Topology (H) Lecture 5 Lecturer: Zuoqin Wang Time: March 22, 2021

Home , Box topology, Final topology, Induced topology, Initial topology, Natural topology

BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES

1. Bases and sub-bases of a topology ¶ Topology deﬁned via a basis.

If you stare at the deﬁnitions of the topologies Tmetric, TSorgenfrey, Tp.c. and TX×Y ,

Tmetric = {U ⊂ X | ∀x ∈ U, ∃r > 0 s.t. B(x, r) ⊂ U},

TSorgenfrey = {U ⊂ R | ∀x ∈ U, ∃ε > 0 s.t. [x, x + ε) ⊂ U},

TX×Y = {W ⊂ X × Y | ∀(x, y) ∈ W, ∃U ∈ TX and V ∈ TY s.t. (x, y) ∈ U × V ⊂ W },

Tp.c. = {U ⊂ X | ∀f0 ∈ U, ∃x1, ··· , xn ∈ [0, 1] and ε > 0 s.t. ω(f0; x1, ··· , xn; ε) ⊂ U}, you can easily ﬁnd a common nature: They are all of the form

(1.1) TB := {U ⊂ X | ∀x ∈ U, ∃B ∈ B s.t. x ∈ B ⊂ U} for a collection of sets B ⊂ P(X). There must be a general rule behind these deﬁnitions! Let’s try to ﬁnd it! Let B ⊂ P(X) be a collection of subsets of X, like B = all metric balls in a given metric space in the case of metric topology. Note: B itself is not a topology.

Question : Under what assumption on B, the family TB deﬁned by (1.1) is a topology?

- By construction, ∅ ∈ TB. - We want X ∈ TB, so we need (B1) ∀x ∈ X, ∃B ∈ B s.t. x ∈ B.

- Suppose U1,U2 ∈ TB, we want U1 ∩ U2 ∈ TB, i.e.

(?) ∀x ∈ U1 ∩ U2, ∃B ∈ B s.t. B ⊂ U1 ∩ U2.

Unfortunately this condition involves U1,U2 ∈ TB, i.e. it is NOT a condition on the family B itself. However, by construction, ∀x∈U1 ∩ U2, ∃B1,B2 ∈B s.t.

x ∈ B1 ⊂ U1, x ∈ B2 ⊂ U2. So for (?) to be true, we may assume

(B2) ∀B1,B2 ∈ B, ∀x ∈ B1 ∩ B2, ∃B ∈ B s.t. x ∈ B ⊂ B1 ∩ B2.

1 2 BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES

Note: (B2) is NOT only a suﬃcient condition for (?), but also a necessary condition, since by construction, any element B ∈ B is also an element B ∈ TB. - Finally suppose Uα ∈ TB. Then we automatically have ∪αUα ∈ TB, since

∀x ∈ ∪αUα, ∃α0 s.t. x ∈ Uα0 . so ∃B ∈ B s.t. x ∈ B ⊂ Uα0 , which implies x ∈ B ⊂ ∪αUα i.e. ∪αUα ∈ TB

Answer : A necessary and sufficient condition for TB defined by (1.1) to be a topology on X is that the collection B satisfies conditions (B1) and (B2). Definition 1.1 (Basis of a topology). (1) A collection B ⊂ P(X) is called a basis (or a base) for a topology if it satisfies conditions (B1) and (B2). (2) The topology TB defined by (1.1) is called the topology generated by the basis B. Remark 1.2. Different bases could generate the same topology. For example, each of the following three collections is a basis that generates the Euclidean topology on R2,: 2 B1 = {B(x, r) | x ∈ R , r > 0}, 2 B2 = {B(x, r) | x ∈ Q , r ∈ Q>0}, B3 = {(a, b) × (c, d) | a, b, c, d ∈ R}.

Note that B2 is a countable family!

