposition 1.6. Assume X is a topological space, A ⊂ X and x ∈ X. Then x ∈ A¯ if and only if there exists a net in A converging (in X) to x.

Proof. ⇒ If x ∈ A¯, then for every neighbourhood U of x there exists an element xU ∈ U ∩A. By Example 1.5 this gives us a net (xU )U converging to x.

⇐ If xi ∈ A and xi → x, then for every neighbourhood U of x there exists i0 such that xi ∈ U for all ¯ i i0. In particular, A ∩ U 6= ∅. Since this is true for all U, we conclude that x ∈ A. Date: November 23, 2017. 1More often than not, it is assumed that ≺ is only a preorder, or quasiorder, that is, condition (b) is omitted. But I follow [M] here. 2The terms limit point and accumulation point are also used, but then they should not be confused with limit/accumulation points of the set {xi | i ∈ I} ⊂ X. 1 Corollary 1.7. Assume F1 and F2 are two topologies on a set X such that for any net (xi)i in X and any 3 x ∈ X we have xi → x in topology F1 if and only if xi → x in topology F2. Then F1 = F2. Proof. By the previous proposition, for any set A ⊂ X and any x ∈ X, the element x belongs to the closure of A in topology F1 if and only if it belongs to the closure of A in topology F2. It follows that a set is closed in topology F1 if and only if it is closed in topology F2. Hence the same is true for open sets, so F1 = F2. Proposition 1.8. A map f : X → Y between topological spaces is continuous at a point x ∈ X if and only if for every net (xi)i in X converging to x we have f(xi) → f(x).

Proof. ⇒ Take a neighbourhood V of f(x). We have to ﬁnd i0 ∈ I such that f(xi) ∈ V for all i i0. By continuity of f at x, there exists a neighbourhood U of x such that f(U) ⊂ V . As xi → x, there exists i0 ∈ I such that xi ∈ U for all i i0. Then f(xi) ∈ V for all i i0. ⇐ Suppose that f is not continuous at x. This means that there exists a neighbourhood V of f(x) such that for every neighbourhood U of x we have f(U) 6⊂ V . Hence, for every such U, there exists an element −1 xU ∈ U \ f (V ). Then, again by Example 1.5, we get a net (xU )U converging to x, yet f(xU ) 6∈ V for all U, so the net (f(xU ))U does not converge to f(x).

2. Subnets

Subnets are analogues of subsequences. An immediate idea is to say that a subnet of a net (xi)i∈I is determined by a subset J ⊂ I with some nice properties. This turns out to be too naive. For example, it would be impossible to prove Proposition 2.2 below using such a notion. The correct deﬁnition is less straightforward.

Deﬁnition 2.1. A net (yj)j∈J is called a subnet of a net (xi)i∈I if there exists a map ϕ: J → I such that 4 yj = xϕ(j) and for every i0 ∈ I there exists j0 ∈ J such that ϕ(j) i0 for all j j0.

Proposition 2.2. Assume x is a cluster point of a net (xi)i∈I in a topological space X. Then there exists a subnet of (xi)i converging to x.

Proof. Consider the set J of pairs (i, U), where i ∈ I and U is a neighbourhood of x such that xi ∈ U. Deﬁne a partial order on J by

(i1,U1) ≺ (i2,U2) iﬀ i1 ≺ i2 and U2 ⊂ U1.

Let us check that J is directed. Take two points (i1,U1) and (i2,U2) in J. As I is directed, there exists 0 0 0 0 k ∈ I such that i1 ≺ k and i2 ≺ k . Since x is a cluster point, there exists k k such that xk ∈ U1 ∩ U2. Then

(i1,U1) ≺ (k, U1 ∩ U2) and (i2,U2) ≺ (k, U1 ∩ U2). Next, deﬁne a map ϕ: J → I by ϕ(i, U) = i. Let us check that ϕ has the property required in Def-

inition 2.1. Take i0 ∈ I and a neighbourhood U0 of x. We can ﬁnd k0 i0 such that xk0 ∈ U0. Then j0 = (k0,U0) is an element of J with the property that ϕ(j) i0 for all j j0.

