Exam 2 extra practice problems

Exam 2 extra practice problems (1) If (X, d) is connected and f : X R is a continuous function such that f(x) = 1 for all x X, show that f must be constant. Solution: Since f(x) = 1 for every x X, either f(x) = 1 or f(x) = 1. Consider = {x X : f(x) = 1} and B = {x X : f(x) = 1}. Since f(x) = 1, x X, B = X, and clearly B =. Moreover, and B are clopen: is closed since = f 1 ({1}) and {1} is closed in R and f is continuous. is also open since = f 1 (( 3/2, 0)), f is continuous and ( 3/2, 0) is open in R. Since X is connected, either = or B =, hence f is constant. (2) Let X be a metric space. If X, define the characteristic function of as the function χ : X R such that χ (x) = 1 when x and χ (x) = 0 when x /. Show that is clopen χ is continuous. Solution: Notice that given U R, either χ 1 (U) = X (if 0, 1 U), or χ 1 (U) = (if 1 U but 0 / U), or χ 1 (U) = X \ (if 0 U but 1 / U), or χ 1 (U) = (if 0, 1 / U). ssume that is clopen. To show that χ is continuous we will show that the inverse image of an open set in R by χ is open. Let U be open. By the remark above, χ 1 (U) is either,, X or c, which are all open, since is clopen. ((1/2, 3/2)) = must (( 1/2, 1/2)) = c must be open, hence is Reciprocally, if χ is continuous, then χ 1 be open, and also χ 1 clopen. (3) Give the example of two connected sets whose intersection is not connected. Solution: Consider the unit circle x 2 +y 2 = 1 and the line x = 0. Both are connected, but their intersection is {(0, 1), (0, 1)}, which is not connected. (4) Let (X, d X ) and (Y, d Y ) be metric spaces. function f : X Y is called Lipschitz if there exists C 0 such that d Y (f(a), f(b)) Cd X (a, b), a, b X. Show that any Lipschitz function is uniformly continuous. 1

2 Solution: Given ε > 0, let δ = ε C. Then if x, y X and d X (x, y) < δ, d Y (f(x), f(y)) Cd X (x, y) < ε. (5) Let (X, d) be a metric space. Show that if F = {x 1,..., x m } X for some m N, then F is closed. Solution: Let (y n ) n be a sequence of elements of F converging to y X. We want to show that y F. Since F is finite, one of the x i s (say x j ) must appear infinitely many times in the sequence (y n ) n, hence we have a constant subsequence, which must converge to x j. Since any subsequence of a converging sequence must converge to the same limit, we have y = x j F. (6) Let M be a set and define the metric d on M by letting { 1, if x y d(x, y) = 0, if x = y. Describe what are the open sets of (M, d). Solution: Given x M, notice that {x} is open, since {x} = M 1/2 x (the ball of radius 1/2 centered at x). Since unions of open sets are open, all subsets of M are open. (Which means all subsets are also closed). (7) Let (X, d X ), (Y, d Y ) be metric spaces. Show that if f : X Y is continuous, then f(e) f(e). Solution: Let x E. We want to show that f(x) f(e). Since x E, there exists a sequence x n E with x n x. Since f is continuous, f(x n ) f(x) and f(x n ) f(e). This shows that f(x) f(e). (8) Let (X, d X ), (Y, d Y ) be metric spaces. Show that if f : X Y is uniformly continuous, then for any (x n ) n Cauchy sequence in X, (f(x n )) n is Cauchy in Y. Solution: Let ε > 0 and δ > 0 such that d X (x, y) < δ implies d Y (f(x), f(y)) < ε, from the uniform continuity of f. Let N N be such that d X (x n, x m ) < δ, n, m N (since (x n ) n is Cauchy). Then by the uniform continuity of f, d Y (f(x n ), f(x m )) < ε, m, n N, therefore (f(x n )) n is Cauchy.

(9) Let (X, d) be a metric space. Show that a finite union of compact sets in X is compact. 3 Solution: Let K 1,..., K m be compact subsets of X. Let (x n ) n be a sequence in K = m i=1 K i. Each element of the sequence is in one of the K i s. Since there are only finitely many of them, there exists j such that infinite many terms of the sequence must belong to the same K j. Look at the subsequence made of those elements of (x n ). Since K j is compact, there must be a subsubsequence converging to x K j K. This shows that K is compact. (10) Let M be a set and define the metric d on M by letting { 1, if x y d(x, y) = 0, if x = y. Show that (M, d) is compact M is finite. Solution: ssume M is compact by contradiction that it has infinitely many elements. Then we can create a sequence (x n ) n of distinct elements in M Since M is compact, there must be a subsequence converging to x M. This means that there exists N N such that d(x n, x) < 1 for all n N. However, by the definition of distance this implies x n = x, n N, a contradiction since we chose the x is to be distinct. Reciprocally, assume M is finite. Then for any sequence (x n ) n, there must be a subsequence whose elements are all the same (since M is finite). This subsequence converges to this constant element, so M is compact. (11) Let E be a normed space with norm, and define a metric d from. Let U E be open and S E. Show that is open in E. S + U = {s + u : s S, u U} Solution: Let s + u S + U. Since u U and U is open, there exists r > 0 such that M r u U. We claim that M r (s + u) S + U. Indeed, if z M r (s + u), then d(z, s + u) = z (s + u) = z s u < r. So z s M r u U. Call z s = a U. Then z = s + a, where s S, a U, hence z S + U. (12) Let M be a set. Two metrics d 1, d 2 on M are said to be equivalent if there exist constants a, b > 0 such that d 1 (x, y) ad 2 (x, y), and d 2 (x, y) bd 1 (x, y), x, y M. ssume d 1 and d 2 are two equivalent metrics on M. Show that

