MATH 140A - HW 6 SOLUTIONS Problem 1 (WR Ch 3 #21). Prove the following analogue of Theorem 3.10(b): If {En} is a sequence of closed nonempty and bounded sets in a complete metric space X , if En En 1, and if ⊃ Å lim diamEn 0, n !1 Æ T then 1 En consists of exactly one point. Solution. If TE has two or more points, say x, y TE with x y, then x, y E for all n, so n 2 n 6Æ 2 n diamE d(x, y) 0, n ¸ È contradicting the fact that diamE 0. Therefore, we know it has at most one point; all that is left n ! is to prove it is nonempty. For each n, pick a point xn in En (which is nonempty). Since En En 1, we know {xn,xn 1,...} ⊃ Å Å ½ E . The fact that diamE 0 implies that for any ² 0 we can pick some N N such that n n ! È 2 diamE ² for n N, which means for any n,m N we have n Ç ¸ ¸ xn xm diam({xN ,xN 1,xN 2,...}) diamEN ², j ¡ j · Å Å · Ç so {x } is a Cauchy sequence. Since X is complete, that means that {x } converges, say x x. n n n ! En is closed and it contains {xn,xn 1,...}, (i.e. the tail of the sequence {xn}) so x En for all n. Å 2 Therefore, x TE , so it is nonempty, finishing the proof. 2 n Problem 2 (WR Ch 3 #22). Suppose X is a nonempty complete metric space, and {Gn} is a se- T quence of dense open subsets of X . Prove Baire’s theorem, namely, that 1 Gn is not empty. (In fact, it is dense in X .) T Solution. Let A 1 G . We will prove A is dense, which will show it is nonempty (since if A Æ 1 n Æ; and A is dense then X A , contradicting the fact that X is nonempty). To do that, we’ll first Æ Æ; prove the following claim: Claim. A X is dense iff N A for every neighborhood N X . ½ \ 6Æ ; ½ Proof of claim. Assuming A X is dense, then A A0 A X . Then for any x N, there are two possible Æ) ½ [ Æ Æ 2 cases: (a) x A, in which case x N A, so N A . 2 2 \ \ 6Æ ; (b) x A and x A0. Then since x is a limit point of A, and N is a neighborhood of x, there ∉ 2 is some y N such that y x and y A. But then y N A, so N A . 2 6Æ 2 2 \ \ 6Æ ; 1 Assuming N A for every neighborhood N X , we want to show that A X . Let x X . (Æ \ 6Æ ; ½ Æ 2 If x A we are done. Otherwise, assume x A, and we must show that x is a limit point of 2 ∉ A. For any neighborhood N of x, we know that N A , so there must be some y N A. \ 6Æ ; 2 \ However, since x A, that means that y x. Therefore x is a limit point of A, finishing the ∉ 6Æ proof of the claim. Now, using the claim, let N be any neighborhood in X , and we’ll show that N A . Since 0 0 \ 6Æ ; G is a dense open subset of X , we know that N G , so take any x N G . Since N G 1 \ 1 6Æ ; 1 2 0 \ 1 0 \ 1 is open, there is a neighborhood N1 of x1, and we can shrink N1 if necessary to guarantee that diamN1 1. Let E1 N1. We repeat this process, so that if xn 1, Nn 1 and En 1 are already · Æ ¡ ¡ ¡ defined, we define the next terms in the sequence as follows: since Gn is a dense open subset of X , we know that Nn 1 Gn , so take any xn Nn 1 Gn. Since Nn 1 Gn is open, there is a ¡ \ 6Æ ; 2 ¡ \ ¡ \ neighborhood N of x , and we can shrink N if necessary to guarantee that diamN 1 . Let n n n n · n E N . Then {E } is a collection satisfying the conditions of the previous exercise, so TE {p} n Æ n n n Æ for some point p N , and since E G , that means there is a point p in each G , and thus 2 0 n ½ n n p A TG . Therefore, N A . 