3A bit more informally we can say that a topology is completely determined by specifying convergence of nets. A natural question then is whether we can start with a rule of convergence instead of open sets to develop a meaningful theory. This is indeed possible and leads to the notion of convergence spaces. Convergence spaces are more general than topological spaces, so not every rule of convergence is defined by a topology, but they have found much fewer applications. 4This is only one of possible definitions, due to Kelley. The definition in [M], due to Willard, is different and less convenient to work with. More precisely, a Willard subnet is a Kelley subnet, but not every Kelley subnet is a Willard subnet. If you are bothered by the fact that there are different notions of subnets, the good news is that they have the same functionality in the following rigorous sense. Two nets (yj )j∈J and (zk)k∈K are said to be AA-equivalent (AA stands for Aarnes and Andenæs - two mathematicians from Trondheim), if for every j0 ∈ J there exists k0 ∈ K such that {zk | k k0} ⊂ {yj | j j0} and, conversely, for every k0 ∈ K there exists j0 ∈ J such that {yj | j j0} ⊂ {zk | k k0}. It is easy to see that AA-equivalent nets converge to the same elements and have the same cluster points. Now, it is possible to show that any Kelley subnet of a net (xi)i is AA-equivalent to a Willard subnet of (xi)i. 2 Thus, by letting yj = xϕ(j) we get a subnet (yj)j∈J of (xi)i∈I . It remains to show that yj → x. Let U be a neighbourhood of x. There exists i0 ∈ I such that xi0 ∈ U. Consider j0 = (i0,U) ∈ J. Then for any j = (i, V ) j0 we have yj = xϕ(j) = xi ∈ V ⊂ U. Hence yj → x.

3. Compactness Theorem 3.1. For any topological space X the following conditions are equivalent: (i) X is compact; (ii) every net in X has a cluster point; (iii) every net in X has a convergent subnet. For the proof we need the following useful observation.

Lemma 3.2. If (xi)i∈I is a net in a topological space X, then the set of its cluster points coincides with \ {xj | j i}. i∈I

Proof. If x is a cluster point of (xi)i, then, by definition, for any index i and any neighbourhood U of x there exists j i such that xj ∈ U. It follows that x is in the closure of {xj | j i}. This proves that the set of cluster points is contained in ∩i∈I {xj | j i}. Conversely, suppose x ∈ ∩i∈I {xj | j i}. Take a neighbourhood U of x and an index i. As x lies in the closure of {xj | j i}, there exists j i such that xj ∈ U. Hence x is a cluster point. Proof of Theorem 3.1. (i)⇒(ii) We will use the formulation of compactness in terms of collections of closed sets with finite intersection property, see [M, Theorem 26.9]. Let (xi)i∈I be a net in X. Consider the closed sets Ai = {xj | i i} (i ∈ I). The collection {Ai}i∈I has the finite intersection property, since for any i1, . . . , in ∈ I there exists i ∈ I such that ik ≺ i for all k = 1, . . . , n, and then

Ai1 ∩ · · · ∩ Ain ⊃ Ai 6= ∅.

By compactness it follows that ∩i∈I Ai 6= ∅, which by the previous lemma means exactly that the set of cluster points of (xi)i is nonempty.

(ii)⇒(iii) If (xi)i is a net, then by assumption it has a cluster point x. Then by Proposition 2.2 there exists a subnet converging to x.

(iii)⇒(ii) If (xi)i is a net, then by assumption it has a subnet converging to a point x. But then x must be a cluster point of (xi)i.

(ii)⇒(i) Let {Ui}i∈I be an open cover of X. Assume it does not have a finite subcover. We can then construct a net in X without cluster points as follows, which contradicts our assumption in (ii). As an index set we take the set J of all finite subsets of I ordered by inclusion (so F1 ≺ F2 iff F1 ⊂ F2). For every F ∈ J choose a point xF ∈ X \ (∪i∈F Ui). Take x ∈ X. Since {Ui}i∈I is an open cover of X, there exists an index i such that x ∈ Ui. Then Ui is a neighbourhood of x, but xF 6∈ Ui for any F {i}. Hence x is not a cluster point of (xF )F . The contradiction implies {Ui}i∈I has a finite subcover, so X is compact. Using the characterization of compact spaces in terms of nets we now prove the following, cf. [M, Theo- rem 26.7]. Theorem 3.3. If X and Y are compact spaces, then so is X × Y .