4 (a) x n x w.r.t. d 1 x n x w.r.t. d 2. (b) f : M R is continuous w.r.t. d 1 f is continuous w.r.t. d 2. Solution: (a) If x n x w.r.t. d 1, then given ε > 0, there exists N N such that d 1 (x n, x) < ε b, n N. Then d 2(x n, x) bd 1 (x n, x) < ε, n N, hence x n x w.r.t. d 2. similar argument shows the reciprocal. (b) Follows from (a) and from the definition of sequential continuity. (13) Let (X, d) be a metric space. n open cover of K X is a collection G of subsets of X such that K {G : G G} and each G G is open. subcover of K is a subset G 1 of G that is also a cover of K. Show that if K is a subset of X with the property that every open cover of K has a finite subcover, then K is (sequentially) compact. Solution: Let (x n ) n be a sequence of elements in K. Define F n = {x k : k n} and let U n = K \ F n. Since F n is closed, U n is open in K. We claim that n=1 F n. Indeed, if n=1 F n =, then n=1 U n = K and {U n, n N} is an open cover of K. By assumption this implies that there is m N such that K U 1... U m = U m. Hence F m =, which is not possible. Therefore there exists x n=1 F n. By definition of the F i s, this implies that given l N, (M 1/l x) {x k : k l}. Pick x 1 (M 1 x) {x k : k 1}; this corresponds to a x n1. Pick x 2 (M 1/(n1 +1)x) {x k : k n 1 + 1} and so on. This defines a subsequence which converges to x. (14) Let (X, d) be a metric space and K X. collection F of subsets of K has the finite intersection property (FIP) if for every n N, whenever F 1,..., F n F, then n k=1 F k. Show that if K has the property that every open cover of K has a finite subcover, then if F is a collection of closed subsets of K having the (FIP) property, then F. F F Solution: Let F be a collection of closed subsets of K having the (FIP) property. ssume by contradiction F F F =. Define G = {X \ F : F F}. It follows that G is an open cover of X, therefore of K. By assumption this implies ( that there ) are F 1,... F m in F such that K m j=1 (X \ F j) = X \ m j=1 F j. Since each F j is a subset of K, this implies m j=1 F j =, a contradiction since F has the (FIP) property. (15) Let (X, d) be a metric space and 1, 2,... X. Show that

5 (a) If B n = n i=1 i, then B n = n i=1 i. (b) If B = i=1 i, then B i=1 i. Solution: (a) Let x B n. Then there exists a sequence (x m ) m of elements in B n such that x m x. By the definition of B n, there exists j such that infinite many of the x i s belongs to j (since there are only finitely many j s). The subsequence made of those x i s must also converge to x, hence x j n i=1 i. Reciprocally, if x n i=1 i, say x j. Then there exists a sequence (x m ) m of elements in j converging to x. In particular the sequence is also of elements in B n, hence x B n too. This argument also proves (b). (16) Is the interior of a connected set always connected? Solution: Not necessarily. Consider two closed balls that touch tangentially and let M be their union. Then M is connected, but the interior of M is disconnected. (17) Let (X, d) be a metric space and f : X R be continuous. Show that {x X : f(x) = 0} is closed. Solution: Notice that {x X : f(x) = 0} = f 1 ({0}). Since f is continuous it must be closed. (18) Let I = [0, 1] R. Suppose f : I I is continuous. Prove that f(x) = x for at least one x. Solution: If f(0) = 0 or f(1) = 1 we are done. Otherwise consider g(x) = f(x) x. Since f(0) > 0 and f(1) < 1, g(0) > 0 and g(1) < 0. By the intermediate value theorem (since g is continuous), there must be x I such that g(x) = 0, that is f(x) = x. (19) Show that if X, Y, Z are metric spaces and f : X Y, g : Y Z are uniformly continuous, then g f is uniformly continuous. Solution: Let ε > 0. Since g is uniformly continuous there exists δ 1 > 0 such that if d Y (x, y) < δ 1, then d Z (g(x), g(y)) < ε. Since f is uniformly continuous, there exists δ > 0 such that if d X (a, b) < δ, then d Y (f(a), f(b)) < δ 1. Then d Z (g(f(a)), g(f(b)) < ε, concluding the proof.