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SELECTED SOLUTIONS TO HOMEWORK 7 2. Let O be an open subset of a topological space X . Prove that a subset A of O is relatively open in O if and only if it is an open subset of X . Solution. By denition of subspace topology, we have A open in O ⇐⇒ A = O ∩ B for some open subset B in X. Now, if A is relatively open in O, then A = O ∩ B for some B open in X . Since O and B are both open in X , their interesection A is also open in X . Conversely, if A is open in X , then O ∩ A is open in O by denition. But O ∩ A = A because A ⊂ O, so A is open in O. 4. Prove that a subspace of a Hausdor space is a Hausdor space. Solution. Let X be a topological space which is Hausdor and Y ⊂ X a subspace. Let a, b ∈ Y with a 6= b. Since a, b ∈ X and X is Hausdor, there exist neighborhoods A of a and B of b in X such that A ∩ B = ∅. Now, by Theorem 6.3 A0 := A ∩ Y and B 0 := B ∩ Y are neighborhoods of a and b in Y , respectively. Since A0 ∩ B 0 ⊂ A ∩ B = ∅, we see that A0 ∩ B 0 = ∅ and so Y is Hausdor, as claimed. 2. Let X = α∈I Xα be given the product topology. Prove that a function f : Y → X is continuous if and only if fα = pα f is continuous for each α ∈ I . Solution. First assume that f is continuous. Since pα is continuous (from p. 99 on the textbook), by Theorem 5.6 we know that fα = pα f is continuous for each α ∈ I also. Conversely, assume that fα is continuous for each α ∈ I . By Theorem 5.3, f will be continuous if f −1 (O) is open in Y whenever O is open in X . By denition 7.6, every open O in X is a union of sets of the form Q −1 B = p−1 α1 (Oα1 ) ∩ · · · ∩ pαk (Oαk ) (∗), where Oαi is open in Xαi . Since f −1 preserves union, and an arbitrary union of open sets is again open, it suces to show that f −1 (B) is open for any set B of the form (∗). Now, since f −1 preserves intersection −1 −1 f −1 (B) = (f −1 p−1 pαk )(Oαk ) α1 )(Oα1 ) ∩ · · · ∩ (f = fα−1 (Oα1 ) ∩ · · · ∩ fα−1 (Oαk ), 1 k where each fα−1i (Oαi ) is open because fα is continuous for each α ∈ I by hypothesis. Since nite intersection of open sets is again open, we see that f −1 (B) is indeed open, whence f is continuous. 6. Prove that the family of open intervals with rational and points is a basis for the topology of R. 1 SELECTED SOLUTIONS TO HOMEWORK 7 2 Let O be an open set in R. By Denition 7.3, we need to show that O is a union of open intervals (a, b) with a, b ∈ Q. Now, since O is open, for each x ∈ O there exists δx > 0 such that Solution. (x − δx , x + δx ) ⊂ O. Since Q is dense in R, there exist ax , bx ∈ Q such that and x − δx < ax < x x < bx < x + δx . Then x ∈ (ax , bx ) ⊂ (x − δx , x + δx ) ⊂ O and so [ (ax , bx ) ⊂ O. x∈O It is clear that y ∈ S x∈O (ax , bx ) for each y ∈ O so in fact O= [ x∈O Thus, indeed {(a, b) | a, b ∈ Q} is a basis for R. (ax , bx ).