Proof. Let (zi)i∈I be a net in X × Y . We want to show that it has a converging subnet. Denote by pX : X × Y → X and pY : X × Y → Y the projection maps. By compactness of X the net (pX (zi))i has a convergent to a point x ∈ X subnet given by a map ϕ: J → I as in Definition 2.1. Then by compactness of Y the net (pY (zϕ(j)))j has a convergent to a point y subnet given by a map ψ : K → J. Then 3 η = ϕ ◦ ψ : K → I defines a subnet of (zi)i. This subnet converges to (x, y), since it converges coordinate- wise: pX (zη(k)) → x (since any subnet of the convergent net (pX (zϕ(j)))j converges to the same point x) and pY (zη(k)) → y. This theorem is just a warm-up. The main result is the following. Q Theorem 3.4 (Tychonoff). Let {Xs}s∈S be any collection of compact spaces. Then s∈S Xs is compact. Every proof of this theorem involves a clever application of the axiom of choice in one form or another. The following is probably the simplest among such proofs. Later we will give an even shorter proof, which, however, relies on several nontrivial auxiliary concepts. Q Proof of Theorem 3.4. We will show that every net (zi)i∈I in Xs has a cluster point. Q s∈S Q For a subset T ⊂ S denote by pT the projection map s∈S Xs → s∈T Xs. By slightly abusing the Q Q 0 notation we will use the same symbol pT for the projection map s∈T 0 Xs → s∈T Xs for any set T containing T . Consider the set U of pairs (T, xT ), where T ⊂ S and xT is a cluster point of the net (pT (zi))i. Define a partial order on U by 0 0 (T, xT ) ≺ (T , xT 0 ) iff T ⊂ T and pT (xT 0 ) = xT . It is proved in a standard way that every chain in U has an upper bound. Namely, given such a chain A 0 0 Q 0 we put T = ∪(T,xT )∈AT and define xT ∈ s∈T 0 Xs such that the sth coordinate of xT equals the sth 0 coordinate of xT for any (T, xT ) ∈ A with s ∈ T . We have to check that (T , xT 0 ) ∈ U. By the definition 0 of the product topology, xT 0 is a cluster point of (pT 0 (zi))i if and only if for any finite subset F ⊂ T the 0 point pF (xT 0 ) is a cluster point of (pF (zi))i. Since A is a chain, for any finite subset F ⊂ T there exists (T, xT ) ∈ A such that F ⊂ T . But then pF (xT 0 ) = pF (xT ) is indeed a cluster point of (pF (zi))i. By Zorn’s lemma we conclude that U has a maximal element (T, xT ). It remains to check that T = S. Q Assume there is s ∈ S \ T . Then arguing as in the proof of Theorem 3.3 (for X = t∈T Xt and Y = Xs) we can choose a subnet (zη(k))k∈K such that pT ∪{s}(zη(k)) → (xT , xs) for some xs ∈ Xs. Namely, there is a subnet of (zi)i given by ϕ: J → I such that (pT (zϕ(j)))j converges to xT , then, by compactness of Xs, there is a subnet of (zϕ(j))j given by ψ : K → J such that (ps(zϕ(ψ(k))))k converges to some xs, so we can put η = ϕ ◦ ψ : K → I. But then (T ∪ {s}, (xT , xs)) ∈ U, which contradicts maximality of (T, xT ). Hence T = S, so xT is a cluster point of (zi)i. 4. Universal nets and ultrafilters

Deﬁnition 4.1. A net (xi)i∈I in a set X is called a universal net, or an ultranet, if for every subset A ⊂ X it is eventually contained either in A or in X \ A, that is, there exists i0 ∈ I (depending on A) such that either xi ∈ A for all i i0 or xi ∈ X \ A for all i i0. Example 4.2. Take a point x ∈ X. Consider the set I of all subsets of X containing x. Order I by the inverse inclusion as in Example 1.2. For every A ∈ I choose a point xA ∈ A. Then (xA)A∈I is a universal net. The following property of universal nets reminds of Cauchy sequences.

