A Course in Real Analysis. Saeed Zakeri

Transcription

1 A Course in Real Analysis Saeed Zakeri

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3 Contents Chapter 1. Topology of Metric Spaces I Basic definitions Convergence Continuity Completeness Compactness Connectedness 23 Problems 28 Chapter 2. Topology of Metric Spaces II Some extension results More on complete spaces Separability and local compactness Contraction maps The Hausdorff distance The Cantor set 53 Problems 60

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5 CHAPTER 1 Topology of Metric Spaces I 1.1. Basic definitions DEFINITION 1.1. A metric on a set X is a function d W X X! R which satisfies the following conditions for all p; q; x 2 X: d.p; q/ > 0 if p q; d.p; p/ D 0; d.p; q/ D d.q; p/; d.p; q/ d.p; x/ C d.x; q/. The pair.x; d/ is called a metric space. We think of d.p; q/ as the distance between p and q. The third condition of a metric is known as the triangle inequality. For convenience, we often call X itself a metric space, assuming a metric d is given on X. EXAMPLE 1.2. On the real line R, the usual distance d.p; q/ D jp qj is clearly a metric. EXAMPLE 1.3. More generally, on the n-dimensional Euclidean space R n, the Euclidean distance d.p; q/ D kp qk D! 1=2 nx.p i q i / 2 id1 between p D.p 1 ; : : : ; p n / and q D.q 1 ; : : : ; q n / defines a metric. The triangle inequality here follows from kx C yk kxk C kyk, which can be verified, for example, using the Cauchy-Schwarz inequality. Unless otherwise stated, we always assume that the metric on R n is the Euclidean distance. EXAMPLE 1.4. One can equip R n with other metrics such as d.p; q/ D nx id1 jp i q i j or d.p; q/ D max 1in jp i q i j: The resulting metric spaces are different, but in a sense very similar; see problem 10.

6 6 1 Topology of Metric Spaces I p q FIGURE 1. Two points p; q in a tree and the unique arc joining them. Here 4 < d.p; q/ < 5. EXAMPLE 1.5. Any non-empty set X can be made into a metric space by setting ( 1 if p q d.p; q/ D 0 if p D q: This is called the discrete metric on X. EXAMPLE 1.6. Let T be a topological tree, i.e., a connected graph without cycles. Identify each edge E of T with the unit interval Œ0; 1 R using some homeomorphism h E W E! Œ0; 1 which allows to define a distance on E: d.p; q/ D jh E.p/ h E.q/j for p; q 2 E: In general, for p; q 2 T define d.p; q/ as the length of the unique arc in T that joins p and q. It is easy to see that this is a metric on T. Note that for any choice of the h E the metric d satisfies the inequality n 1 < d.p; q/ n, where n is the number of distinct edges encountered in the unique arc between p and q (Fig. 1). EXAMPLE 1.7. For any set X, let B.X/ denote the space of all bounded functions X! R. Then d 1.f; g/ D sup jf.x/ x2x defines a metric on B.X/, often known as the uniform metric. g.x/j Observe that the second metric in Example 1.4 is the uniform metric if we identify R n with B.f1; 2; : : : ; ng/. EXAMPLE 1.8. Let C Œa; b denote the space of all continuous functions Œa; b! R. Then C Œa; b BŒa; b, so the uniform metric defined above turns C Œa; b into a metric space. As another example,

7 1.1 Basic definitions 7 consider d.f; g/ D Z b a jf.x/ g.x/j dx for f; g 2 C Œa; b which also defines a metric on C Œa; b. Notice that d.f; g/ D 0 implies f D g since the integral of a non-negative continuous function is positive unless the function is identically zero. On the other hand, on the space of all Riemann integrable functions Œa; b! R, the latter does not define a metric: A non-negative integrable function can have zero integral without being identically zero. In what follows,.x; d/ is a given metric space. All the drama is understood to be happening in this metric space. DEFINITION 1.9. Let p 2 X and r > 0. The set B.p; r/ D fx 2 X W d.x; p/ < rg is called the ball of radius r centered at p. A ball of radius r is also called an r-ball for short. The triangle inequality has the following useful formulation in terms of balls: LEMMA If q 2 B.p; r/ and 0 < s r d.p; q/, then B.q; s/ B.p; r/. In fact, if x 2 B.q; s/, then so x 2 B.p; r/. d.x; p/ d.x; q/ C d.q; p/ < s C d.p; q/ r; DEFINITION Let E X. A point p is called an interior point of E if B.p; r/ E for some r > 0. The set of all interior points of E is called the interior of E and is denoted by VE. We call E an open set if VE D E, i.e., if every point of E is an interior point of E. THEOREM Every ball is open. This is immediate from Lemma THEOREM For every set E, the interior VE is open. Moreover, VE is the largest open set that is contained in E: If G E and if G is open, then G VE.

