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

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Topology (H) Lecture 5 Lecturer: Zuoqin Wang Time: March 22, 2021 Topology (H) Lecture 5 Lecturer: Zuoqin Wang Time: March 22, 2021 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 = fU ⊂ X j 8x 2 U; 9r > 0 s.t. B(x; r) ⊂ Ug; TSorgenfrey = fU ⊂ R j 8x 2 U; 9" > 0 s.t. [x; x + ") ⊂ Ug; TX×Y = fW ⊂ X × Y j 8(x; y) 2 W; 9U 2 TX and V 2 TY s.t. (x; y) 2 U × V ⊂ W g; Tp:c: = fU ⊂ X j 8f0 2 U; 9x1; ··· ; xn 2 [0; 1] and " > 0 s.t. !(f0; x1; ··· ; xn; ") ⊂ Ug; you can easily find a common nature: They are all of the form (1.1) TB := fU ⊂ X j 8x 2 U; 9B 2 B s.t. x 2 B ⊂ Ug 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, ; 2 TB. - We want X 2 TB; so we need (B1) 8x 2 X; 9B 2 B s.t. x 2 B: - Suppose U1;U2 2 TB; we want U1 \ U2 2 TB, i.e. (?) 8x 2 U1 \ U2; 9B 2 B s.t. B ⊂ U1 \ U2: Unfortunately this condition involves U1;U2 2 TB, i.e. it is NOT a condition on the family B itself. However, by construction, 8x2U1 \ U2, 9B1;B2 2B s.t. x 2 B1 ⊂ U1; x 2 B2 ⊂ U2: So for (?) to be true, we may assume (B2) 8B1;B2 2 B; 8x 2 B1 \ B2; 9B 2 B s.t. x 2 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 2 B is also an element B 2 TB: - Finally suppose Uα 2 TB. Then we automatically have [αUα 2 TB, since 8x 2 [αUα; 9α0 s.t. x 2 Uα0 . so 9B 2 B s.t. x 2 B ⊂ Uα0 , which implies x 2 B ⊂ [αUα i.e. [αUα 2 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 = fB(x; r) j x 2 R ; r > 0g; 2 B2 = fB(x; r) j x 2 Q ; r 2 Q>0g; B3 = f(a; b) × (c; d) j a; b; c; d 2 Rg: 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 = f(xα) j xα 2 Xαg: α As in the case of the Cartesian product of two topological spaces, we may choose ¨ « Y B = Uα j Uα 2 Tα : α It is easy to check that B satisfies (B1), (B2), so we get a topology Y Y TBox = fU ⊂ Xα j 8(xα) 2 U; 9Uα 2 Tα s.t. (xα) 2 Uα ⊂ Ug: 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 = f B j B ⊂ Bg: B2B0 0 Proof. As we remarked above, B ⊂ TB. So for any sub-family B ⊂ B, we have [ B 2 TB: B2B0 Conversely, for any U 2 STB and any x 2 U, by definition there exists Bx 2 B s.t. x 2 Bx ⊂ U. It follows U = x2U 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 2 Tα; 8α );;X 2 \αTα. • U1;U2 2 Tα; 8α ) U1 \ U2 2 Tα; 8α ) U1 \ U2 2 \αTα. • Uβ 2 Tα; 8α )[βUβ 2 Tα )[βUβ 2 \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 = fB j 9S1; ··· ;Sm 2 S s.t. B = S1 \···\ Smg: S (1) If S2S S = X, thenSB is a basis of TS , i.e. TS = TB. 0 0 (2) In general, if X = S2S ⊂ X, then B is a basis of a topology TB on X , and TS = fXg [ TB. S S Proof. (1) By definition, S ⊂ B. So the condition S2S = X implies B2B = 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), fXg [ 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 S2S 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 = f!(f; x1; ··· ; xn; ") j f 2 M([0; 1]; R); n 2 N; x1; ··· ; xn 2 [0; 1]; " > 0g is a basis, while S = f!(f; x; ") j f 2 M([0; 1]; R); x 2 [0; 1]; " > 0g is a sub-basis. Example 1.12. For the standard Euclidean topology on R, B = f(a; b) j a < bg is a basis, while S = f(−∞; a); (a; +1) j a 2 Rg is a sub-basis. Example 1.13. The the product topology on X × Y , B = fU × V j U 2 TX ;V 2 TY g is a basis, while [ S = fU × Y j U 2 TX g fX × V j V 2 TY g 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 2 T and any x 2 U, there exists B 2 B such that x 2 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 2 T and any x 2 U, there exists S1; ··· ;Sn 2 S such that x 2 n \i=1Si ⊂ U.
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