Lemma 4.3. If (xi)i∈I is a universal net in a topological space X, then it converges to each of its cluster points. In particular, if X Hausdorﬀ, then either (xi)i does not have any cluster points or it converges to a unique point.

Proof. Assume x is a cluster point of (xi)i. Take a neighbourhood U of x. Then by universality of the net there exists i0 ∈ I such that either xi ∈ U for all i i0 or xi ∈ X \ U for all i i0. Since x is a cluster point, the second alternative is impossible. Thus xi ∈ U for all i i0. As U was arbitrary, this shows that xi → x. The analogy with Cauchy sequences should not be taken too far. Cauchy sequences are deﬁned in terms of a metric, while universal nets do not even need a topology.5 But as a result universal nets have a number of properties which make them even better than Cauchy sequences. For example, we have the following.

5True analogues of Cauchy sequences can be deﬁned for topological spaces with some additional structure, such as uniform spaces. 4 Lemma 4.4. If (xi)i is a universal net in a set X, then for any map f : X → Y the net (f(xi))i in Y is universal. Note that we do not assume that X and Y are topological spaces or that f is continuous. −1 −1 Proof. For any set A ⊂ Y , the net (xi)i is eventually contained either in f (A) or in f (Y \ A) = −1 X \ f (A). This means that (f(xi))i is eventually contained either in A or in Y \ A.

Theorem 4.5. Every net (xi)i∈I in a set X has a universal subnet, that is, a subnet which is itself a universal net. In order to prove this theorem we will introduce yet another concept, which is less intuitive but more convenient to work with. Definition 4.6. A collection U of nonempty subsets of X is called an ultrafilter on X if it is closed under finite intersections and for every subset A ⊂ X we have either A ∈ U or X \ A ∈ U. Remark 4.7. Every ultrafilter U also has the property that if A ∈ U and A ⊂ B, then B ∈ U. Indeed, we have either B ∈ U or X \ B ∈ U. But in the second case we would get ∅ = A ∩ (X \ B) ∈ U, which is a contradiction. Hence B ∈ U.

There is a close connection between universal nets in X and ultrafilters on X. Namely, let (xi)i∈I be a universal net. Consider the collection U of all subsets A of X such that {xi | i i0} ⊂ A for some i0 (depending on A). This is an ultrafilter. Indeed, if A, B ∈ U, then {xi | i i1} ⊂ A and {xi | i i2} ⊂ B for some i1, i2 ∈ I. We can find k i1, i2. Then {xi | i k} ⊂ A∩B, so A∩B ∈ U. Therefore U is closed under finite intersections. Next, for any subset A ⊂ X we have either {xi | i i0} ⊂ A or {xi | i i0} ⊂ X \ A for some i0, so either A ∈ U or X \ A ∈ U. Conversely, given an ultrafilter U, consider U as a directed set ordered by the inverse inclusion. For every A ∈ U choose xA ∈ A. Then (xA)A∈U is a universal net, since for every A ⊂ X we have either A ∈ U or 6 X \ A ∈ U, so either xB ∈ A for all B A or xB ∈ X \ A for all B X \ A. We will first prove an analogue of Theorem 4.5 for ultrafilters. Theorem 4.8. Every collection A of subsets of X with finite intersection property is contained in an ultrafilter U on X. Proof. Consider the set of all collections C of subsets of X such that A ⊂ C and C has the finite intersection property. Order this set by inclusion. Every chain in this set has an upper bound, namely, the union of collections in the chain. Hence, by Zorn’s lemma, there exists a maximal collection U. We claim that U is an ultrafilter. Take a set A ⊂ X. Assume A 6∈ U. Then, by maximality of U, the collection U ∪ {A} does not have the finite intersection property. So there exist A1,...,An ∈ U such that

A1 ∩ · · · ∩ An ∩ A = ∅.

