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Point Set Topology Point Set Topology A. Topological Spaces and Continuous Maps Definition 1.1 A topology on a set X is a collection T of subsets of X satisfying the following axioms: T 1. ∅;X ∈T. T2. {Oα | α ∈ I}⊆T =⇒ α∈IOα ∈T. T3. O; O0 ∈T =⇒ O∩O0 ∈TS. A topological space is a pair (X; T )whereXis a set and T is a topology on X. When the topology T on X under discussion is clear, we simply denote (X; T )by X.Let(X; T ) be a topological space. Members of T are called open sets. (T 3) implies that a finite intersection of open sets is open. Examples 1. Let X be a set. The power set 2X of X is a topology on X and is called the discrete topology on X. The collection I = {∅;X} is also a topology on X and is called the indiscrete topology on X. 2. Let (X; d) be a metric space. Define O ⊆ X to be open if for any x ∈ O, there exists an open ball B(x; r) lying inside O. Then, Td = {O ⊆ X | O is open} ∪ {∅} is a topology on X. Td is called the topology induced by the metric d. 3. Since Rn is a metric space with the usual metric: n d((x ; :::; x ); (y ; :::; y )) = (x − y )2; 1 n 1 n v i i ui=1 uX t Rn has a topology U induced by d. This topology on Rn is called the usual topology. 4. Let X be an infinite set. Then, T = {∅;X}∪{O⊆X | X\O is finite} is a topology on X. T is called the complement finite topology on X. 5. Define O ⊆ R to be open if for any x ∈ O, there exists δ>0 such that [x; x + δ) ⊆ O. Then, the collection T 0 of all such open sets is a topology on R. T 0 is called the lower limit topology on R.NotethatU⊂ T0 but U6=T0. We shall denote this topological space by Rl. 1 Proposition 1.2 Let (X; T ) be a topological space and A a subset of X. Then, the collection TA = {O ∩A | O ∈T}is a topology on A. TA is called the subspace topology on A and (A; TA) is called a subspace of (X; T ). Proof Exercise. Let (X; T ) be a topological space. Definition 1.3 E ⊆ X is said to be closed if X \ E is open. (i.e. X \ E ∈T.) Proposition 1.4 (1) ∅ and X are closed in X. (2) If E;F are closed in X,thenE∪F is closed in X. (3) If {Eα | α ∈ I} is a collection of closed sets in X,then α∈IEα is closed in X. Proof Exercise. T Definition 1.5 A basis for the topology T on X is a subcollection B of T such that any open set in X is a union of members of B. Definition 1.6 A subbasis for the topology T on X is subcollection S of T such that the collection of all finite intersections of members of S is a basis for T . Hence, any open subset of X is a union of finite intersections of members in S. Examples 1. {(a; b) | a<b}is a basis for the usual topology on R. {(a; ∞); (−∞;b) | a; b ∈ R} is a subbasis for the usual topology on R. 2. Let (X; d)beametricspaceandTd the topology on X induced by d.The collection B of all open balls of X is a basis for Td. 3. {[a; b) | a<b}is a basis for the lower limit topology on Rl. What would be a subbasis for the lower limit topology on Rl? 4. If (X; T ) is a topological space, then T itself is a basis for T . Proposition 1.7 Let X be a set and B a collection of subsets of X such that B1. U∈B U = X, B2. forS any U1;U2 ∈Band x ∈ U1 ∩ U2, there exists U ∈Bsuch that x ∈ U ⊆ U1 ∩ U2. Then the collection TB of all unions of members of B is a topology on X and B is a basis for TB. Proof T 1andT2 are clearly satisfied. B2 implies that if Uα;Uβ ∈B,then Uα∩Uβ∈TB.LetO1=∪Uαand let O2 = ∪Uβ be in TB,whereUα;Uβ ∈B.Then 2 O1 ∩ O2 = (Uα ∩ Uβ)isinT.ThatBis a basis for TB follows from the definition of TB. S Proposition 1.8 Let X be a set and S be a collection of subsets of X such that X = V ∈S V . Then the collection TS of all unions of finite intersections of