Solutions to Tutorial 3 (Week 4)

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1 The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 3 (Week 4) MATH3961: Metric Spaces (Advanced) Semester 1, 2018 Web Page: Lecturer: Florica C. Cîrstea Material covered (1) The interior, closure and boundary of a set in a topological space (2) Convergence of sequences in a metric space; (3) Cauchy sequences in a metric space; (4) Limit points of a set in a metric space. Outcomes This tutorial helps you to (1) work with the concept of interior, closure and boundary of a set in a topological space; (2) work with the concept of limit points of a set in a metric space; (3) work with the closure of a set, interior of a set and bounded sets in a metric space. Summary of essential material Let (X, τ) be a topological space and A X. The interior of A, in short int(a), is given by The closure of A, denoted by A, is defined by int (A) := {Ω X : Ω τ, Ω A}. A := {F X : F c τ, A F }. The boundary of A, denote by A, is the closure of A without the interior of A, that is A := A \ int(a). Let A X, where (X, d) is a metric space. A point x X is called a limit point of A if for every r > 0, we have (B d (x; r) \ {x}) A. The set of all limit points of A is called the derived set of A and is denoted by A. Let (X, d) be a metric space. A sequence {x n } n 1 in X is called a Cauchy sequence if for every ε > 0, there exists n ε 1 such that d(x m, x n ) < ε for all m, n n ε. A set A in a metric space (X, d) is called bounded if and only if diam A <, where we define diam A := sup {d(x, y) : x, y A}. Copyright c 2018 The University of Sydney 1

2 Solutions to Tutorial 3 MATH3961: Metric Spaces Semester 1 1. Sketch (where possible) the following sets A, anddecidewhethera is an Lecturer: Laurentiu Paunescu subset, or a closed subset, or neither, of the appropriate space R n.thenforeach Questions A, findinta, toa complete and FrA. during the tutorial 1. (a) Sketch A = (where n N (n, possible) n +1), the N following = {0, 1, sets 2,...} 1. Sketch (where possible) the following sets A, anddecidewhethera A determine whether is ana is an subset, subset, or or Solution: aclosed closedsubset, subset, Theor or set neither, neither, A is as of of shown: the theappropriate appropriatespace spacer R n n (with.thenforeach the usual metric). Then, A, findinta, for each A A, andfind FrA. int(a), A and A. (a) A = (a) A = n N (n, n + 1), where N = {0, 1, 2,...} denotes the set of natural numbers. n N (n, n +1), N = 0 {0, 11, 2, 2...} Solution: TheSolution: set A is the setthe union A isset as of Ashown: is shown intervals below: and so it is. The complement X \ A is given by 0 X 1\ A =(, 2 3 0] 4{1, 2, 53,...}, 6 Theand set Ait is the not union since of anyintervals ball andcontaining, so it is. for The example, complement 1contains X \ points A is The given inset A, by Ai.e.,pointsoutsideX is as a union of \ A. Hence intervals. A is not Theclosed. complement of A is given by Since R \ AA is=, (, IntA X 0] \ A = N, =(, A. and TheclosureofA it0] is not {1, 2, 3, is...