2 Æ n 0 \ 6Æ ; Problem 3 (WR Ch 3 #23). Suppose {pn} and {qn} are Cauchy sequences in a metric space X . Show that the sequence {d(pn,qn)} converges. Solution. For any m,n, d(p ,q ) d(p ,p ) d(p ,q ) d(q ,q ) d(p ,q ) d(p ,q ) d(p ,p ) d(q ,q ). n n · n m Å m m Å m n Æ) n n ¡ m m · n m Å m n Doing the same argument but switching m and n, we also have d(p ,q ) d(p ,q ) d(p ,p ) d(q ,q ), m m ¡ n n · m n Å n m so putting these two together we have d(p ,q ) d(p ,q ) d(p ,p ) d(q ,q ). j n n ¡ m m j · n m Å m n For any ² 0, since {p } is Cauchy, we can find an N N such that for m,n N , d(p ,p ) ²/2, È n 1 2 ¸ 1 n m Ç and since {q } is Cauchy, we can find an N N such that for m,n N , d(q ,q ) ²/2. Therefore, n 2 2 ¸ 2 n m Ç for n,m N max(N ,N ) we have ¸ Æ 1 2 d(p ,q ) d(p ,q ) d(p ,p ) d(q ,q ) ² ² ², j n n ¡ m m j · n m Å m n Ç 2 Å 2 Æ so {d(pn,qn)} is a Cauchy sequence, and since R is a complete metric space and {d(pn,qn)} is a Cauchy sequence of real numbers, it converges. Problem 4 (WR Ch 4 #1). Suppose f is a real function defined on R1 which satisfies lim[f (x h) f (x h)] 0 h 0 Å ¡ ¡ Æ ! for every x R1. Does this imply that f is continuous? 2 2 Solution. It does not imply that f is continuous. The following is a counterexample. ( 1 if x=0 f (x) . Æ 0 otherwise Notice that for ² 1 , for any ± 0 and any x such that 0 0 x ± (i.e. x 0), we have Æ 2 È Ç j ¡ j Ç 6Æ f (x) f (0) 0 1 1 1 , j ¡ j Æ j ¡ j Æ È 2 so f is discontinuous at 0. However, for any x 0, note that for any h x , x h 0 and x h 0, so 6Æ Ç j j Å 6Æ ¡ 6Æ f (x h) f (x h) 0 0 0, Å ¡ ¡ Æ ¡ Æ and thus lim[f (x h) f (x h)] 0. If x 0, then for any h 0, h 0 Å ¡ ¡ Æ Æ È ! f (x h) f (x h) 0 0 0, Å ¡ ¡ Æ ¡ Æ and thus lim[f (x h) f (x h)] 0 again. This finishes the proof of the counterexample. h 0 Å ¡ ¡ Æ ! Problem 5 (WR Ch 4 #7). If E X and f is a function defined on X , the restriction of f to E is ½ the function g whose domain of definition is E, such that g(p) f (p) for p E. Define f and g Æ 2 on R2 by: f (0,0) g(0,0) 0, f (x, y) x y2/(x2 y4), g(x, y) x y2/(x2 y6) if (x, y) (0,0). Prove Æ Æ Æ Å Æ Å 6Æ that f is bounded on R2, that g is unbounded in every neighborhood of (0,0), and that f is not continuous at (0,0); nevertheless, the restrictions of both f and g to every straight line in R2 are continuous! Solution. f is bounded on R2 because ¯ 2 ¯ 1 ¯ x y ¯ ( x y2)2 0 x2 2 x y2 y4 0 x2 y4 2 x y2 ¯ ¯. j j ¡ ¸ () ¡ j j Å ¸ () Å ¸ j j () 2 ¸ ¯ x2 y4 ¯ Å g is unbounded in every neighborhood of (0,0) because if we let x 1 and y 1 then (x , y ) n Æ n3 n Æ n n n ! (0,0) and 1 1 1 n3 n2 n5 n g(xn, yn) . Æ 1 1 Æ 2 Æ 2 ! 1 n6 Å n6 n6 This is because (x , y ) (0,0) implies that for any neighborhood U of (0,0), there is some N n n ! 2 N such that (x , y ) U for n N, and g(x , y ) gets arbitrarily large. Next, we show f is not n n 2 ¸ n n continuous at p (0,0) simply by using Theorems 4.2 and Theorem 4.6 to say Æ 4.6 4.2 f is continuous at p lim f (x) f (p) f (pn) f (p) for every sequence pn p. x p () ! Æ () ! ! So to show f is not continuous, we only need to find one sequence p p such that f (p ) f (p). n ! n 6! Let x 1 and y 1 , and put p (x , y ), so that p p (0,0). But then n Æ n2 n Æ n n Æ n n n ! Æ 1 1 1 2 2 4 f (p ) f (x , y ) n n n 1 0 f (0,0) f (p), n Æ n n Æ 1 1 Æ 2 Æ 2 6Æ Æ Æ n4 Å n4 n4 3 so f is not continuous at (0,0).