Remark 1.3. Let’s emphasis again: By deﬁnition, B ⊂ TB, i.e. every element in B is an open set in the topology TB. Usually the converse is not true.

¶ Example: The box topology. Example 1.4 (The box topology). Suppose we have a family of topological spaces (Xα, Tα). We want to define a topology on the Cartesian product (maybe infinite/uncountable) Y Xα = {(xα) | xα ∈ Xα}. α As in the case of the Cartesian product of two topological spaces, we may choose ¨ « Y B = Uα | Uα ∈ Tα . α It is easy to check that B satisfies (B1), (B2), so we get a topology Y Y TBox = {U ⊂ Xα | ∀(xα) ∈ U, ∃Uα ∈ Tα s.t. (xα) ∈ Uα ⊂ U}. α α Q This is called the box topology on X = α Xα. BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES 3

¶ Topology deﬁned via a basis: Minimality.

To understand the relation between B and TB, we give an alternative explanation of the sentence “the basis B generates the topology TB”:

Proposition 1.5. If B is a basis for a topology TB, then [ 0 TB = { B | B ⊂ B}. B∈B0

0 Proof. As we remarked above, B ⊂ TB. So for any sub-family B ⊂ B, we have [ B ∈ TB. B∈B0

Conversely, for any U ∈ STB and any x ∈ U, by deﬁnition there exists Bx ∈ B s.t. x ∈ Bx ⊂ U. It follows U = x∈U Bx, i.e. U is of the given form. As a consequence,

0 0 Corollary 1.6. Let B be a basis for a topology TB, and T is a topology s.t. B ⊂ T . 0 Then TB ⊂ T .

It follows that TB is the “smallest” topology so that all sets in B are open: \ 0 (F) TB = T . B⊂T 0 T 0 is a topology

¶ The topology generated by an arbitrary family of subsets. It turns out that the formula (F) can be used to construct topology from any family of subsets that is not necessary a basis. To see this, we ﬁrst observe: T Proposition 1.7. Given any family Tα of topologies on X, α Tα is a topology.

Proof. •∅ ,X ∈ Tα, ∀α ⇒ ∅,X ∈ ∩αTα. • U1,U2 ∈ Tα, ∀α ⇒ U1 ∩ U2 ∈ Tα, ∀α ⇒ U1 ∩ U2 ∈ ∩αTα. • Uβ ∈ Tα, ∀α ⇒ ∪βUβ ∈ Tα ⇒ ∪βUβ ∈ ∩αTα. Now let S ⊂ P(X) be any collection of subset in X. Deﬁnition 1.8. The topology generated by S is \ 0 TS := T . S⊂T 0 T 0 is a topology

So in other words, TS is the weakest topology so that all sets in S are open.

A natural question is: What is TS ? 4 BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES

Proposition 1.9. Let S ⊂ P(X) be a family of subsets, and denote

(1.2) B = {B | ∃S1, ··· ,Sm ∈ S s.t. B = S1 ∩ · · · ∩ Sm}. S (1) If S∈S S = X, thenSB is a basis of TS , i.e. TS = TB. 0 0 (2) In general, if X = S∈S ⊂ X, then B is a basis of a topology TB on X , and TS = {X} ∪ TB. S S Proof. (1) By definition, S ⊂ B. So the condition S∈S = X implies B∈B = X, which is equivalent to say that the family B satisfies the condition (B1). By construction, B also satisfies the condition (B2). So B is a basis. Obviously for any topology T 0,

T 0 ⊃ S ⇐⇒ T 0 ⊃ B. T T 0 0 So S⊂T 0 T = B⊂T 0 T , i.e. the topology generated by B is exactly TS . (2) By (1), {X} ∪ TB is a topology on X. By construction, it is the weakest topology on X so that all sets in S are open.

¶ The topology defined via a sub-basis. In view of Proposition 1.9, it is natural to define S Definition 1.10. If a family S ⊂ P(X) satisfies S∈S S = X, we will call S a sub-basis (or a sub-base) of the topology TS .

