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MATH 671 : Topology Homework Solutions

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MATH 671 : Topology Homework Solutions Byeongho Ban [email protected] MATH 671 : Topology Homework Solutions :Taught by R. Inanc Baykur Lastly updated on May 10, 2019 Mathematics & Statistics University of Massachusetts, Amherst Byeong Ho Ban 1 MATH 671 Editor : Byeongho Ban Due date : September 18th, 2018 2.4 Let X be a topological space and let A be a collection of subsets of X. Prove the following containment. (d) ! [ [ Int A ⊇ IntA A2A A2A Proof. Suppose that [ x 2 IntA =) 9A 2 A such that x 2 IntA A2A =) 9U 3 x such that U is open and U ⊂ A: [ =) 9U 3 x such that U is open and U ⊂ A A2A ! [ =) x 2 Int A : A2A Therefore, ! [ [ Int A ⊇ IntA: A2A A2A Page 2 MATH 671 Editor : Byeongho Ban 2.1 Let X be an infinite set. (a) Show that T1 = fU ⊆ X : U = ; or X n U is finite.g is a topology on X, called the finite complement topology. (c) Let p be an arbitrary point in X, and show that T3 = fU ⊆ X : U = ; or p 2 Ug is a topology on X, called the particular point topology. (e) Determine whether T5 = fU ⊆ X : U = X or X n U is infiniteg is a topology on X. Proof. (a) Clearly ; 2 T1. Observe that X n X = ; is finite, so X 2 T1. Suppose that fUαgα2A ⊆ T1, where A is an arbitrary set. If Uα = ; for every α 2 A, [ Uα = ; 2 T1: α2A Suppose 9β 2 A such that Uβ 6= ;, then observe that ! [ \ X n Uα = (X n Uα) ⊂ X n Uβ : finite. α2A α2A Thus, [ Uα 2 T1 α2A and T1 is closed under arbitrary union. N Lastly, suppose fUngn=1 ⊂ T1. If X n Un is finite for all n = 1; 2;:::;N, observe that N ! N \ [ X n Un = (X n Un) : finite n=1 n=1 since finite union of finite sets is finite. Thus, N \ Un 2 T1: n=1 If 9j 2 f1; 2;:::;Ng such that Uj = ;, then N \ Un = ; 2 T1: n=1 Thus, T1 is closed under finite intersection. Therefore, T1 is a topology. Page 3 MATH 671 Editor : Byeongho Ban Proof. (c) Clearly ; 2 T3 and X 2 T3 since p 2 X. Let fUαgα2A ⊆ T3 where A is an arbitrary set be given. If Uα = ; 8α 2 A, [ Uα = ; 2 T3: α2A If 9β 2 A such that Uβ 6= ;, since Uβ 2 T3, we have p 2 Uβ. Thus, [ [ p 2 Uβ ⊆ Uα =) Uα 2 T3: α2A α2A Thus, T3 is closed under arbitrary union. N Let fUngn=1 ⊆ T3 be given. If Un 6= ; 8 n = 1; 2;:::;N, then p 2 Un 8n, thus N N \ \ p 2 Un =) Un 2 T3: n=1 n=1 If 9j 2 f1; 2;:::;Ng such that Uj = ;, N \ Un = ; 2 T3: n=1 Thus, T3 is closed under finite intersection. Therefore, T3 is a topology. (e) It is not a topology. Let y 2 X be given, and note that for each x 2 X, fxg 2 T5 since X n fxg is infinite. However, [ fxg = X n fyg 62 T5 x2Xnfyg since X n fyg 6= X and X n (X n fyg) = fyg is not infinite set. Thus, T5 is not closed under arbitrary union so T5 is not a topology. Page 4 MATH 671 Editor : Byeongho Ban 2.5 For each of the following properties, give an example consisting of two subsets X; Y ⊆ R2, both considered as topological spaces with their Euclidean topologies, together with a map f : X ! Y that has the indicated property. (a) f is open but neither closed nor continuous. (b) f is closed but neither open nor continuous. (c) f is continuous but neither open nor closed. (d) f is continuous and open but not closed. (e) f is continuous and closed but not open. (f) f is open and closed but not continuous. Proof. (a) Consider fa : A = B2(0; 0) n @B1(0; 0) ! B = B4(0; 0) defined by ( (x; y) if (x; y) 2 B1(0; 0); fa(x; y) = (2x; 2y) if (x; y) 2 A n B1(0; 0): Clearly, fa is not continuous. Note that fa(B 3 (0; 0)) = B1(0; 0) [ B3(0; 0) n B2(0; 0) is not closed. Thus, fa is not closed. 