Solution Sangchul Lee March, 018 1 Solution of Homework 7 Problem 1.1 For a given k N, Consider two sequences (a n ) and (b n,k ) in R. Suppose that a n b n,k for all n,k N Show that limsup a n B k := limsup B n,k for all k N. Consequently it follows that lim supa n liminf B k. k Proof. Recall Theorem 3.17: Theorem. If (s n ) and (t n ) are two sequences in R such that s n t n for all n N, then we always have lim sup s n limsup t n and liminf s n liminf t n. Now fix k and take limsup to the inequality a n b n,k as n. Writing A = limsup a n, it follows from Theorem 3.19 that A B k. Now applying Theorem 3.19 again, we obtain that A = liminf k A liminf k B k. Remark. In the last step of the proof, we regarded A as a constant array. Since the theorem does not cover the case A {,+ }, however, we have to treat those cases separately. But if A = then the conclusion is trivial, and if A = + then we also have B k = + for all k N and again we have no issue concluding the same inequality. Problem 1. Fix α > 1. Take x 1 > α and define x n+1 = α + x n 1 + x n = x n + α x n 1 + x n. (1) Prove that x 1 > x 3 > x 5 >. () Prove that x < x 4 < x 6 <. (3) Prove that lim x n = α. 1
Proof. Define f : [0, ) [0, ) by f (x) = α + (1 + α)x (1 + α) + x. Then it is easy to check that x n+ = f (x n ) for all n N. Also, by writing f (x) in two different ways f (x) = x (x α) (1 + α) + x = α + (x α)( α 1) (1 + α) + x we easily find that if x > α, then x > f (x) > α, if x < α, then x < f (x) < α. Since x 1 > α and x = α (x 1 α)( α 1) 1+x 1 < α, both (1) and () immediately follow by the mathematical induction. Moreover, since both (x n ) and (x n+1 ) are monotone and bounded, they converge to some x [0, ) and y [0, ), respectively. Moreover, x = lim x n = lim x n+ = lim f (x n ) = f (x) and likewise y = f (y). Since the unique non-negative fixed point of f is α, it follows that x = α = y. Finally, notice that this is enough to conclude that (x n ) converges and its limit is also α as desired. Problem 1.4 Suppose f is a real function defiend on R 1 which satisfies lim[ f (x + h) f (x h)] = 0 h 0 for every x R 1. Does this imply that f is continuous? Proof. Consider the example f (x) = { 1, if x = 0, 0, otherwise. Then it is easy to check that f satisfies the condition. Indeed, with this function we easily check that lim f (x + h) = lim f (x h) = lim f h 0 h 0 x x (x ) = 0 for any x R. Therefore f satisfies the condition while f is not continuous at 0. Problem 1.5 Let f be a continuous real function on a metric space X. Let Z( f ) (the zero set of f ) be the set of all p X at which f (p) = 0. Prove that Z( f ) is closed. Proof. We provide two proofs. 1 st proof. Notice that Z( f ) = f 1 ({0}). Since {0} is a closed subset of R, its inverse image under the continuous function f is also closed as desired.
nd proof. Let p Z( f ). Then there exists a sequence (p n ) in Z( f ) that converges to p in X. By the continuity of f, we must have f (p) = lim f (p n ) = lim 0 = 0. So it follows that p Z( f ) as well. This proves that Z( f ) Z( f ) and therefore Z( f ) is closed. Problem 1.6 Let f and g be continuous mappings of a metric space X into a metric space Y, and let E be a dense subset of X. Prove that f (E) is dense in f (X). If g(p) = f (p) for all p E, prove that g(p) = f (p) for all p X. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Proof. Let y f (X). Then there exists p X such that y = f (p). Since E is dense in X, there exists a sequence (p n ) in E such that p n p. Then ( f (p n )) is a sequence in f (E) such that f (p n ) f (p) = y. So it follows that every point of f (X) is in the closure of f (E) and therefore f (E) is dense in f (X). For the second part, let p X be arbitrary and pick a sequence (p n ) in E that converges to p. Then f (p) = lim f (p n ) = lim g(p n ) = g(p). Remark. In fact, the above proof of the second part immediately generalizes to prove the following claim: If f and g are continuous mappings between metric spaces X and Y, then the set is a closed subset of X. {x X : f (x) = g(x)} Solution of Homework 8 Problem.1 A subset A of a metric space X is pathwise connected if for every x,y A there is a continuous function f : [0, 1] A such that f (0) = x and f (1) = y. Prove that every pathwise connected set is connected. Proof. We prove the contrapositive. Assume that a subset E of X is not connected. Let A,B be non-empty sets that separate E. Since E = A B and both A and B are non-empty, we can pick x A and y B. Now we claim that there is no continuous function f : [0, 1] E satisfying f (0) = x and f (1) = y. Assume otherwise that there is such a function f. Then U = f 1 (A) and V = f 1 (B) satisfies 0 U and 1 V, hence both U and V are non-empty. U V = f 1 (A B) =. f 1 (Ā) is a closed set which contains U, hence Ū f 1 (Ā) and hence Ū V f 1 (Ā B) =. Likewise we have U V =. Combining altogether, we find that U and V separate the interval [0, 1], which is impossible because we already know that [0,1] is connected. Therefore it follows that E is not pathwise connected as required. 3