}, since A =[0, any ). Since interval X \ containing, A is for closed, example, it follows 1 contains that X points \ A = in X A, \ Ai.e., andpoints so outside R \ A. Hence A is not closed. and it is Since not A is, since we any have int A ball = containing, A. The closure for of example, A is A = 1contains [0, ). The boundary points in of A, A is i.e.,pointsoutsidex FrA = A (X given by A = A \ int(a) \ A. \ A) Hence ={0, = [0, ) A1, is 2, \ not 3, A = closed. }. {0, 1, 2, 3, } = N. (b) Since A A=setofallintegersinR is, IntA = A. TheclosureofA is A =[0, ). Since X \ A is closed, (b) Ait follows is the set that Z of X all \ A integers = X \ Ainand R. so Solution: Solution: The set A is sketched below: The set A is asfra sketched: = A (X \ A) ={0, 1, 2, 3, }. (b) A =setofallintegersinr Solution: The Since set A any is as sketched: interval containing, for example, 0 contains points outside A, itfollowsthata The set A is not is not since. any The complement of A is 3 2 X 1 \ A = interval containing, for example, 0 contains points outside A. The complement 0 of (n, 1An is+1), 2R\A 3= n Z (n, n+1), which is the union of intervals (n, n + 1). Thus Since any interval containing, forn Z R \ A is so that A is closed. Thus, we have example, 0 contains points outside A = A. We have int(a) = since any interval centered at any point x A A, itfollowsthata which is the union is not of. intervals The complement of A is X \ A (n, n +1). ThusX \ A is so that contains points outside A. The boundary of A is given by A = A \ int(a) = A. A is closed. (c) (n, n +1), Clearly A = IntA {(x 1, = x 2. ) SinceA R 2 : x 1 isx 2 closed, = 0}. A = A. Again it is clear that X \ A = R n Z andsolution: so FrA = A (X \ A) =A. which is the union of intervals (n, n +1). ThusX \ A is so that (c) A is A closed. = { Note that x 1 x (x 1,x 2 ) R 2 x 1 x 2 =0 } 2 = 0 implies either x 1 = 0 or x 2 = 0. Thus the set A is as shown below: Clearly Solution: IntA =. Note SinceA that is xclosed, 1 x 2 =0implieseitherx A = A. Again it is clear 1 =0orx that X 2 \ =0. A = RThus andthe so FrA set A= is A as(x shown \ A) below. =A. Clearly, any ball containing (0, 0) in A contains (c) A = { points (x 1,x 2 ) R 2 outside A. x 1 x 2 =0 } Hence A is not. Next A is the union of two closed subsets { (x 1,x 2 ) R 2 x 1 =0 } and { (x 1,x 2 ) R 2 x 2 =0 }. Solution: Therefore Note A is closed. that x 1 x 2 =0implieseitherx 1 =0orx 2 =0. Thus the set A is as shown below. Clearly, any ball containing (0, 0) in A contains points outside A. Hence A is not. Next A is the union of two Copyright c 2014 The University closed subsets { of Sydney 1 (x 1,x 2 ) R 2 x 1 =0 } and { (x 1,x 2 ) R 2 x 2 =0 }. Therefore A is closed. Clearly IntA =. Since A is closed, A = A. ItiseasilyseenthatX \ A = R 2 Copyright c 2014 The University of Sydney and so 1 Clearly, any ball containing (0, 0) in A contains points outside A. Hence A is FrA = A (X \ A) =A. (d) A = { not. Moreover, the interior (x 1,x 2 ) R 2 x 1 rational } of A is empty as any ball in R 2 centered at any point of A contains points not in A. Next, A is the union of two closed subsets { (x 1, x 2 ) R 2 x 1 = 0 } and { (x 1, x 2 ) R 2 x 2 = 0 }. Therefore, Solution: The set A cannot be sketched. Since any ball containing (x 1,x 2 )withx 1 rational, contains points (y 1,y 2 )withy 1 irrational (i.e. the set A is closed, which implies that A = A. The boundary of A is given by A = A \ int (A) = A. points outside A), it follows that A is not. Similarly, the complement X \ A of A is not and hence A is not closed. Using the same reason, we see that IntA 2 = and A = R 2. Also X \ A = R 2. Hence FrA = A (X \ A) =R 2.

