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Math 535 - General Topology Fall 2012 Homework 2 Solutions Problem 1. (Brown Exercise 2.4.5) Consider X = [0, 2] \ {1} as a subspace of the real line R. Show that the subset [0, 1) ⊂ X is both open and closed in X. Solution. [0, 1) is open in X because we can write [0, 1) = (−8, 1) ∩ X and (−8, 1) is open in R. On the other hand, [0, 1) is closed in X because we can write [0, 1) = [0, 1] ∩ X and [0, 1] is closed in R. 1 Problem 2. (Bredon Exercise I.3.8) Let X be a topological space that can be written as a union X = A ∪ B where A and B are closed subsets of X. Let f : X → Y be a function, where Y is any topological space. Assume that the restrictions of f to A and to B are both continuous. Show that f is continuous. Solution. Lemma. Let A ⊆ X be a closed subset. If C ⊆ A is closed in A, then C is also closed in X. e ∩ A for some closed subset C e ⊆ X. Proof. Since C is closed in A, it can be written as C = C Therefore C is an intersection of closed subsets of X, and thus is closed in X. Let C ⊂ Y be a closed subset. Its preimage under f is the union f −1 (C) = f −1 (C) ∩ A ∪ f −1 (C) ∩ B = (f |A )−1 (C) ∪ (f |B )−1 (C). Since the restriction f |A : A → Y is continuous, (f |A )−1 (C) is closed in A, and thus closed in X by the lemma. Likewise, (f |B )−1 (C) is closed in X. Therefore their union f −1 (C) = (f |A )−1 (C) ∪ (f |B )−1 (C). is closed in X, so that f in continuous. Remark. The same proof shows that the statement still holds if A and B are both open in X. 2 Problem 3. A map between topological spaces f : X → Y is called an open map if for every open subset U ⊆ X, its image f (U ) ⊆ Y is open in Y . a. (Munkres Exercise 2.16.4) Let X and Y be topological spaces. Show that the projection maps pX : X × Y → X and pY : X × Y → Y are open maps. Solution. Lemma. A map f : X → Y is open if and only if f (B) ⊆ Y is open in Y for every B ∈ B belonging to some basis B of the topology on X. Proof. (⇒) Each member B ∈ B is open in X. (⇐) Let U ⊆ X be open in X. Then U is a union U = Its image under f is ! [ f (U ) = f Bα S α Bα of basic open subsets Bα ∈ B. α = [ f (Bα ) α where each f (Bα ) is open in Y by assumption. Thus f (U ) is a union of open subsets and hence open. Take an “open box” U × V ⊆ X × Y , where U ⊆ X is open and V ⊆ Y is open. Its projection onto the first factor is pX (U × V ) = U ⊆ X which is open in X. Since open boxes form a basis of the topology on X × Y , the lemma guarantees that pX is an open map, and likewise for pY . 3 b. Find an example of metric spaces X and Y , and a closed subset C ⊆ X × Y such that the projection pX (C) ⊆ X is not closed in X. In other words, the projection maps are (usually) not closed maps. Solution. Take X = Y = R and consider the hyperbola in R × R 1 C = {(x, ) | x 6= 0} = {(x, y) ∈ R × R | xy = 1}. x Its projection onto the first factor is pX (C) = R \ {0} which is not closed in R. To show that C is closed in R×R, note that the function f : R×R → R defined by f (x, y) = xy is continuous, and C is the preimage C = f −1 ({1}). Since the singleton {1} is closed in R, C is closed in R × R. 4 Problem 4. (Munkres Exercise 2.19.7) Consider the set of sequences of real numbers Y RN = {(x1 , x2 , . . .) | xn ∈ R for all n ∈ N} ∼ R = n∈N and consider the subset of sequences that are “eventually zero” R∞ := {x ∈ RN | xn 6= 0 for at most finitely many n}. a. In the box topology on RN , is R∞ a closed subset? Solution. Yes, R∞ is closed in the box topology. Let x ∈ RN \ R∞ , which means that the sequence x has infinitely many non-zero entries xn 6= 0. For all those indices n, pick an open neighborhood UQ n of xn ∈ R which does not contain 0. For other values of n, take