members ofSS is a topology on X and S is a subbasis for TS . Proof Similar to 1.7. Definition 1.9 A subset N containing a point x in X is called a neighbourhood of x if there exists an open set O in X such that x ∈ O ⊆ N. Definition 1.10 Let A be a subset of X. (1) x ∈ A is called an interior point of A if there exists a neighbourhood U of x inside A. (2) x ∈ X is called a limit point of A if (O \{x})∩A6=∅, for any neighbourhood O of x. o (3) A = {x ∈ A | x is an interior point of A} is called the interior of A. (4) A = A ∪{x∈X | xis a limit point of A} is called the closure of A. Proposition 1.11 o o (1) A is an open set and A ⊆ A. o o (2) X \ A =(X\A) and X \ A = X \ A. (3) A is a closed set and A ⊇ A. o (4) A is open ⇐⇒ A = A . (5) A is closed ⇐⇒ A = A . (6) Ao is the largest open set contained in A. (7) A is the smallest closed set containing A. o (8) A = {O | O is open and O ⊆ A}. (9) A = S{E | E is closed and E ⊇ A}. (10) Aoo T= Ao and A = A. o o o (11) (A ∩ B) = A ∩ B . (12) A ∪ B = A ∪ B. (13) Let B ⊆ A ⊆ X.ThenBis closed in A if and only if B = E ∩ A for some closed set E in X. A A (14) Let B ⊆ A ⊆ X and let B be the closure of B in A.ThenB =B∩A. o A (15) Let B ⊆ A ⊆ X and let B be the interior of B in A.IfAis open, then Bo A = Bo . 3 Proof Exercise. Definition 1.12 Let (X; T )and(Y;T 0) be topological spaces and f : X −→ Y be amap.Letx∈X.fis said to be continuous at x if for any open neighbourhood U of f(x), there exists an open neighbourhood O of x such that f[O] ⊆ U. f is said to be continuous on X if f is continuous at each point of X. Proposition 1.13 Let (X; T )and(Y;T 0) be topological spaces and f : X −→ Y be a map. The following statements are equivalent. (1) f is continuous on X. (2) For any open set U ⊆ Y , f −1[U]isopeninX. (3) For any closed set E ⊆ Y , f −1[E]isclosedinX. (4) For any A ⊆ X, f[A] ⊆ f[A]. Proof Exercise. Proposition 1.14 Let f : X −→ Y and f : Y −→ Z be maps. If f and g are continuous, then f ◦ g is continuous. Proof This follows easily by 1.13(2). Definition 1.15 Let (X; T )and(Y;T 0) be topological spaces and f : X −→ Y a map. (1) f is called an open map if for any open set U in X, f[U]isopeninY. (2) f is called a closed map if for any closed set E in X, f[E]isclosedinY. (3) f is called a homeomorphism if f is bijective, continuous and f −1 is continuous. Proposition 1.16 Let f : X −→ Y be a bijective continuous mapping. The following statements are equivalent. (1) f −1 : Y −→ X is a homeomorphism. (2) f : X −→ Y is an open map. (3) f : X −→ Y is a closed map. Proof Exercise. Proposition 1.17 (The Combination Principle) Let A1;:::;An be a finite collec- n tion of closed subsets of X such that i=1 Ai = X.Thenf:X→Yis continuous if and only if f | Ai is continuous forS each i =1;:::;n. Proof The ‘only if’ part is obvious. Suppose f | Ai is continuous for each −1 n −1 i =1;:::;n.LetEbe a closed subset of Y .Thenf [E]= i=1(f [E] ∩ Ai)= n −1 −1 i=1(f | Ai) [E]. Since f | Ai is continuous, (f | Ai) [E]isclosedinS Ai.AsAi S 4 −1 −1 is closed in X,(f|Ai) [E]isclosedinX. Hence f [E]isclosedinX.This shows that f is continuous on X. B. Product and Sum Topologies Definition 1.18 Let {Xλ}λ∈I be a collection of topological spaces and let pα : λ∈I Xλ −→ X α be the natural projection. QThe product topology on λ∈I Xλ is defined to be the one generated by the subbasis −1 {pα [Oα] | Oα is open inQXα,α∈I}. Note that when λ∈I Xλ is given the product topology the projection pα is con- tinuous and open.Q In the case that the product is finite the collection { α∈I Oα | Oα is open in Xα} is a basis for the product topology on λ∈I Xλ.
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