3 (d) A = { (x 1,x 2 ) R 2 x 1 rational } Solution: The set A cannot be sketched. Since any ball containing (x 1,x 2 )withx 1 rational, contains points (y 1,y 2 )withy 1 irrational (i.e. points (d) A outside = {(x 1, A), x 2 ) it Rfollows 2 : x 1 that is rational}. A is not. Similarly, the complement X \ ASolution: of A is not The set anda hence cannot A is benot sketched. closed. Since any ball in R 2 centered Usingat the (x 1 same, x 2 ) reason, with x 1 we rational, see that contains IntA = points and (y A 1 =, yr 2 ) 2. with Alsoy 1 X irrational \ A = R 2 (i.e. points. Hence outside A), it follows that int (A) = and thus A is not. Similarly, any ball in R 2 centered FrA at (x= 1, Ax 2 ) (X with\ A) x 1 =R R 2 \. Q, contains points (y 1, y 2 ) with y 1 (e) A = { rational (x (i.e. points in A) so that the interior of A c is empty. This yields that A is 1, 0) R 2 0 <x 1 < 4 } not closed and A = (int(a c )) c = R 2. The boundary of A is A = A \ int (A) = R 2. Solution: The set A is as shown on the right. A is clearly not. The (e) complement A = {(x 1, 0) X R \ 2 A : of 0 A < x contains 1 < 4}. the point (0, 0). Since any ball containing Solution: (0, 0) contains The set Apoints is shown in A, below. itfollowsthatx \ A is not so that A is not closed. 0 4 Clearly IntA =, and The interior of A is empty = { since any (x 1, 0) R 2 ball centered 0 x 1 4 } at any point of A will contain. points outside A. The set A is clearly not since A int(a). The complement of A contains Now X \ A = R 2 the point (0, 0). Since any ball in R and so 2 centered at (0, 0) contains points in A, we see that R 2 \ A is not. Hence, A is not closed and FrA = A (X \ A) { (x 1, 0) R }. A = {(x 1, 0) R 2 : 0 x 1 4}. 2. Let (X, d) beametricspace. Therefore, we find that the boundary of A is A = A \ int (A) = A. (a) Let Y X be an subset in X. IfG Y is in Y,provethatG is 2. Let (X, inτ) X. be a topological space and let A X. Show that the following hold: Solution: (a) A is Since if and G only is if A in= Y int(a).,thereisansetu in X such that G = Solution: U Y. SinceIfYA is is, inthen X, AitfollowsthatG, is the biggest asanintersectionof subset of X contained in A, two that sets is A in = int X, (A). isinx. Conversely, if A = int(a), then A is since int (A) is. (b) Let (b) YA isxclosed be aif closed and only subset if A in= X. A. IfH Y is closed in Y,provethatH is closed in X. Solution: If A is closed, then A is the smallest closed subset of X containing A. Hence, A = A. Conversely, if A = A, then A is closed since A is closed. (c) A = A. 2 Solution: The closure of any set is a closed set, that is A is closed. Hence, from (b), we have A = A. (d) ( A c) c = int(a) and (int(a c )) c = A. Solution: By definition, we have A c = {F X : F c τ, A c F }. By taking the complement and using De Morgan s laws (denoting F c = Ω), we find that ( ) A c c = {Ω : Ω τ, Ω A} = int (A). (1) If we replace A by A c in (1), we obtain that int (A c ) = ( A ) c, or equivalently by applying the complement, we get (int(a c )) c = A. (e) A = A A c. Solution: By definition, we have A = A \ int (A) = A (int (A)) c. Using (d), we find that A = A A c as claimed. 3

4 3. Let (X, τ) be a topological space. Let A and B be subsets of X. Show the following: (a) A B = A B and int(a B) = int(a) int(b); Solution: We first prove that A B A B. By definition, A B is the smallest closed subset of X containing A B. As a union of two closed sets, A B is a closed set, containing A B since A A and B B. This yields A B A B. We show the other inclusion. Since A (respectively, B) is the smallest closed subset of X containing A (respectively, B), we have A A B and B A B. This proves that A B A B. For the second identity, recall that int (A B) is the largest subset of X contained in A B. As an intersection of sets, int (A) int (B) is an set contained in both A and B since int (A) A and int (B) B. Thus, we have int(a) int(b) int (A B). For the remaining inclusion, int (A B) is an set contained in both A and B. Since int (A) (resp., int (B)) is the