Problem. Let us consider {( S = x,sin 1 ) } : 0 < x < 1 x and E = S in R. (a) Show that E is connected. (b) Show that E is not pathwise connected. Remark. E is called the topologist s sine curve which looks like Proof. (a) Notice that S is pathwise connected. Indeed, for any two points (a, sin(1/a)) and (b, sin(1/b)) in S, the function ( ( )) 1 f (t) = (1 t)a +tb,sin (1 t)a +tb is a continuous function from [0,1] to S that joins those two points. Then by the previous problem, S is connected. Then its closure E = S is also connected by one of the previous homework problems. (b) For each y [ 1,1] we can find θ [0,π] such that sinθ = y. Then with x n = 1 (x n,sin(1/x n )) (0,y). This implies that {0} [ 1,1] E. θ+πn, we find that Now for the sake of argument, assume otherwise that E is pathwise connected. Then there exists a continuous function f : [0,1] E such that f (0) = (0,0) and f (1) = ( π,1). Write f (t) = (x(t),y(t)) and define t = supz(x), where Z(x) = {t [0,1] : x(t) = 0} is the zero-set of x. Since x : [0,1] R is continuous, Z(x) is closed and hence t Z(x). In other words, x(t ) = 0. Now we define K = f ([t,1]) and investigate the property of this set. (1) Since [t,1] is compact, the same is true for K. () Since [t,1] is an interval, so is x([t,1]) by the intermediate value theorem (which is the statement of 3.(b) below). In particular, [0, π ] = [x(t ),x(1)] x([t,1]) and hence {(x,sin(1/x)) : 0 < x π } K. (3) Notice that x(t) > 0 if t > t. So f (t ) is the unique point in K whose 1 st coordinate is 0. This is enough to derive a contradiction. Indeed, ( ) p n := (n + 1)π,( 1)n K 4
for all integers n 0 and (p n ) has two limit points (0,1) and (0, 1). So both (0,1) K and (0, 1) K must hold by the closedness of K, which is impossible by the property (3) above. Problem.3 Let (X,d) be a metric space. (a) Show that X is connected if and only if every continuous function f : X N is a constant. (b) Consider a continuous function g : X Y, where Y is another metric space. Using (a), show that if X is connected then g(x) is a connected subset of Y. In particular, if X is an interval in R and Y = R, then g(x) is also an interval. Proof. (a) We prove that X is disconnected if and only if there exists a non-constant continuous function f : X N. ( ) Let A,B be non-empty subsets of X that separate X. Since Ā B = A B =, we know that A = B c and B = Ā c are both open. Then the function { 1, x A f (x) =, x B is continuous since for each open subset U of N, the inverse image f 1 (U) is one of, A, B, and X, all of which are open. ( ) Assume that there exists a non-constant continuous function f : X N. Fix an element n f (X) and notice that both {n} and N \ {n} are simultaenously open and closed in N. (Indeed, both {n} = N (n 1,n + 1 ) and N \ {n} = N ((,n) (n, )) are open relative to N. Similar idea shows that they are both closed relative to N.) Now write A = f 1 ({n}), B = f 1 (N \ {n}). By the assumption, both A and B are non-empty open and closed subsets of X such that A B = and A B = X. So A and B separates X and hence X is disconnected. (b) Assume that X is connected. Let h : g(x) N be any continuous function. Then h g is well-defined and continouous, hence is constant by (a). Since g is onto g(x), it follows that h is also constant. Therefore by (a) again, it follows that g(x) is connected. Now the last claim follows by noting that the only non-empty connected subsets of R are intervals. Problem.4 If f is defined on E, the graph of f is the set of points (x, f (x)), for x E. In particular, if E is a set of real numbers, and f is real-valued, the graph of f is a subset of the plane. Suppose that E is compact, and prove that f is continuous on E if and only if its graph is compact. Proof. ( ) If f : E R is continuous, then γ(x) = (x, f (x)) is also continuous by Theorem 4.10. Now notice that γ(e) is compact and is exactly the graph of f. 5