Un = R. Then the open box n Un is an open neighborhood of x which does not intersect R∞ . Q Indeed, for any y ∈ n Un and every index n such that xn 6= 0, we have yn ∈ Un so that yn 6= 0 by construction. Because there are infinitely many such indices, we conclude y ∈ / R∞ . b. In the product topology on RN , is R∞ a closed subset? Solution. No, R∞ is not closed in the product topology. Q Let x ∈ RN \ R∞ and consider any open neighborhood U = n Un of x which is a “large box”, i.e. Un ⊆ R is open for all n and Un = R except for finitely many n. In particular, there is a number N such that Un = R for all n ≥ N . Consider a sequence y with yn = 0 for all n ≥ N and yn ∈ Un for 1 ≤ n < N . Then we have y ∈ U ∩ R∞ . Because “large boxes” form a basis of the product topology, every open neighborhood of x intersects R∞ . Therefore R∞ is not closed. Remark. In fact, the argument shows that x is not an interior point of RN \ R∞ , so that the interior of RN \ R∞ is empty. Equivalently, the closure of R∞ is all of RN , i.e. R∞ is dense in RN . 5 Problem 5. Let X be a topological space, S a set, and f : X → S a function. Consider the collection of subsets of S T := {U ⊆ S | f −1 (U ) is open in X}. a. Show that T is a topology on S. Solution. 1. The preimage f −1 (S) = X is open in X, so that the entire set S is in T . Likewise, f −1 (∅) = ∅ is open in X, so that the empty set ∅ is in T . 2. Let Uα be a family of members of T . Then we have ! [ [ f −1 Uα = f −1 (Uα ) α α S where each f −1 S (Uα ) is open in X by assumption. Thus f −1 ( α Uα ) is also open in X, so that the union α Uα is in T . 3. Let U and U 0 be members of T . Then we have f −1 (U ∩ U 0 ) = f −1 (U ) ∩ f −1 (U 0 ) where f −1 (U ) and f −1 (U 0 ) are open in X by assumption. Thus f −1 (U ∩ U 0 ) is also open in X, so that the finite intersection U ∩ U 0 is in T . b. Show that T is the largest topology on S making f continuous. Solution. Note that T makes f continuous by construction: for all U ∈ T , the preimage f −1 (U ) ⊆ X is open in X. Let T 0 be a topology on S making f continuous. Then for every U ∈ T 0 , the preimage f −1 (U ) is open in X, which means U ∈ T . This proves T 0 ≤ T . c. Let Y be a topological space. Show that a map g : S → Y is continuous if and only if the composite g ◦ f : X → Y is continuous. Solution. (⇒) The maps f and g are continuous, hence so is their composite g ◦ f . (⇐) Assume g ◦ f is continuous; we want to show that g is continuous. Let U ⊆ Y be open and take its preimage g −1 (U ) ⊆ S. To check that this subset is open, consider its preimage f −1 g −1 (U ) = (g ◦ f )−1 (U ) ⊆ X which is open in X since g ◦ f is continuous. By definition of T , g −1 (U ) is indeed open in S. 6 d. Show that T is the smallest topology on S with the property that a map g : S → Y is continuous whenever g ◦ f is continuous. Solution. Let T 0 be a topology on S with said property. We know that f : X → (S, T ) is continuous, but it can be written as the composite f id X→ − (S, T 0 ) − → (S, T ). By the property of T 0 , the composite id ◦ f being continuous guarantees that the identity id : (S, T 0 ) → (S, T ) is continuous, i.e. T ≤ T 0 . 7 Problem 6. Consider the subset X = {0} ∪ { n1 | n ∈ N} ⊂ R viewed as a subspace of the real line R. As a set, X is the disjoint union of the singletons {0} ` and { n1 } for all n ∈ N. However, show that X does not have the coproduct topology on {0} q n∈N { n1 }. Solution. In the coproduct topology on {0} q topology), the summand {0} is open. ` 1 n∈N { n } (which happens to be the discrete However, in the subspace topology on X, the singleton {0} is not open. Indeed, any open ball Br (0) around 0 will contain other points n1 ∈ Br (0), for all n such that n1 < r. 8