2 Observe that, for any U ⊂ A open, f(U) = (U \ B1(0; 0)) [ (2U \ (2A n B2(0; 0)) is open. So, fa is open. (b) 2 2 Consider fb(x; y): R ! R defined by ( (0; 0) x ≥ 0; fb(x; y) = (1; 0) x < 0: Clearly, fb is not continuous. Note that only possible images of fb are f(0; 0); (1; 0)g, f(0; 0)g, f(1; 0)g and ; which are closed. 2 2 2 Thus, fb is closed. However, since fb(R ) = f(0; 0); (1; 0)g is not open in R even if R is open, fb is not open. (c) 2 Consider the function fc : R ! R × f0g defined by 1 f (x; y) = ; 0 : c 1 + x2 + y2 2 2 Clearly, fc is continuous. However, note that fc(R ) = (0; 1] × f0g is neither open nor closed even if R is open and closed set. Thus, fc is neither open nor closed. (d) Consider X = B1(0; 0) and Y = B2(0; 0) and let fd : X,! Y be an inclusion map. Then clearly, fd is continuous open map. However, it is not closed since fd(X) = B1(0; 0) which is not closed even if X is closed and open. (e) 2 2 2 Consider fe : R ! R defined by fe(x; y) = (a; b) where (a; b) is a constant point in R . Then, clearly fe is continuous. 2 2 Also, since any possible image of fe is f(a; b)g or ;, it is closed map. However, since fe(R ) = f(a; b)g is not open even if R is open, fe is not open. (f) 2 Consider ff : R ! f(0; 0); (1; 0)g = A defined by ff = fb. Clearly, ff is not continuous. Note that any images of ff are open and closed in A at the sametime. Thus, ff is open and closed. Page 5 MATH 671 Editor : Byeongho Ban 2.6 Prove Proposition 2:30 (characterization of continuity, openness, and closedness in terms of closure and interior.) Proposition 2.30 Suppose X and Y are topological spaces, and f : X ! Y is any map. (a) f is continuous if and only if f(A) ⊆ f(A) for all A ⊆ X. (b) f is closed if and only if f(A) ⊇ f(A) for all A ⊆ X. (c) f is continuous if and only if f −1(B˚) ⊆ f −1˚(B) for all B ⊆ Y . (d) f is open if and only if f −1(B˚) ⊇ f −1˚(B) for all B ⊆ Y . Proof. (a) Suppose that f is continuous. Let A ⊆ X be given. Then, f −1 f(A) is closed. Also, observe that A ⊆ f −1 f(A) , so A ⊆ f −1 f(A) since f −1 f(A) is closed. Thus, observe that f(A) ⊆ f f −1 f(A) = f(A): Conversely, suppose that f(A) ⊆ f(A) 8A ⊆ X. And let a closed subset C ⊆ Y be given. Now, observe that f f −1(C) ⊆ f (f −1(C)) = C = C and that f −1(C) ⊆ f −1(C) ⊆ f −1 f f −1(C) = f −1(C): Therefore, f −1(C) is closed for any closed subset C ⊆ Y and so f is continuous. (b) Suppose that f is closed and let A ⊆ X be given. Observe that A ⊆ A, so f(A) ⊆ f(A): Note that f is closed map so f(A) is closed. Thus, f(A) ⊆ f(A): Conversely, suppose that f(A) ⊇ f(A) 8A ⊆ X and let a closed subset C ⊆ X be given. Now, note that f(C) ⊆ f(C) = f(C) ⊆ f(C): Thus, f(C) = f(C) and f(C) is closed. Therefore, f is closed map. Page 6 MATH 671 Editor : Byeongho Ban Proof. (c) Suppose that f is continuous and let a subset B ⊆ X be given. Since B˚ ⊆ B, note that f −1(B˚) ⊆ f −1(B): Since f −1˚(B) is a union of all open subsets of f −1(B) and f −1(B˚) is open, f −1(B˚) ⊆ f −1˚(B): Conversely, suppose that f −1(B˚) ⊆ f −1˚(B) 8B ⊆ X. Let an open subset U ⊆ X be given. And observe that f −1(U) = f −1(U˚) ⊆ f −1˚(U) ⊂ f −1(U): Therefore, f −1(U) = f −1˚(U), so f −1(U) is open, It implies that f is continuous. (d) Suppose that f is open and let B ⊆ Y be given. Then note that f(f −1˚(B)) is open and f(f −1˚(B)) ⊆ f(f −1(B)) = B, so f(f −1˚(B)) ⊆ B:˚ Since f is open, observe that f −1˚(B) ⊆ f −1(f(f −1˚(B))) ⊆ f −1(B˚): Conversely, suppose that f −1˚(B) ⊆ f −1(B˚) 8B ⊆ Y and let an open subset U ⊆ X be given. Now observe that U = U˚ ⊆ f −1(˚f(U)) ⊆ f −1(f(˚U)): And it implies that f(˚U) ⊆ f(U) ⊆ f(f −1(f(˚U))) = f(˚U): Thus, f(U) = f(˚U) and it means that f(U) is open. Therefore, f is open. 2.8 Let X be a Hausdorff space, let A ⊆ X and let A0 denote the set of limit points of A.
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