MTG 6316 HOMEWORK Spring 2017 153. Let U n be a finite open cover of X and f : U Y be continuous for each { k}k=1 k k ! k =1,...,n. Show that if f (x)=f (x) for all x U U , then the function k j 2 k \ j F : X Y defined by ! F (x)=f (x) for all x U k 2 k is continuous. Proof. First of all, F is well defined on X. Now for any fixed x X,thereexists 2 some U ,1 K n, such that x U . Note that f is continuous over K 6 6 2 K K U . Then for any neighborhood V Y of y = F (x)=f (x), there is some K ⇢ K neighborhood W U of x such that f (W ) V . Note that W is open in U ⇢ K K ⇢ K and that UK is open in X,soW is open in X,i.e.,W is a neighborhood of x in X.ButF (W )=f (W ) V .Sincex and V were arbitrary, we conclude that F K ⇢ is continuous. 1 1 2 n 1 Consider a = 0, , ,..., − , 1,q ,q ,...,q A and let g = h(a). Then k, where n n n 0 1 n 2 9 0 k n 1and k x< k+1 . In particular, − n n k g g (x) n − ✓ ◆ k+1 k k g n g n k k = g k+1 − k x + g n − − n n ✓ ◆ n − n ✓ ◆ ✓ ◆ k+1 k g n g n k = 1− x − n n ✓ ◆ k +1 k g g n − n ✓ ◆ ✓ ◆ k +1 k +1 k +1 k k k 3✏ g f + f f + f g < . n − n n − n n − n 5 ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ Then k k k k ✏ ✏ 3✏ f(x) g(x) f(x) f + f g + g g(x) < + + = ✏ < 1. | − | − n n − n n − 5 5 5 ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ As x was arbitrary, ⇢(f,g)=sup d(f(x),g(x)) : x I =sup f(x) g(x) : x I ✏ { 2 } {| − | 2 } proves that D is dense in (I,R). C 2 MTG 6316 HOMEWORK Spring 2017 154. (a) (Section 31, #1) Show that if X is regular, every pair of points of X have neigh- borhoods whose closures are disjoint. Proof. Recall the following: Definition 1. Suppose that one-point sets are closed in X. Then X is said to be regular if for every pair of a point x X and a closed set A X disjoint from x, 2 ⇢ there exist disjoint open sets containing x and A, respectively. The space X is said to be normal if for each pair A, B of disjoint closed sets of X, there exist disjoint open sets containing A and B, respectively. Lemma 2. Let X be a topological space. Let one-point sets in X be closed. (i) X is regular if and only if given a point x X and a neighborhood U of x, 2 there is a neighborhood V of x such that Cl(V ) U. ⇢ (ii) X is normal if and only if given a closed set A and an open set U containing A, there is an open set V containing A such that Cl(V ) U. ⇢ Then, for any pair of points x, y X, by regularity of X, there are disjoint 2 open sets U, V containing x, y, respectively. However, by (i) of Lemma 2, there are neighborhoods U of x and V of y such that Cl(U ) U and 0 0 0 ⇢ Cl(V ) V .SinceU, V are disjoint, so are Cl(U ) and Cl(V ). 0 ⇢ 0 0 (b) (Section 31, #2) Show that if X is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint. Proof. Let A, B X be disjoint closed sets. By normality of X, there are ⇢ disjoint open sets U A and V B. However, by (ii) of Lemma 2, there are ⊃ ⊃ open sets U A and V B such that Cl(U ) U and Cl(V ) V .Since 0 ⊃ 0 ⊃ 0 ⇢ 0 ⇢ U, V are disjoint, so are Cl(U 0) and Cl(V 0). 1 Let p ⇥ X → Y be a closed continuous surjective map. Show that if X is normal, then so is Y . Let A and B be disjoint closed sets in Y . By continuity, p−1 A and p−1 B are disjoint closed in X.Since −1 −1 X is normal, there exist disjoint open sets U1 and U2 in X such that p A L U1 and p B L U2. Now as U1 and U2 are open, then X − U1 and X − U2 are closed.