MTG 6316 HOMEWORK Spring 2017 101. (Section 26, #12) Let p : X Y be a closed continuous surjective map such that p 1(y) ! − is compact, for each y Y . (Such a map is called a perfect map.) Show that if Y is 2 1 compact, then X is compact. [Hint: If U is an open set containing p− (y), there is a 1 neighborhood W of y such that p− (W ) is contained in U.] Proof. Suppose that Y is compact and there exists a perfect map p : X Y . ! Let U↵ ↵ I be an open cover of X. Fix an arbitrary point y Y .Sincep is 2 { } 1 2 a perfect map, then p− (y) is a nonempty compact subspace of X. By Lemma y Ny 1 26.1 (on page 164), there is a finite subcover Ui i=1 U↵ ↵ I of p− (y). Let { } ⇢ { } 2 U y = Ny U y.ThenU y is open in X, and p 1(y) U y.ThusX U y is closed i=1 i − ⇢ \ in X, and p 1(y) (X U y)= ,soy/p (X U y). Since p is a closed map, S − \ \ ; 2 \ p (X U y) is closed in Y . Consider \ V y = Y [p (X U y)] , \ \ y y y then V is open in Y and y V .Thus, V y Y forms an open cover for Y , and 2 { } 2 since Y is compact, there exists a finite subcover V yi N of Y . We want to show { }i=1 that U yi N covers X. Consider any subset A X. For any b Y p(A), we { }i=1 ⇢ 2 \ have p 1(b) A = ,sop 1(b) X A.Thenwehave − \ ; − ⇢ \ 1 p− (Y p(A)) X A. \ ⇢ \ Take A = X U y,wehave \ 1 y 1 y p− (V )=p− (Y p (A)) X A = U , \ ⇢ \ for any y Y .Thuswehave 2 N N N 1 1 y 1 y y X = p− (Y ) p− V i = p− (V i ) U i . ⇢ ! ⇢ i[=1 i[=1 i[=1 Furthermore, N N Nyi X U yi = U yi . ⇢ 0 j 1 i[=1 i[=1 j[=1 @ A N yi Nyi Therefore, i=1 Uj j=1 is a finite subcover of U↵ ↵ I for X. { } { } 2 S 1 Let X be a metric space and f : X X be a homeomorphism. Given N X, define the ! ⇢ maximal invariant set in N by 1 Inv(N)= f k(N)= x N : f k(x) N, k Z . { 2 2 8 2 } k= \1 Definition: N X is a trapping region for f if ⇢ 1. N is compact and forward invariant, i.e. f(N) N,and ⇢ 2. there exists K>0sothatf K (N) int(N). ⇢ (a) Show that if N is a trapping region, then 1 1 j kn Inv(N)= cl f (N) = x N : xn N,kn > 0,kn , such thatf (xn) x as n . ! { 2 9 2 !1 ! !1} k\=0 j[=k (b) Show that if N is a trapping region and nonempty, then Inv(N) is compact and nonempty. (c) Show that if N is a trapping region, Inv(N) int(N). ⇢ Proof. (a) Since X is a metric space, let ⇢ denote the metric. Since N is a trapping region j and X is Hausdor↵, N is closed. Let A = 1 cl 1 f (N) and B = x N : x k=0 j=k { 2 9 n 2 N,k > 0,k such thatf kn (x ) x as n ⇣ . We will show⌘ that n n !1 n ! T !1S} B Inv(N) A B. ⇢ ⇢ ⇢ Let x B. Then 2 x N,k > 0,k such thatf kn (x ) x as n . 9 n 2 n n !1 n ! !1 Let k Z. Since kn , M N so that m M implies km + k>0. By forward 2 !1 9 2 ≥ invariance of N, f k+km (x ) N. Since f is a homeomorphism, f k is continuous, being the m 2 composition of continuous functions. Then f k+kn (x ) f k(x) N as N is closed. As k n ! 2 was arbitrary, this proves B Inv(N). ⇢ Let us see that Inv(N) A.Foranyk Z ⇢ 2 1 1 f k(N) f j(N) cl f j(N) . ⇢ ⇢ ! j[=k j[=k 1 Then 1 1 1 1 Inv(N)= f k(N) f k(N) cl f j(N) = A. ⇢ ⇢ k= k=0 k=0 j=k ! \1 \ \ [ Now we show A B by first showing A N. k 0, ⇢ ⇢ 8 ≥ 1 1 f k(N) N f j(N) cl f j(N) N, ⇢ ) ⇢ ! ⇢ j[=k j[=k since N is a trapping region (closed and forward invariant). Thus A N. Let x A and ⇢ 2 j n 0. Since x cl 1 f (N) , ≥ 2 j=n ⇣S ⌘ 1 x N, k n, so that ⇢ f kn (x ),x < , (Munkres, Lemma 21.2). 