Example 1.11. For the pointwise convergence topology Tp.c. on M([0, 1], R),

B = {ω(f; x1, ··· , xn; ε) | f ∈ M([0, 1], R), n ∈ N, x1, ··· , xn ∈ [0, 1], ε > 0} is a basis, while

S = {ω(f; x; ε) | f ∈ M([0, 1], R), x ∈ [0, 1], ε > 0} is a sub-basis.

Example 1.12. For the standard Euclidean topology on R, B = {(a, b) | a < b} is a basis, while S = {(−∞, a), (a, +∞) | a ∈ R} is a sub-basis. Example 1.13. The the product topology on X × Y ,

B = {U × V | U ∈ TX ,V ∈ TY } is a basis, while [ S = {U × Y | U ∈ TX } {X × V | V ∈ TY } is a sub-basis. BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES 5

¶ Characterization of a basis/sub-basis. A natural questions is: Give a family B, how can we tell whether it is a basis of a given topology T ? Here is a simple criterion: Proposition 1.14. Let (X, T ) be a topological space. A collection B ⊂ P(X) is a basis for T if and only if (1) B ⊂ T , (2) for any U ∈ T and any x ∈ U, there exists B ∈ B such that x ∈ B ⊂ U.

Proof. By deﬁnition, if B is a basis for T , then (1), (2) holds. Conversely, it is obvious that (2) implies (B1), and (1), (2) together implies (B2). So B is a basis. Moreover, (2) implies T ⊂ TB. But by minimality, TB ⊂ T . So the topology generated by B is T . Similarly we have Proposition 1.15. Let (X, T ) be a topological space. A collection S ⊂ P(X) is a basis for T if and only if (1) S ⊂ T , (2) for any U ∈ T and any x ∈ U, there exists S1, ··· ,Sn ∈ S such that x ∈ n ∩i=1Si ⊂ U. ¶ Continuity via a basis/sub-basis. We have seen that a map f : X → Y between topological spaces is continuous if and only if f −1(open) = open. It is not hard to see that we only need to check the openness for some subsets, namely, the openness of the pre-images of those open sets in a basis, or a sub-basis:

Theorem 1.16. Suppose B is a basis of TY , and S a sub-basis of TY . Then

a map f :(X, TX ) → (Y, TY ) is continuous ⇐⇒f −1(B) is open in X for any B ∈ B ⇐⇒f −1(S) is open in X for any S ∈ S.

Proof. It is enough to prove −1 −1 f (S) ∈ TX , ∀S ∈ S =⇒ f (V ) ∈ TX , ∀V ∈ TY , which is a consequence of the fact that f −1 preserves unions and intersections: [ n\(α) [ n\(α) −1 −1 f ( Sα,i) = f (Sα,i). α i=1 α i=1 6 BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES

¶ Example: The order topology. It turns out that the construction in Example 1.12 works for any totally ordered set. Deﬁnition 1.17. A partially ordered set (also known as poset)(X, ≤) is a set X with a partial order relation ≤, namely • x ≤ x, • if x ≤ y, y ≤ z, then x ≤ z, • if x ≤ y, y ≤ x, then x = y. A totally ordered set is a poset (X, ≤) such that for any x, y, either x ≤ y or y ≤ x. Note that given any order relation ≤, we may deﬁne < via x < y ⇐⇒ x ≤ y and x 6= y. Now we extend Example 1.12 to any totally ordered set:

Deﬁnition 1.18. The order topology Torder on a totally ordered set (X, ≤) is the topology generated by the sub-basis S that consists all sets of the form {x | x < a}, {x | x > a}, for a ∈ X.

It is not hard to see that a basis B of the order topology Torder consists of all sets of the form {x | x < a}, {x | x > a}, {x | a < x < b}, for a, b ∈ X. ¶ Example: The product topology.

