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 (E )= n 1 n=1 X and