( And) as p is( a closed) map, then p X − U1 and p X − U2 are also closed. Thus V1 = Y − p X − U1 and V2 = Y − p X(− )U2 are open with( )A L V1 and B L V2. Now ( ) ( ) ( ) ( ) V1 = V2 = Y − p X − U1 = Y − p X − U2 = Y − p X − U1 < X − U2 . We claim now that Y = p( X −U(1 < X −))U2 (. Suppose( y " Y)).Thensince( ( p is surjective,) ( there)) is some x " X such that p x = y.SinceU1 = U2 = o,thenx " X − U1 or x " X − U2.Thusy " p X − U1 < X − U2 , implying Y = p X − U1 (< X −)U2(. Therefore,) V1 = V2 = o.ThusY is normal. ( ) ( ) ( ) ( ) ( ) 1 Not officially assigned: 159. Show that a closed subspace of a normal space is normal. P: Let Y be a closed subspace of the normal space X. Then Y is Hausdorff by Thm 17.11. Let A and B be disjoint closed subspaces of Y. Since A and B are closed also in X, they can be separated in X by disjoint open sets U and V. Then U ∩Y and V ∩ Y are open sets in Y separating A and B. Y is normal. ∎ Show that every locally compact Hausdor↵ space is regular. Since X is locally compact and Hausdor↵, by theorem 29.1 there exists a compact Hausdor↵ space Y such that X is a subspace of Y and Y X is a single point. Moreover, by theorem 32.3 \ Y is normal since it is compact and Hausdor↵. But every normal space is regular, thus Y is regular. Finally, by theorem 31.2 a subspace of a regular space is also regular. Thus the space X is regular, as desired. MTG 6316 HOMEWORK Spring 2017 162. (Section 32, #4) Show that every regular Lindel¨ofspace is normal. Proof. Recall: Definition 3. A space for which every open covering contains a countable subcovering is called a Lindel¨of space. The idea of the proof is similar to proving Lemma 32.1 (on page 200). Let X be regular and Lindel¨of. Let A, B X be an arbitrary pair of disjoint closed sets. ⇢ Then, for every p =(x, y) C = A B, there are disjoint open sets U ,V such 2 ⇥ p p that x U , y V .SinceA, B are closed, we may assume that each U is disjoint 2 p ⇢ p p from B and that each Vp is disjoint from A (if not, subtract A or B from Vp or Up, respectively, and we still keep all the properties mentioned above). Moreover, by part (a) of Lemma 31.1 (on page 196), we may assume that Cl(U ) B =Cl(V ) A = . (1) p \ p \ ; Let Wp = Up Vp. Note that Wp p C X (A B) forms an open covering [ { } 2 [ { \ [ } of X. Then by the Lindel¨ofcondition, there is a countable subcovering W { n}n1=0 of X,whereW0 = X (A B). Thus Wn n1=1 Wp p C is a countable open \ [ { } ⇢ { } 2 covering of A B. By earlier assumption, we have that Un n1=1 Up p C covers [ { } ⇢ { } 2 A and that Vn n1=1 Vp p C covers B. The rest of the proof is identical to the { } ⇢ { } 2 proof of Lemma 32.1. For every n,define n n U 0 = U Cl(V ) and V 0 = V Cl(U ). (2) n n − i n n − i i[=1 i[=1 Note that by (1), U and V are open coverings of A and B,respectively. { n0 }n1=1 { n0}n1=1 Define 1 1 U 0 = U 0 A and V 0 = V 0 B, n ⊃ n ⊃ n=1 n=1 [ [ both are open. Claim: U V = . Suppose not, then there is z U V ,so 0 \ 0 ; 2 0 \ 0 there are i, j N such that z U 0 V 0. Without loss of generality, assume that 2 2 i \ j i j. Then by (2), we have z U but z/Cl(U ) U , which is a contradiction, 6 2 i 2 i ⊃ i completing the proof of the claim. Therefore X is normal by definition. 2 Theorem (Strong form of the Urysohn Lemma). Let X be a normal space. There is a continuous function f : X [0, 1] such that f(x)=0forx A,andf(x)=1forx B, ! 2 2 and 0 <f(x) < 1otherwise,ifandonlyifA and B are disjoint closed Gδ sets in X. Definition A is a Gδ set in X if A is the intersection of a countable collection of open sets of X. Proof. ( )Letusseethat ) 1 1 1 G := f − 0, = A. (1) A n n=1 \ ✓ ◆◆ Note that 0, 1 are open in [0, 1] so by continuity of f, G is a G set. Let x G and n A δ 2 A 1 1 1 suppose towards a contradiction x/A.