9 n 2 9 n ≥ n n Then f kn (x ) x and k as n showing that A B. n ! n !1 !1 ⇢ (b) f is a bijection, therefore f k(N) = , k Z. By (a), Inv(N)=A implies Inv(N) 6 ; 8 2 is closed as A is the intersection of closed sets. Since Inv(N) N and N is compact, ⇢ j Inv(N)iscompact.LetC =cl 1 f (N) for k 0. Since N is a trapping region, k j=k ≥ N C C ... C ... The C⇣ are non-empty⌘ since f is a bijection and N = imply ⊃ 0 ⊃ 1 ⊃ ⊃ k Sk 6 ; j j j f (N) = , j Z. The Ck are nested because a b 1 f (N) 1 f (N). The Ck 6 ; 8 2 ) j=b ⇢ j=a are compact, being closed subsets of N. By Munkres, TheoremS 26.9, p.170S and the nested sequence property, it follows that Inv(N) = . 6 ; (c) Let N be a trapping region. Then K>0sothatf K (N) int(N). Then 9 ⇢ 1 Inv(N)= f k(N) f K (N) int(N). ⇢ ⇢ k= \1 Exercise 28. Give an example of E [0, 1] so that n ✓ 1 λ(E )= n 1 n=1 X and λ lim En =0. n ⇣ !1⌘ 2 Problem 104. Recall that RK denotes R in the K-topology. (a)Showthat[0, 1] is not compact as a subspace of RK . (b)ShowthatRK is connected [Hint: ( , 0) and (0, )inherittheirusualtopologies 1 1 as subspaces of RK .] (c)ShowthatRK is not path connected. Proof. The K topology on R is described in Munkres: let K denote the set of all numbers of the form 1/n, for n Z+, and let B00 be the collection of all open intervals (a, b), along with 2 all sets of the form (a, b) K. The topology generated by B00 will be called the K-topology − on R. The open sets 1 1 ( , 1+ ) 1 and ( 1, 2) K { n +1 n }n=1 − − form an open cover of [0, 1] with no possible finite subcover. For part (b)thehintstatesthat( , 0) and (0, ) inherit their usual topologies as a 1 1 subspace of RK . Therefore both of these intervals are connected as they are in R. Let U, V be open sets of RK such that U V = RK and U V = . Then [ \ ; (1) ( , 0) = (U ( , 0)) (V ( , 0)) 1 \ 1 [ \ 1 (2) (0, )=(U (0, )) (V ( , 0)). 1 \ 1 [ \ 1 Since (1) holds, and ( , 0) is connected, it must be true that U ( , 0) = ( , 0) or V ( , 0) = ( , 0).1 A similar situation occurs in (2). Without loss\ 1 of generality1 suppose that\ 1U ( , 0)1 = ( , 0) which implies ( , 0) U and similarly (0, ) V = V which implies\ (01, ) V . It1 must be the case that1U =(✓ , 0] and V =(0, 1)or\ U =( , 0) 1 ✓ 1 1 1 and V =[0, ). In either case, V or U fails to be open. Therefore RK is connected. 1 RK is strictly finer than the standard topology on R. Then a continuous function f : R Y , when Y = RK is still continuous when Y = R. Thus f satisfies the condi- ! tions for the Intermediate Value Theorem to hold. Let f be a path from 0 to 1 in RK where f(0) = 0 and f(1) = 1. The Intermediate Value Theorem holds so it must be the case that [0, 1] f([0, 1]) RK . ✓ ✓ f is continuous and [0, 1] R is compact which implies f([0, 1]) must also be compact. By ⇢ Theorem 27.1, [0, 1] f([0, 1]) RK must also be compact because [0, 1] RK is a closed ✓ ⇢ ⇢ interval. But this is impossible since (a)showed[0, 1] RK is not compact. Therefore RK is not path-connected. ⇢ 1 Show that a connected metric space having more than one point is uncountable. Let X be a connected space with metric d having at least two points, say a and b. Suppose there is some number c between 0 and d(a, b) such that no point x X satisfies d(a, x)=c.Thenthesets x X d(a, x) <c and x X d(a, x) >c form a separation2 of X, which is a contradiction since X is assumed{ 2 | to be connected.} { Then2 for| every real} number c between 0 and d(a, b) there is a point x X such 2 that d(a, x)=c. Then the image of the function f : X R defined by f(x)=d(a, x) must contain the entire interval [0,d(a, b)], and thus X must be uncountable.! 1 Let X be a compact Hausdor↵ space.