Let (Xα, Tα) be a familyQ of topological spaces. Now we define the product topology on the CartesianQ product α Xα. We have seen that by generalizingQ the basis B in Example 1.13 to α Xα, we can define the box topology on α Xα. NowQ we generalize the sub-basis S in Example 1.13 to construct the product topology on α Xα. Note that for X × Y , if we denote πX : X ×QY → X be the canonical projection, then −1 U × Y = πX (U). In general, we let πβ : α Xα → Xβ be the canonical projection. Q Definition 1.19. The product topology Tproduct on α Xα is the topology generated by the sub-basis [ −1 S = {πβ (Vβ) | Vβ ∈ Tβ}. β

By deﬁnition, it is obvious that Tproduct is weaker than Tbox. Remark 1.20. Although it seems that the product topology is not as natural as the box topology, it turns out that the product topology is much more important: it has many nice properties. On the other hand, the box topology has many bad properties and is widely used as counterexamples. (The box topology play the role of negative character!) Of course for ﬁnite products, these two topologies are the same. BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES 7 Q Proposition 1.21. For anyQ β, the projection map πβ : α Xα → Xβ is continuous and open, where we endow α Xα with either the product topology or the box topology. Proof. Since the product topology is weaker than the box topology, it is enough to prove the continuity for Tproduct, and prove the openness for TBox:

• For Tproduct, πβ is continuousQ since the pre-image of any open set Vβ in (Xβ, Tβ) −1 is an open set πβ (Vβ) in ( α Xα, Tproduct). • For TBox, πβ is open since forQ any open set W ⊂ TBox and any x ∈ W , there exists Uα ∈ Tα such that x ∈ α Uα. As a result, πβ(x) ∈ Uβ ⊂ πβ(W ).

In fact, the product topology can be characterized via the projection map πβ’s: Q Proposition 1.22. The product topology Tproduct is the weakest topology on α Xα so that each πβ is continuous.

Proof. We have seen that each πβ is continuous with respectQ to Tproduct. Conversely if −1 each πβ is continuous with respect to some topology T on α Xα, then each πβ (Vβ) is open in T , so Tproduct is weaker than T .

¶ Universal property of the product topology. The product topology can be characterized by the following universal property:

Theorem 1.23 (Universal property of the productQ topology). Let X,Xα be topological spaces, and fα : X → Xα be maps. Endow α Xα with the product topology. Then the map Y f : X → Xα, x 7→ (fα(x)) α is continuous if and only if eachQ fα = πα ◦ f is continuous. Moreover, the product topology is the only topology on α Xα that satisﬁes this property. Proof.

• (only if) If f is continuous, then fβ = πβ ◦ f is continuous. • (if) Suppose each fα is continuous. To prove f is continuous, it is enough to −1 −1 prove each f (πβ (Vβ)) is open in X, where Vβ is open in Xβ. However, −1 −1 −1 −1 f (πβ (Vβ)) = (πβ ◦ f) (Vβ) = fβ (Vβ).

So by the continuity of fβ, they are open in X. • Finally we show that the product topology is characterized by the universal property: Suppose T is a topology on X that satisﬁes the universalQ property. – By the universality of T andQ the continuity of eachQ πβ :( α Xα, Tproduct) → Xβ, the identity map Id : ( α Xα, Tproduct) → ( α Xα, T ) is continuous. 8 BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES

– Moreover,Q by the universalityQ of T and the continuity of theQ identity map Id : ( α Xα, T ) → ( α Xα, T ), we see each projection πβ :( α Xα, T ) → Xβ is continuous. Q – The continuity of πβ :( α XQα, T ) → Xβ andQ the universality of Tproduct imply the identity map Id : ( α Xα, T ) → ( α Xα, Tproduct) is continuous. So we must have T = Tproduct. As a consequence, we have

CorollaryQ 1.24. Fix an xβ Q∈ Xβ for each β 6= α, and deﬁne an “embedding map” gα : Xα → β Xβ of Xα into β Xβ by requiring ¨ x , β 6= α π (g (x)) = β β α x, β = α.