largest subset of X contained in A (resp., B), we infer that int (A B) is a subset of both int (A) and int (B). This shows that int (A B) int (A) int (B). (b) A B A B and int(a) int(b) int (A B); Solution: As an intersection of closed sets, we have A B is a closed set, containing A B since A B A A and A B B B. Recall that A B is the smallest closed subset of X containing A B. This means that A B A B. In general, this inclusion is strict, see (c) below. The set int (A B) is the largest subset of X contained in A B. As a union of subsets, int(a) int(b) is an set, which is contained in A B since int (A) A and int (B) B. It follows that int(a) int(b) int (A B). In general, this inclusion is strict, see (c) below. (c) The following two statements are false. Find counterexamples for them. A B = A B and int(a) int(b) = int(a B). Solution: In general, A B A B and int(a B) int(a) int(b). To see this, take for example A = Q and B = R \ Q. Then A B =, which being a closed set yields that A B =, whereas A B = R R = R using that both Q and R \ Q are dense in R. Moreover, we have int (A) = int (B) = so that int (A B) = int (R) = R whereas int (A) int (B) =. (d) If A B, then A B and int(a) int(b). Solution: Let A B X. The set A is the smallest closed subset of X that contains A. Since B is a closed set containing B and thus A, it follows that A B. By definition, int (B) is the largest subset of X contained in B. Since int (A) is an set contained in A and thus in B, we deduce that int(a) int(b). 4. Let (X, d) be a metric space and let A X. Show that (a) int (A) = {x A : r > 0 such that B d (x; r) A}. Solution: By definition, int (A) = {Ω : Ω is and Ω A}. Let x int (A) be arbitrary. Hence, there exists Ω X an set such that x Ω A. Since Ω is an set in a metric space (X, d), there exists r > 0 such that B d (x; r) Ω A. Conversely, let x A be such that B d (x; r) A for some r > 0. Then, since B d (x, r) is an set and x B d (x; r), we find that x int (A). 4

5 (b) The set A is if and only if it is the union of balls in (X, d). Solution: If A is a union of balls in (X, d), then A is from the properties of sets and the fact that an ball is an set. Conversely, we assume that A X is. Then A = int (A) and using (a), we can write int (A) as x int (A) {B d (x; r) : r > 0 such that B d (x; r) A}. This shows that A is a union of balls in (X, d). (c) A = {x X : r > 0, B d (x; r) A }. Solution: Recall that A = (int (A c )) c, see Exercise 2(d). Using A c instead of A in (a), we get int (A c ) := {x A c : r > 0 such that B d (x; r) A c }. Applying the complement, we obtain that (int (A c )) c = A {x A c : r > 0, B d (x; r) A c }, which yields that (int (A c )) c = {x X : r > 0, B d (x; r) A } = A. 5. Let (X, d) be a metric space and let A X. Show the following: (a) A point x X is a limit point of A if and only if there exists a sequence of pairwise distinct points in A which converges to x in (X, d). Solution: Let x X. Assume that there exists a sequence {x n } n 1 of pairwise distinct points in A which converges to x in (X, d). Then, for every ε > 0, there n ε 1 such that x n B d (x; ε) for all n n ε. Observe that at most one element in the sequence {x n } can be x so that (B d (x; ε) \ {x}) A for all ε > 0. This shows that x is a limit point of A. Let x X be a limit point of A. Then, we have (B d (x; ε) \ {x}) A for every ε > 0. For ε 1 = 1, there exists x 1 (A \ {x}) B d (x; ε 1 ). By taking ε 2 = min{1/2, d(x 1, x)} > 0, we find x 2 (A \ {x}) B d (x; ε 2 ). Proceeding inductively, we construct a sequence {x n } n 1 in A \ {x} such that x n B d (x; ε n ) for ε n := min{1/n, d(x n 1, x)} for all n 2. This yields a sequence {x n } n 1 of pairwise distinct points in A which converges to x in (X, d) since ε n 0 as n. (b) A = A A, where A denotes the derived set of A. Solution: Define F := {x X : r > 0, B d (x; r) A }. Recall that A = F by Exercise 4(c). We show that F = A A. If x A, then for any r > 0, the set B d (x; r) A is non-empty since it contains x. Hence, we have A F. If x X \A, then x F