In a similar way, if X \ A 6∈ U, then there exist B1,...,Bm ∈ U such that

B1 ∩ · · · ∩ Bm ∩ (X \ A) = ∅. But then A1 ∩ · · · ∩ An ∩ B1 ∩ · · · ∩ Bm = ∅, which contradicts the ﬁnite intersection property of U. Hence either A ∈ U or X \ A ∈ U.

6It is not true that these two constructions - from universal nets to ultrafilters and back - are inverse to each other. More precisely, if we start with an ultrafilter U, construct a universal net (xA)A∈U , and then consider the corresponding ultrafilter U 0, then we do get that U 0 = U. The reason is that by construction we have U ⊂ U 0. But then if there exists A ∈ U 0 \U, we have X \ A ∈ U as U is an ultrafilter, and this gives ∅ = A ∩ (X \ A) ∈ U 0, which is a contradiction. Hence U 0 = U. On the other hand, if we start with a universal net (xi)i∈I , consider the ultrafilter U defined by it, and then construct a universal net (xA)A∈U from U, then this new net is in general different from (xi)i. Nevertheless, by the previous paragraph, both universal nets produce the same ultrafilter U. This exactly means that they are AA-equivalent (see footnote 4). We can therefore conclude that there is a one-to-one correspondence between the ultrafilters on X and the AA-equivalence classes of universal nets in X. 5 Next, take A, B ∈ U. We cannot have X \ (A ∩ B) ∈ U, since A ∩ B ∩ (X \ (A ∩ B)) = ∅. It follows that A ∩ B ∈ U, so U is closed under finite intersections. Therefore U is indeed an ultrafilter. An example of an ultrafilter closely related to Example 4.2 of a universal net can be constructed as follows. Take a point x ∈ X. Then the collection of all subsets of X containing x is an ultrafilter on X. Such ultrafilters are called principal. The ultrafilters that are not principal are called free. Corollary 4.9. Free ultrafilters exist on any infinite set X. Later we will show much more, see Proposition 5.10. Proof. Consider the collection A of all subsets of X of the form X \ F , where F ⊂ X is finite. As X is infinite, A has the finite intersection property. Hence, by Theorem 4.8, there is an ultrafilter U containing A. This ultrafilter cannot be principal, since for every x ∈ X we have X \{x} ∈ U.

Proof of Theorem 4.5. We will deduce the existence of a universal subnet of (xi)i∈I from Theorem 4.8 applied to the set I rather than X. Consider the collection A of subsets of I of the form {j | j i}, i ∈ I. It has the finite intersection property, hence it is contained in an ultrafilter U on I. Order U by the inverse inclusion. Take any map ϕ: U → I such that ϕ(A) ∈ A for every A ∈ U. This map has the property from Definition 2.1. Indeed, given i0 ∈ I, consider the set A0 = {i | i i0}. Then, for any A A0, we have ϕ(A) ∈ A ⊂ A0, so ϕ(A) i0. Therefore ϕ: U → I defines a subnet of (xi)i. Since U is an ultrafilter on I, by the discussion before Theorem 4.8 the net (ϕ(A))A∈U in I is universal. Hence, by Lemma 4.4, the net (xϕ(A))A∈U is universal as well, being the image of (ϕ(A))A∈U under the map f : I → X, f(i) = xi. We can now establish one more characterization of compactness. Theorem 4.10. A topological space X is compact if and only if every universal net in X converges to some point.