( ) Assume that the graph of f is compact. In order to prove that f is continuous, the following lemma will be useful: Lemma. Let (X,d) be a metric space and p, p n X. Then the followings are equivalent: (1) (p n ) converges to p. () Every subsequence of (p n ) has a further subsequence that converges to p. Proof of Lemma. (1) () is trivial, since every subsequence of (p n ) converges to p. For the other direction, we prove the contrapositive (1) (). If (p n ) does not converge to p, then there exists ε > 0 such that for any N N, there exists n N such that d(p n, p) ε. Use this statement recursively to define n 1 < n < such that d(p nk, p) ε for all k N. Then no subsequence of (p nk ) can converge to p, hence () follows. Now let us return to the original proof. Let x E and (x n ) E converges to x. Since p n = (x n, f (x n )) is a sequence in the compact set graph( f ), any subsequence of (p n ) has a further subsequence that converges in graph( f ). Writing (p nk ) for this further subsequence and (x,y ) = lim k p nk, we know that x nk x and hence x = x. Now the fact that (x,y ) graph( f ) translates to the equality y = f (x ) = f (x), from which we deduce that p nk (x, f (x)) as k. So p n (x, f (x)) by the lemma above. In particular, f (x n ) f (x). Since this is true for any x E and for any (x n ) x, it follows that f is continuous. Problem.5 Let f be a real uniformly continuous function on the bounded set E in R 1. Prove that f is bounded on E. Show that the conclusion is false if boundedness of E is omitted from the hypothesis. Proof. Since f is uniformly continuous, there exists δ > 0 such that f (x) f (y) < 018 whenever x,y E satisfies x y < δ. Now consider the family U = {N δ (x) : x E}. Since E is bounded, its closure Ē is compact. Then it is easy to check that U is an open cover of Ē, so there is a finite subcover {N δ (x 1 ),,N δ (x n )} of Ē. Notice that this is also a cover of E. So if we set M = max{ f (x i ) + 018 : i = 1,,n}, then for each x E we can find i such that x N δ (x i ) and hence Therefore f is bounded. f (x) f (x i ) + f (x) f (x i ) f (x i ) + 018 M. To show that boundedness of E is essential, simply check that id : R R defined by id(x) = x is uniformly continuous but not bounded. Problem.6 Every rational x can be written in the form x = m/n, where n > 0, and m and n are integers without any 6
common divisors. When x = 0, we take n = 1. Consider the function f defined on R 1 by { 0, (irrational) f (x) = 1 n, ( ) x = m n. Prove that f is continuous at every irrational point, and that f has a simple discontinuity at every rational point. Proof. For both claims, it suffices to show that lim y x f (y) = 0 for all x R. To this end, fix x R and consider an arbitrary ε > 0. Find N N such that 1 N < ε and define δ by δ = inf{ x (p/q) : p,q Z, 1 q N and 0 < x (p/q) 018}. It is easy to check that the set of all rational numbers p/q that satisfy the above condition is non-empty and finite. So δ is the minimum of finitely many positive reals, hence δ > 0. Then whenever y R satisfies 0 < x y < δ, we find that either y is an irrational number so that f (y) = 0, or y = m/n is a rational number whose denominator n in lowest term is larger than N, so that f (y) = 1 n < 1 N < ε. So we always have f (y) < ε. Therefore in view of the ε δ definition of the limit, we have lim y x f (y) = 0 as desired. Problem.7 IF E is a non-empty subset of a metric space X, define the distance from x X to E by ρ E (x) = inf z E d(x,z). (a) Prove that ρ E (x) = 0 if and only if x Ē. (b) Prove that ρ E is a uniformly continuous function on X, by showing that for all x X, y X. ρ E (x) ρ E (y) d(x,y) Proof. (a) If x Ē, then there exists a sequence (x n ) E such that x n x. Then ρ E (x) d(x,x n ) 0 and hence ρ E (x) = 0. If x / Ē, then there exists a neighborhood N δ (x) such that N δ (x) Ē =. Then d(x,z) δ for all z E and hence ρ E (x) δ > 0. (b) For any x X, y Y and z E, we have ρ E (x) d(x,z) d(x,y) + d(y,z). So taking infimum over z E it follows that ρ E (x) d(x,y) + ρ E (y), from which we obtain ρ E (x) ρ E (y) d(x,y). Then interchanging the role of x and y yields ρ E (y) ρ E (x) d(x,y) holds as well. Therefore the desired inequality follows. 7
Problem.8 Assume that f is a continuous real function defined in (a,b) such that ( ) x + y f (x) + f (y) f for all x,y (a,b). Prove that f is convex. Proof. Fix x,y (a,b) and define ϕ : [0,1] R by ϕ(t) = t f (x) + (1 t) f (y) f (tx + (1 t)y). Then it suffices to show that ϕ(t) 0 for all t [0,1], or equivalently, ϕ 1 ([0, )) = [0,1]. To this end, we make the following observations. ϕ(0) = ϕ(1) = 0. If s,t [0,1] satisfy ϕ(s) 0 and ϕ(t) 0, then u = s+t also satisfy ϕ(u) 0. This easily follows from [s f (x) + (1 s) f (y)] + [t f (x) + (1 t) f (y)] u f (x) + (1 u) f (y) = and ( ) [sx + (1 s)y] + [tx + (1 t)y] f (ux + (1 u)y) = f ϕ is continuous. f (sx + (1 s)y) + f (tx + (1 t)y). From these observations, it is easy to deduce that ϕ(k/ n ) 0 for any n N and k Z with 0 k n. But the set of all rationals of the form k/ n is dense is [0,1], hence it follows that ϕ 1 ([0, )) is a dense subset of [0,1]. Since ϕ 1 ([0, )) is closed by the continuity of ϕ, it follows that and hence the desired claim follows. ϕ 1 ([0, )) = ϕ 1 ([0, )) = [0,1] 8