if and only if (B d (x; r) \ {x}) A for every r > 0, or equivalently if and only if x is a limit point of A. This shows that F = A A. (c) The set A is closed. Solution: A set in a metric space is closed if and only if it contains all its limit points. To prove that A is closed, we show that whenever x X is a limit point of A, then x A. Indeed, x (A ) implies that (B d (x; r) \ {x}) A for any r > 0 fixed. Let y (B d (x; r) \ {x}) A. Since y A (that is, y is a limit point of A), we have (B d (y; ε) \ {y}) A for all ε > 0. Choose ε > 0 small such that B d (y; ε) B d (x; r) \ {x} (the latter is an set containing y). It follows that (B d (x; r) \ {x}) A. Since r > 0 was arbitrary, we find x A. Extra questions for further practice 6. Let (X, τ) be a topological space and let A X. (a) Show that the interior of A is the largest set contained in A in the following sense: int(a) is the unique V τ such that V A and if W τ with V W A, then V = W. 5

6 Solution: As a union of sets contained in A, the interior of A is an set contained in A. If W τ with int (A) W A, then we must have W int (A), which proves that int (A) is the (unique) largest subset of X contained in A. (b) Show that in general, A X does not have a largest closed set contained in A by proving the following: On (X = R, d(x, y) = x y ) for any closed set F in (0, 1) there exists a closed set G F with F G (0, 1). Solution: Let F be any closed set contained in (0, 1). Then clearly, 0 F and 1 F. Thus, 0 and 1 are not limit points of F. Hence, there exist r 1 > 0 and r 2 > 0 such that ( r 1, r 1 ) (F \ {0}) = and (1 r 2, 1 + r 2 ) (F \ {1}) =. Then, G = [r 1 /2, 1 r 2 ] satisfies G is closed and F G (0, 1). 7. Let (X, d) be a metric space and let A X. Then the distance from x to A is defined as dist (x, A) := inf {d(x, a) : a A}. Prove the following: (a) A = {x X : dist (x, A) = 0}; Solution: Denote F := {x X : dist (x, A) = 0}. Then, x X belongs to F if and only if for every ε > 0, there exists a ε A such that d(x, a ε ) < ε, or in other words if and only B d (x; ε) A for every ε > 0. Hence, F = A by Exercise 4(c). (b) int (A) = {x X : dist (x, A c ) > 0}. Solution: Recall that int (A) = ( A c) c, see Exercise 2(d). Using A c instead of A in (a), we find that A c = {x X : dist (x, A c ) = 0}. By taking the complement in both sides, we get that ( A c) c = {x X : dist (x, A c ) > 0}, proving the claim. 8. A set A in a metric space (X, d) is called bounded if and only if diam A <, where we define diam A := sup {d(x, y) : x, y A}. Prove the following: (a) A is bounded if and only if for every x A, there exists r > 0 so that A B d (x, r); Solution: Assume that for any x A fixed, there exists r > 0 so that A B d (x, r). Then for every y A, we have d(y, x) < r. Hence, the triangle inequality yields that d(y, z) d(y, x) + d(x, z) < 2r for all y, z A. Hence, diam A = sup {d(y, z) : y, z A} 2r <, proving that A is bounded. If A is bounded, then there exists r > 0 such that diam A < r, which yields that d(x, y) diam A < r for all x, y A. Hence, A B d (x; r) for every x A. (b) If A is a finite set, then A is bounded; Solution: If A is a finite set, then A = {a 1,..., a N } for some N 1. Then, we find that diam A = max {d(a i, a j ) : 1 i j N} <. (c) Every Cauchy sequence in (X, d) is a bounded set. Solution: Let {x n } n 1 be a Cauchy sequence in (X, d). Then, for every ε > 0, there exists n ε 1 such that d(x m, x n ) < ε for all m, n n ε. In particular, for ε = 1, there exists n 1 1 such that d(x m, x n ) < 1 for every m, n n 1. This shows that B := {x n : n n 1 } is bounded with diam B 1. Denote A := {x j : 1 j n 1 } so that {x n : n 1} = A B. The set A is finite and thus bounded by using (b). If x i, x j A, then d(x i, x j ) diam A. If x i, x j B, then d(x i, x j ) diam B. If x i A and x j B, then using the triangle inequality and x n1 A B, we get d(x i, x j ) d(x i, x n1 ) + d(x j, x n1 ) diam A + diam B. Thus, d(x i, x j ) diam A+diam B for all i, j 1. Then, diam (A B) diam A+diam B. Hence, {x n } n 1 is a bounded set. 6