Then gα is continuous. Remark 1.25. The universal property fails for the box topology. For example, if we let Y X = RN = R n∈N and consider the map f : R → RN, t 7→ (t, t, t, ··· ). Then f is NOT continuous if we endow the target space RN with the box topology, 1 1 1 1 since the Cartesian product (−1, 1)×(− 2 , 2 )×(− 3 , 3 )×· · · is open in the box topology, 1 1 1 1 f −1 (−1, 1) × (− , ) × (− , ) × · · · = {0}, 2 2 3 3 but {0} is NOT open in R.

2. Topologies defined by maps ¶ The induced topology. We start with two observations:

• WeQ have just showed that the product topologyQ Tproduct is the weakest topology on α Xα so that all the projections πβ : α Xα → (Xβ, Tβ) are continuous. • Similarly if (X, T ) is a topological space, and A ⊂ X a subset, then the subspace topology TA on A is the weakest topology on A so that the inclusion map ι : A,→ X is continuous. In general, we can use maps to construct new topologies: BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES 9

Deﬁnition 2.1. Let {(Yα, Tα)} be a family of topological spaces, and let

F = {fα : X → (Yα, Tα)} be a family of maps. The F-induced topology (which is also known as initial topology or weak topology or limit topology)on X, denoted by TF , is deﬁned to be the weakest topology such that all the maps fα’s are continuous. Remark 2.2. We need to explain the deﬁnition a little bit. • Firstly, as we have seen last time, if we endow X with the strongest topology, namely the discrete topology, then any fα is continuous. This is not interesting. • If there are topologies Tβ on X so that each fα is continuous, then as we have seen, T := ∩βTβ is a topology on X. Moreover, by construction fα is continuous with respect to T . As a consequence, there exists a unique weakest topology on X so that each fα is continuous.

It is not hard to ﬁnd this topology. By deﬁnition, Tf is the topology generated by the sub-basis [ −1 SF = {fα (Vα) | Vα ∈ Tα}. α

Note that in the special case where we only have one target space (Y, TY ) and only one map f : X → (Y, TY ), the sub-basis −1 Sf = {f (V ) | V ∈ TY }. itself is a topology on X. So in this case, −1 Tf = {f (V ) | V ∈ TY }. By repeating the proof of Theorem 1.23, one has Proposition 2.3. Let Z be a topological space. Endow X with the F-induced topology. Then a map f : Z → X is continuous if and only if each fα ◦f : Z → Yα is continuous. Moreover, the F-induced topology is the only topology satisfying this property.

¶ More examples of induced topology. We have seen that both the subspace topology and the product topology can be explained as an induced topology. Here we list a couple more: Example 2.4. (The pointwise convergence topology revisited) Let X = M([0, 1], R) be the space of all real-valued functions on [0, 1]. For any x ∈ [0, 1], let evx : X → R be the evaluation map

evx(f) := f(x).

Fact: The topology on X generated by {evx | x ∈ [0, 1]} is the pointwise convergence topology. (Why?) 10 BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES

Example 2.5. (The weak and weak-* topology) Let X be a topological vector space1. As usual we denote by X∗ the dual space, i.e. X∗ = the space of all continuous linear functionals on X = {f : X → K | f is linear and continuous (w.r.t. the original topology on X)}. Then we can deﬁne a new topology on X and deﬁne a natural topology on X∗: • The the weak topology on X is the topology induced by X∗, i.e. the weakest topology on X so that all f ∈ X∗ is continuous. [As a consequence, the weak topology on X is weaker than the original topology on X.] ∗ • The weak-∗ topology on X is the topology induced by {evx | x ∈ X}, i.e. the weakest topology on X∗ so that all the evaluations maps ∗ evx : X → R, l 7→ l(x) are continuous. The weak topology and weak-∗ topology is widely used in functional analysis and PDE. Those who are curious may have a look at my functional analysis notes, lecture 14 and lecture 15.

