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Solutions to Additional Exercises Exercise (3.4) Prove the following: Let X be a topological space. (1) X and ∅ are closed sets. (2) The intersection of any collection of closed sets in X is closed. (3) The union of any finite collection of closed sets in X is closed. Proof: We shall apply Theorem (3.5) on page 39. Theorem (3.5) Let X be a topological space. (1) X and ∅ are open sets. (2) The union of any collection of open sets in X is open. (3) The intersection of any finite collection of open sets in X is open. We also need Definition (3.7): A subset C in X is closed if X − C is open. Now we are ready to prove Exercise (3.4). Since by Theorem 3.5(1), ∅ is open, and since ∅ = X − X, by Definition (3.7), X is closed. Similarly, since by Theorem 3.5(1), X is open, and since X = X − ∅, by Definition (3.7), ∅ is closed. This proves Exercise (3.4)(1). To prove Exercise (3.4)(2), let C1 , C2 , · · · be a collection of closed sets in X. By Definition (3.7), X − C1 , X − C2 , · · · is a collection of open sets in X. By DeMorgan Law, [ \ X − Ci = (X − Ci ). i i Since each X − Ci is open and since, by Theorem (3.5)(2), the union of any collection T S of open sets in X is open, X − i Ci = i (X − Ci ) is open in X. Now it follows by T Definition (3.7) that i Ci is closed in X. This proves Exercise (3.4)(2). To prove Exercise (3.4)(3), let C1 , C2 , · · · , Cm be a finite collection of closed sets in X. By Definition (3.7), X − C1 , X − C2 , · · · , Cm is a finite collection of open sets in X. By DeMorgan Law, X− m [ i=1 m \ Ci = (X − Ci ). i=1 Since each X −Ci is open and since, by Theorem (3.5)(3), the intersection of any finite S Tm collection of open sets in X is open, X − m i=1 Ci = i=1 (X − Ci ) is open in X. Now S it follows by Definition (3.7) that m i=1 Ci is closed in X. This proves Exercise (3.4)(3). 1 Exercise (3.10) Let X be a topological space with closed subsets A and B such that X = A ∪ B. Let f : A 7→ Y and g : B 7→ Y e continuous functions such that for each x ∈ A ∩ B, f (x) = g(x). Define a new function F = f ∪ g : X 7→ Y by f (x) F (x) = (f ∪ g)(x) = g(x) for x ∈ A . for x ∈ B (1) Prove that F is continuous. (2) Give an example to show that the condition that A and B must be closed is necessary. Proof: (1) We shall first show that, for any set U ⊆ Y , F −1 (U ) = f −1 (U ) ∪ g −1 (U ). Firstly, let x ∈ F −1 (U ) ⊆ X. Since X = A ∪ B, either x ∈ A or x ∈ B. If x ∈ A, then f (x) = F (x) ∈ U , and so by the definition of f −1 , x ∈ f −1 (U ). It follows that x ∈ f −1 (U ) ∪ g −1 (U ). If x ∈ B, then g(x) = F (x) ∈ U , and so by the definition of g −1 , x ∈ g −1 (U ). It also follows that x ∈ f −1 (U ) ∪ g −1 (U ). Hence we have F −1 (U ) ⊆ f −1 (U ) ∪ g −1 (U ). Nest, we pick an arbitrary element x0 ∈ f −1 (U )∪g −1 (U ). Then either x0 ∈ f −1 (U ) or x0 ∈ g −1 (U ). If x0 ∈ f −1 (U ), then by the definition of F , F (x0 ) = f (x0 ) ∈ U , and so by the definition of F −1 , x0 ∈ F −1 (U ). Similarly, if x0 ∈ g −1 (U ), then by the definition of F , F (x0 ) = g(x0 ) ∈ U , and so by the definition of F −1 , x0 ∈ F −1 (U ) also. This proves that F −1 (U ) ⊇ f −1 (U ) ∪ g −1 (U ), and so we have established the equality F −1 (U ) = f −1 (U ) ∪ g −1 (U ). Now we are ready to complete the proof of Exercise (3.10). Since we are given the condition that both A and B are closed in X, we will make use of these two known results Exercise (3.9) Let X and Y be topological spaces. A function f : X 7→ Y is continuous if and only if