Proof. ⇒ If (xi)i is a universal net in X, then by compactness it has a cluster point x. Hence xi → x by Lemma 4.3. ⇐ If (xi)i is a net in X, then by Theorem 4.5 it has a universal subnet. By assumption this subnet converges to a point x. But then x is a cluster point of (xi)i. Thus every net in X has a cluster point. Hence X is compact. As an application of Theorem 4.10 we can now give a very short proof of Tychonoﬀ’s theorem. Q Another proof of Theorem 3.4. It suﬃces to show that every universal net (zi)i in s Xs converges to some point. For every s ∈ S, denote by ps the projection onto Xs. The net (ps(zi))i in Xs is universal by Lemma 4.4, hence it converges to a point xs ∈ Xs by compactness of Xs. Then the net (zi)i converges to x = (xs)s∈S, since it does so coordinate-wise. An attentive reader should see that this proof is, in fact, very close to that of [M, Theorem 37.3], but with all the nontrivial arguments pushed into developing the theory of universal nets.

5. Stone–Cechˇ compactification Let X be a Hausdorff topological space. Definition 5.1. The Stone–Cechˇ compactification of X is a Hausdorff compactification Y satisfying the following universal property: any continuous map f of X into a compact Hausdorff space Z extends to a continuous map f˜: Y → Z. Note that since by definition of a compactification the space X is assumed to be dense in Y , the map f˜ is uniquely determined by f. The universality implies that the Stone–Cechˇ compactification of X is unique up to isomorphism if it exists: 0 if Y is another compactification with the same property, then the identity map idX extends to continuous 6 maps Y → Y 0 and Y 0 → Y , which are then inverse to each other. The Stone–Cechˇ compactification of X is denoted by βX or β(X). It is proved in [M, §38] that the Stone–Cechˇ compactification of X exists if and only if X is completely regular.7 Our goal is to show that its construction is closely related to the theory of universal nets/ultrafilters. We will concentrate on the discrete spaces, for which the relation is particularly nice. Assume from now on that X is a discrete topological space. Imagine that we do not know that βX exists and want to construct it. Since every universal net in X must converge in any compactification of X, a natural idea is to add to X imaginary limits of all universal nets. Since different nets may converge to the same point, we need to consider certain equivalences classes of nets rather than the nets themselves. In view of the connection between universal nets and ultrafilters discussed in the previous section (and particularly in footnote 6), it seems like not a bad idea to consider ultrafilters instead of universal nets. We thus define the set βX as the set of all ultrafilters on X. For every x ∈ X, consider the principal ultrafilter ωx = {A ⊂ X | x ∈ A} generated by x. Then X can be identified with the subset of βX via the map X 3 x 7→ ωx. Before we turn to defining a topology on βX, let us see how to extend a map f : X → Z, where Z is a compact Hausdorff space, to a map f˜: βX → Z. There is only one natural choice: if a point ω in βX ˜ corresponds to a universal net (xi)i, then we should have f(ω) = limi f(xi). Our discussion of the connection between universal nets and ultrafilters implies that this way we indeed get a well-defined map f˜: βX → Z, but at this point it makes sense to forget about universal nets and do everything needed from scratch for ultrafilters. Definition 5.2. Given an ultrafilter ω on X and a map f from X into a Hausdorff topological space Z, we say that z ∈ Z is the limit of f along ω, and write z = lim f(x) or z = ω- lim f(x), x→ω x if for every neighbourhood U of z we have f −1(U) ∈ ω.

It is easy to see that limx→ω f(x) is unique if it exists.

Lemma 5.3. If Z is compact Hausdorff, then limx→ω f(x) exists for any map f : X → Z and any ultrafilter ω on X. Proof. The collection of closed sets f(A) for A ∈ ω has the finite intersection property. Hence, by compactness of Z, the intersection ∩A∈ωf(A) is nonempty. Take a point z in this intersection and a neighbourhood U of z. If f −1(U) 6∈ ω, then the set A = X \ f −1(U) = f −1(Z \ U) is in ω. But then f(A) ⊂ Z \ U, which −1 contradicts z ∈ f(A). Hence f (U) ∈ ω. Therefore for any map f : X → Z, with Z compact Hausdorff, we can define f˜: βX → Z by f˜(ω) = lim f(x). x→ω Note also that by the uniqueness of limits along ultrafilters and the proof of the above lemma we have \ {f˜(ω)} = f(A). A∈ω Finally, we have to define a topology on βX. Since the maps f˜ as above are supposed to be continuous, it is natural to take the coarsest topology with respect to which all such maps are continuous. This is not very good, as the collection of all possible maps of X into compact Hausdorff topological spaces is huge. As we will see, it suffices to consider the maps X → {0, 1}, that is, the characteristic functions of subsets of X. It follows immediately by the definition of limx→ω that for any A ⊂ X we have ( 1, if A ∈ ω, lim χA(x) = x→ω 0, otherwise.

