STEVENS MA 651 - Lecture 4 Topological Spaces 2

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MA651 Topology. Lecture 4. Topological spaces 2This text is based on the following books:• ”Linear Algebra and Analysis” by Marc Zamansky• ”Topology” by James Dugundgji• ”Fundamental concepts of topology” by Peter O’Neil• ”Elements of Mathematics: General Topology” by Nicolas BourbakiI have intentionally made several mistakes in this text. The first homework assignment is to findthem.25 Continuous MapsWe have been considering topologies on one given set; we now want to relate different topologicalspaces. Given (X, TX) and (Y, TY), note that the a map f : X → Y relates the sets and alsoinduces two mapsˆf : P (X) → P (Y ),ˆf−1: P (Y ) → P(X). Of these,ˆf−1should be used torelate the topologies, since it is the only one that preserves the Boolean operations involved in thedefinition of a topology. Thus the suitable maps f : X → Y are those for which simultaneouslyˆf−1: TY→ P(X). Formally stated,Definition 25.1. Let (X, TX) and (Y, TY) be topological spaces. A map f : X → Y is calledcontinuous if the inverse image of each set open in Y is open in X (that isˆf−1maps TYinto TX).Example 25.1. A constant map f : X → Y is always continuous: The inverse image of any setU open in Y is either Ø or X, which are open.Example 25.2. Let X be any set, T1, T2two topologies on X. The bijective map 1 : (X, T1) →(X, T2) is continuous if and only if T2⊂ T1. Note that a continuous map need not send open setsto open sets, and also that increasing the topology T1preserves continuity.1Example 25.3. A map sending open sets to sets is called an open map. An open map need not becontinuous. 1 : (X, T1) → (X, T2) is open if and only if T1⊂ T2, but is not continuous wheneverT16= T2.Example 25.4. Let Y ⊂ X. The relative topology TYcan be characterized as the smallest topologyon Y for which the inclusion map i : Y → X is continuous. For, if U ∈ T , the continuity ofi requires i−1(U) = U ∩ Y to be open in Y , so that any topology for which i is continuous mustcontain TY.The elementary properties are:Proposition 25.1.1. (Composition) If f : X → Y and g : Y → Z are continuous, so also, is g ◦ f : X → Z.2. (Restriction of domain) If f : X → Y is continuous and A ⊂ X is taken with a subspacetopology, then f |A : A → Y is continuous.3. (Restriction of range) If f : X → Y is continuous and f (X) is taken with the subspacetopology, then f : X → f(X) is continuousProof is left as a homework.The basic theorem on continuity is:Theorem 25.1. Let X, Y be topological spaces, and f : X → Y a map. The following statementsare equivalent:1. f is continuous.2. The inverse image of each closed set in Y is closed in X.3. The inverse image of each member of a subbasis (basis) for Y is open in X (not necessarilya member of a subbasis, or basis for X!).4. For each x ∈ X and each neighborhood W (f(x)) in Y , there exists a neighborhood V (x) inX such that f(V (x)) ⊂ W (f(x )).5. f(A) ⊂ f(A) for every A ⊂ X.6. f−1(B) ⊂ f−1(B) for every B ⊂ Y .Proof.• (1) ⇔ (2), since f−1(Y − E) = X − f−1(E) for any E ⊂ X.2• (1) ⇔ (3). Let {Uα| α ∈ A } be a subbasis for Y . If f is continuous, each f−1(Uα) is open.Converse ly, if each f−1(Uα) is open, then because any open U ⊂ Y can be writtenU =[{Uα1∩ · · · ∩ Uαn| {α1, · · · , αn} ⊂ A },we have thatf−1(U) =[{f−1(Uα1) ∩ · · · ∩ f−1(Uαn)}is a union of open sets and so is open.• (1) ⇔ (4). Since f−1(W (x)) is open, we can use it for V (x).• (4) ⇔ (5). Let A ⊂ X and b ∈ A; we show f(b) ∈ f (A) by proving each W (f(b)) intersectsf(A). For, finding V (b) with f(V (b)) ⊂ W (f(b)),b ∈ A ⇒ Ø 6= V (b) ∩ A⇒ Ø 6= f(V (b) ∩ A) ⊂ f(V (b)) ∩ f (A) ⊂ W (f(b)) ∩ f(A).• (5) ⇔ (6). Let A = f−1(B); then f(A) ⊂ f(A) = f[f−1(B)] = B ∩ f(X) ⊂ B, so thatA ⊂ f−1B, as required.• (6) ⇔ (2). Let B ⊂ Y be closed; then f−1(B) ⊂ f−1(B), and since always f−1(B) ⊂ f−1(B)(by Proposition 23.2 (a): for every set A: A ⊂ A), this shows that f−1(B) is closed (byProposition 23.1).The formulation (4) of Theorem (25.1) shows that continuity is a ”local” matter, a fact havingmany applications. Precisely,Definition 25.2. An f : X → Y is continuous at x0∈ X if for each neighborhood W (f(x0)) inY , there exists a neighborhood V (x0) in X such that f(V (x0)) ⊂ W (f(x0)) (i.e. Theorem (25.1)(4) is satisfied at x0).From this viewpoint, the equivalence of (1) and (4) in Theorem (25.1) asserts: f is continuousaccording to Definition (25.1), if and only if it is continuous at each point of X.26 Ope n Maps and Closed MapsDefinition 26.1. A map f : X → Y is called open (closed) if the image of each set open (closed)in X is open (closed) in Y .3We have already seen (Example (25.1) that a continuous map need not be an open map, and(Example (25.3)) that an open map need not be continuous. The following example shows that,in general, an open map need not be a closed map (even though it is continuous); the concepts”open map”, ”closed map”, and ”continuous map” are therefore independent.Example 26.1. Let A ⊂ X and let i : A → X be the inclusion map a → a. By Example (25.4)i is continuous. Furthermore, i is open (closed) if and only if A is open (closed) in X. P roof for”open”: If A is open, and U ⊂ A open in A, then by Theorem (24.2) (Let Y be a subspace of X.If A ⊂ Y is closed (open) in Y , and Y is closed (open) in X, then A is closed (open) in X.),i(U) = U is open in X. The proof for ”closed” is analogues.Example 26.2. If f : X → Y is bijective, then the conditions ”f closed” and ”f open are in factequivalent. For, if f is open and A ⊂ X is closed, then A = X − U and f(A) = f(X) − f(U) =Y − f(U), so f(A) is also closed. As Examples (25.2) and (25.3) show, ”bijective open” and”bijective continuous” are still distinct notions.The behavior of inverse images further emphasizes the distinction between open maps and closedmaps:Theorem 26.1.1. Let p : X → Y be a closed map. Given any subset S ⊂ Y and any open U containing p−1(S),there exists an open V ⊃ S such that p−1(V ) ⊂ U.2. Let p : X → Y be an open map. Given any subset S ⊂ Y , and any closed A containingp−1(S), …


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