for each closed set C ⊆ Y , f −1 (C) is closed in X. Theorem (3.12) Let A be a subset of a topological space X. A subset B ⊆ A is closed in A if and only if B = A ∩ C for some set C ⊆ X closed in X. Let C be a closed set in Y . Since f and g are continuous functions, by Exercise (3.9), f −1 (C) is closed in A and g −1 (C) is closed in B. Since A is closed and since f −1 (C) is closed in A, it follows by Theorem (3.12) that for some closed set D in X, f −1 (C) = D ∩ A. Then by Exercise (3.4)(2) above that f −1 (C) = D ∩ A is 2 closed in X. Similarly, since B is closed and since g −1 (C) is closed in B, it follows by Theorem (3.12) and Exercise (3.4)(2) that g −1 (C) is closed in X. By Exercise (3.4)(2), and by the equality we just established, F −1 (C) = f −1 (C) ∪ g −1 (C) is also closed in X. Therefore, by Exercise (3.9), F is continuous. (2) Example: Let X = [0, 2] with A = [0, 1) and B = [1, 2], and let Y = [0, 3]. Define f : A 7→ Y by f (x) = x and g : B 7→ Y by g(x) = 3. Then both f and g are continuous. As A ∩ B = ∅, it satisfies the condition that for each x ∈ A ∩ B, f (x) = g(x). But it is well known that from Calculus that f ∪ g has a discontinuity at x = 1. Exercise (3.31) Show that the topology defined on Y in Definition (3.33) satisfies the conclusions of Theorem 3.5. Proof: We first establish the two equalities. Let f : X 7→ Y be a function. For any collections of subsets Ui in Y , ! f −1 \ ! Ui = i \ f −1 (Ui ), f −1 i −1 [ i Ui = [ f −1 (Ui ). i Pick an arbitrary x ∈ f ( i Ui ). By the definition of f −1 , f (x) ∈ i Ui . By the definition of intersection, f (x) ∈ Ui , for any i. By the definition of f −1 , x ∈ f −1 (Ui ) T for any i. By the definition of intersection, x ∈ i f −1 (Ui ). The other equality can be proved similarly. We are now ready to do Exercise (3.31). Let T 0 denote the topology defined on Y by Definition (3.33). Firstly, since f : X 7→ Y is a function, Y = f −1 (X). By Theorem (3.5)(1), and since X is a topological space with topology T , X ∈ T . By Definition (3.33), Y ∈ T 0 as X = f −1 (Y ) ∈ T . Similarly, ∅ ∈ T 0 as ∅ = f −1 (∅) ∈ T . This shows that T 0 satisfies Theorem (3.5)(1). Next, let Ui be a collection of open sets in T 0 . By Definition (3.33), each f −1 (Ui ) S is in T . By Theorem (3.5)(2), i f −1 (Ui ) ∈ T . It now follows by the second equality S S S that f −1 ( i Ui ) = i f −1 (Ui ) ∈ T . By Definition (3.33), and since f −1 ( i Ui ) ∈ T , S 0 0 i Ui ∈ T . This proves that T satisfies Theorem (3.5)(2). Finally, let U1 , U2 , · · · , Um be a finite collection of sets in T 0 . By Definition T −1 (3.33), each f −1 (Ui ) is in T . By Theorem (3.5)(3), m (Ui ) ∈ T . By the i=1 f T T 3 m −1 (Ui ). By Definition (3.33), and since first equality that f −1 ( m i=1 f i=1 Ui ) = Tm 0 −1 Tm f ( i=1 Ui ) ∈ T , i=1 Ui ∈ T . This proves that T 0 satisfies Theorem (3.5)(3). T T Exercise (3.36) Show that if X is compact and ∼ is an equivalence relation on X, then X/ ∼ is compact. Proof: Define a function from X 7→ X/ ∼ by f (x) = [x]. Then by the definition of an onto function, f is onto. According to Definition (3.33), a subset U ⊂ X/ ∼ if open iff f −1 (U ) is open in X. By the definition of continuous functions, f is a continuous function from X onto Y . Now we apply Theorem (3.19) to finish the proof. Theorem (3.19) Let X be a compact topological space and f : X 7→ Y a continuous function from X onto a topological space Y . Then Y is compact. Since f : X 7→ X/ ∼ is a continuous function from X onto X/ ∼. By Theorem (3.19), X/ ∼ is compact. 4