7Sometimes by a compactification of X one means simply a compact space Y together with a continuous map X → Y with dense image. If we adopt this definition, then the Stone–Cechˇ compactification exists for any topological space X. The construction is the same as for completely regular spaces. 7 We then see that the coarsest topology on βX with respect to which all the mapsχ Ã are continuous is given by the subbasis consisting of the sets

UA = {ω ∈ βX | A ∈ ω} for all A ⊂ X.

In fact, since UA ∩ UB = UA∩B, this is a basis of topology. It remains to see that the topology on βX has the right properties. In order to check compactness of βX we will need the following property of ultraﬁlters reminiscent of prime ideals. Lemma 5.4. If ω ∈ βX and A ∪ B ∈ ω for some A, B ⊂ X, then either A ∈ ω or B ∈ ω. Proof. Assume A 6∈ ω. Then X \ A ∈ ω. Hence B \ A = (A ∪ B) ∩ (X \ A) ∈ ω,

and therefore B ∈ ω by Remark 4.7. Lemma 5.5. The space βX is compact Hausdorﬀ. Proof. In order to prove compactness it suﬃces to show that every covering of βX by the basic open sets

UA has a ﬁnite subcovering. Assume {UAi }i∈I is such a covering. We claim that

Ai1 ∪ · · · ∪ Ain = X

for some i1, . . . , in ∈ I. Indeed, assume this is not the case. Then the collection {X \ Ai}i∈I of subsets of X has the ﬁnite intersection property. By Theorem 4.8, this collection is contained in an ultraﬁlter ω. But

then ω 6∈ UAi for all i ∈ I, which is a contradiction. Thus our claim is proved. Now, for every ω ∈ βX, we have

Ai1 ∪ · · · ∪ Ain = X ∈ ω. By the previous lemma it follows that A ∈ ω for some k. Hence the sets U ,...,U form a finite open ik Ai1 Ain covering of βX. In order to check that βX is Hausdorff, take two different ultrafilters ω and ω0. Then one of them, 0 say ω, contains a set A not contained in the other. But then X \ A ∈ ω , so UA and UX\A are disjoint 0 neighbourhoods of ω and ω , respectively. Note that the last part of the proof shows that different ultrafilters cannot be contained in each other (which we already used in footnote 6).

Remark 5.6. We have βX \ UA = UX\A. This shows that the basic open sets UA are closed, hence βX is totally disconnected. In fact, it is possible to show that βX is extremely disconnected, meaning that the closure of every open set is open. Lemma 5.7. The set X is discrete and dense in βX.

Proof. We have U{x} = {ωx}, since if ω ∈ U{x}, then ωx ⊂ ω and hence ωx = ω. Thus X is discrete in βX. In order to see that X is dense, take an ultrafilter ω and a basic neighbourhood UA of ω. Then A ∈ ω, so A 6= ∅, and we have ωx ∈ UA for all x ∈ A. The previous two lemmas show that βX is a Hausdorff compactification of X. In order to finish the proof that βX is the Stone–Cechˇ compactification of X it remains to establish the following. Lemma 5.8. For any map f : X → Z, where Z is compact Hausdorff, the map f˜: βX → Z is continuous. Proof. Take ω ∈ βX and put z = f(ω). Let U be a neighbourhood of z. Choose a smaller neighbourhood V −1 of z such that V¯ ⊂ U. Consider the set A = f (V ). Then UA is a neighbourhood of ω, and for every 0 ω ∈ UA we have f(ω0) ∈ f(A) ⊂ V¯ ⊂ U. ˜ It follows that f is continuous at ω. 8 At a first glance the construction of βX in terms of ultrafilters may seem quite different from that in [M]. But it is not, and if we wanted to understand better the construction in [M] we would see that the appearance of ultrafilters is quite natural. Namely, the construction in [M] goes as follows. We consider the collection {fi}i∈I of all maps X → [0, 1]. I I Then the map X 3 x 7→ (fi(x))i∈I ∈ [0, 1] is an embedding, and as βX we take the closure of X in [0, 1] . It is not difficult to see that since any function X → [0, 1] can be approximated in the supremum- norm by finite linear combinations of characteristic functions of subsets of X, we would obtain the same compactification if instead of all maps X → [0, 1] we considered only the characteristic functions. Thus, we consider the embedding P(X) X → {0, 1} , x 7→ (χA(x))A∈P(X), and define βX as the closure of X in {0, 1}P(X). What is this closure? As we know, as a set, the space {0, 1}P(X) can be identified with P(P(X)), that is, with the set of collections of subsets of X.