¶ The co-induced topology. Not only maps can be used to “pull-back” topologies from the target spaces to the source space, they can also be used to “push-forward” topologies from source spaces to the target space. More precisely, let (Xα, Tα) be a family of topological spaces, Y be a set and fα : Xα → Y be a family of maps. Again we can put a topology on Y so that each fα is continuous. Of course for this purpose, we can’t put too much open sets in Y . On the other hand side we don’t want to use the trivial topology in Y , which is too weak and makes any map from any topological space to Y continuous. So the question is to put a topology which is as strong as possible so that each fα is continuous. It is thus natural to deﬁne

Deﬁnition 2.6. Let (Xα, Tα) be a family of topological spaces, let Y be a set and let F = {fα : Xα → Y } be a family of maps. The strongest topology T on Y making each fα continuous is called the co-induced topology (also known as the ﬁnal topology or the strong topology or the colimit topology) by the family F.

One has to be a little bit careful: Does such a strongest topology exist? Note that in deﬁning the weak topology induced by a family F, we used the fact that the intersection of a family of topologies on X is a still a topology on X (See Proposition 1.7). In general, the union of a family of topologies on X will fail to be a topology on X (ﬁnd an example!). However, if one think about this problem really hard, one may

1If one assume topological vector spaces to be Hausdorﬀ in the deﬁnition, then in studying the dual space one need to assume that the dual space X∗ separate points in X. This is the case, for example, if the topological vector space X is locally convex. BASES AND SUB-BASES, INDUCED AND CO-INDUCED TOPOLOGIES 11

ﬁnd that we are really in a similar situation as the weak topology: we are not taking the union of a family of topologies, but the intersection of a family of topologies (since we have a family of constraints!): The following theorem is almost immediate:

Theorem 2.7. Let (Xα, Tα) be a family of topological spaces, and fα : Xα → Y be a family of maps. Then the co-induced topology on Y induced by {fα} is \ −1 T = {V ⊂ Y | fα (V ) ∈ Tα}. α Proof. One can easily check by deﬁnition that the collection T is a topology on Y , and each fα is continuous with respect to this topology. On the other hand, if we add any other set V0 to this family, then by construction, there exists some α such that −1 fα (V0) 6∈ Tα, so fα is no longer continuous. As in the case of the induced topology, the co-induced topology may be characterized by the following universal property: Proposition 2.8. Let Z be a topological space. Endow Y with the F-induced topology. Then a map f : Y → Z is continuous if and only if each f ◦fα : Xα → Z is continuous. Moreover, the co-induced topology on Y induced by F is the only topology satisfying this property. ¶ Examples.

Example 2.9 (The intersection of topologies). LetT Tα be topologies on X. Now we have a diﬀerent way to look at the topology T = α Tα, where Tα is a family of topology on X: Let Idα :(X, Tα) → X be the identity map. Then T is the co-induced topology on X induced by the family {Idα}.

Example 2.10 (The join of topologies ). Let Tα be topologies on X. We can also induce Þ a topology: consider the identity maps Idα : X → (X, Tα). The join topology is the Þ induced topology on X induced by the familyS {Idα}. What is it? Well, it is nothing else but the weakest topology generated by α Tα.

Example 2.11. (The topological union) Given a family of topological spaces (Xα, Tα), so that the induced subspace topologies agree on each intersection Xα ∩ Xβ, we may deﬁne a topology T on the union X = ∪Xα to be the strongest topology on X so that each ια : Xα ,→ X is continuous. In other words, it is the co-induced topology by the inclusion maps ια’s. The topological space (X, T ) is called the topological union of (Xα, Tα). The most important example of the co-induced topology is the so-called quotient topology which we will study in detail later.