Lemma 5.9. Under the identiﬁcation of {0, 1}P(X) with P(P(X)), the closure of X in {0, 1}P(X) is exactly the set of ultraﬁlters on X.

P(X) Proof. Let ω ∈ {0, 1} be an ultraﬁlter on X. Take a ﬁnite collection of subsets of X. Let A1,...,An be the elements in this collection such that ωAi = 1 and B1,...,Bm be the elements such that ωBk = 0. In other words, A1,...,An ∈ ω and B1,...,Bm 6∈ ω, hence X \ B1,...,X \ Bm ∈ ω. Then the set

C = A1 ∩ · · · ∩ An ∩ (X \ B1) ∩ · · · ∩ (X \ Bm) ∈ ω is nonempty. But then for any x ∈ C we have χA(x) = ωA for all A ∈ {A1,...,An,B1,...,Bm}. This shows that every neighbourhood of ω contains an element of X, so ω belongs to the closure of X. Conversely, assume ω ∈ {0, 1}P(X) belongs to the closure of X. First of all, ω does not contain the empty set, since there is no point x ∈ X such that χ∅(x) = 1. Assume next that for some A, B ∈ ω we have A ∩ B 6∈ ω, that is, ωA = ωB = 1 but ωA∩B = 0. But then there is no x ∈ X such that χC (x) = ωC for all C ∈ {A, B, A ∩ B}. Hence ω does not belong to the closure of X, which is a contradiction. Thus ω is closed under finite intersections. Assume now that there is a subset A such that both A and X \ A are not in ω. But then there is no x ∈ X such that χC (x) = ωC both for C = A and C = X \ A. Hence ω does not belong to the closure of X, which is a contradiction. Thus, for every subset A, we have either A ∈ ω or X \ A ∈ ω. Hence ω is an ultrafilter. Finally, let us note that if X is an infinite discrete topological space, then βX is very large. For example, we have the following. Proposition 5.10. If X is countably infinite, then |βX| = 2c, where c is the cardinality of the continuum. In particular, βX is not first countable, hence it is nonmetrizable. Here by 2|Y | we mean the cardinality of P(Y ).

Proof. We may identify X with N. We have c |βN| ≤ |P(P(N))| = |P(R)| = 2 . On the other hand, we know that the space [0, 1]R is separable. This means that there exists a map N → [0, 1]R with dense image. This map extends to a continuous map βN → [0, 1]R, which then must be surjective. Hence |βN| ≥ 2c. In order to prove the second part of the proposition it suffices to show that if Y is a first countable ∞ separable Hausdorff topological space, then |Y | ≤ c. Assume (yn)n=1 is a dense sequence in Y . By first ∞ countability, for every element y not in this sequence we can find a subsequence (ynk )k=1 (with nk < nk+1 for all k) such that ynk → y. It follows that |Y | ≤ |P(N)| = c.

9 References [M] J.R. Munkres. Topology, 2nd ed. Pearson, 2000.

Sergey Neshveyev, Mathematics institute, University of Oslo, Moltke Moes vei 35, 0851 Oslo, Norway E-mail address: [email protected]