8 8 1 Topology of Metric Spaces I PROOF. If p 2 VE, there is an r > 0 such that B.p; r/ E. For every q 2 B.p; r/ Lemma 1.10 shows that B.q; s/ B.p; r/ E for small enough s > 0, so q 2 VE. This proves B.p; r/ VE and shows VE is open. If G E and if G is open, then G D VG VE. THEOREM Arbitrary unions and finite intersections of open sets are open. The intersection of infinitely many open sets is not necessarily open, as the example T 1 n1 ; 1 n n D f0g on the real line shows. PROOF. Suppose fu g is a collection of open sets and U D S U. If p 2 U, then p 2 U for some index. Since U is open, there is an r > 0 such that B.p; r/ U, so B.p; r/ U. Thus, U is open. Now suppose fu i g 1in is a finite collection of open sets and U D T 1in U i. If p 2 U, then p is in every U i, so for each 1 i n there is an r i > 0 such that B.p; r i / U i. Let r D min 1in r i. Then r > 0 and B.p; r/ D T 1in B.p; r i/ U. This proves U is open. DEFINITION Let E X. A point p is called a limit point of E if B.p; r/ \ E ; for every r > 0. The set of all limit points of E is called the closure of E and is denoted by E. We call E a closed set if E D E, i.e., if every limit point of E is a point of E. THEOREM For every set E, the closure E is closed. Moreover, E is the smallest closed set that contains E: If E F and if F is closed, then E F. PROOF. Suppose p is a limit point of E and r > 0. Take any q 2 B.p; r/ \ E. By Lemma 1.10, B.q; s/ B.p; r/ for small enough s > 0. Since B.q; s/ \ E ;, this shows B.p; r/ \ E ;. Since r was arbitrary, this proves p 2 E. Thus, E is closed. If E F and if F is closed, then E F D F. Theorems 1.13 and 1.16 suggest a kind of duality between the notions of interior and closure. This is easily established by observing that p is an not interior point of E if and only if for every r > 0 the ball B.p; r/ meets the complement E c D X E. Thus, THEOREM The complement of the interior is the closure of the complement:. VE/ c D E c :

9 1.1 Basic definitions 9 COROLLARY E is open if and only if E c D X E is closed. This follows at once from the previous theorem: E is open E c D. VE/ c E c D E c E c is closed: COROLLARY Arbitrary intersections and finite unions of closed sets are closed. This follows from Theorem 1.14, Corollary 1.18, and the fact that for any family fe g of sets, \ c D [ E E c and [ c D \ E E c : THEOREM Let E R be non-empty and bounded above (resp. below). Then sup E (resp. inf E) belongs to E. In particular, if E is closed, it has a largest (resp. smallest) element. PROOF. Suppose E is bounded above, so z D sup E exists. For every r > 0 there is an x 2 E such that z r < x z (otherwise z r would be an upper bound for E smaller than z). This shows every r-ball centered at z meets E. Hence z 2 E. The proof when E is bounded below is similar. DEFINITION Let E X. A point p is called an accumulation point of E if every ball B.p; r/ contains a point of E other than p. The set of all accumulation points of E is called the derived set of E and is denoted by E 0. A point in E E 0 is called an isolated point of E. THEOREM (i) E D E [ E 0. Hence, every limit point of E is either an accumulation point or an isolated point of E. (ii) If p 2 E 0, every ball B.p; r/ contains infinitely many points of E. PROOF. (i) is clear from the definitions. For (ii), suppose B.p; r/ \ E is finite for some r > 0 and let p 1 ; : : : ; p n be the points in this intersection other than p. If s D min 1in d.p; p i /, then s > 0 and B.p; s/ \ E can contain at most one point, namely p. This contradicts p 2 E 0.

10 10 1 Topology of Metric Spaces I DEFINITION The diameter of a non-empty set E X is defined by diam E D sup d.p; q/ 2 Œ0; C1 : p;q2e We say E is bounded if diam E < C1. Equivalently, E is bounded if it is contained in some ball. THEOREM (i) diam E D 0 if and only if E has a single point. (ii) diam E diam F if E F. (iii) diam E D diam E. PROOF. Only (iii) deserves a comment: We have diam E diam E since E E. For the reverse inequality, take any pair p; q 2 E and an arbitrary " > 0. Find x 2 B.p; "/ \ E and y 2 B.q; "/ \ E. Then, d.p; q/ d.p; x/ C d.x; y/ C d.y; q/ diam E C 2": Taking the supremum over all such p; q shows diam E diam E C2". Letting "! 0 then gives diam E diam E. Suppose X is a metric space and Y is a subset of X. The metric d on X restricts to a metric on Y which turns it into a metric space. Sometimes we express this by saying that Y inherits its metric space structure from X. If p 2 Y and r > 0, and if for clarity we use the notations B X.p; r/ and B Y.p; r/ for the respective r-balls in X and Y, then (1) B Y.p; r/ D B X.p; r/ \ Y: The property of being open or closed for a set E Y generally depends on whether E is viewed as a subset of Y or a subset of X. For example, the unit interval.0; 1/ is open as a subset of R but it is not open as a subset of R 2. THEOREM Suppose E Y X. Then E is open (resp. closed) in Y if and only if E D H \ Y for some set H that is open (resp. closed) in X. PROOF. If E is open in Y, each p 2 E is the center of a ball B Y.p; r p / contained in E. The union H D S p2e B X.p; r p / is open in X and by (1), H \ Y D [ p2e.b X.p; r p / \ Y / D [ p2e B Y.p; r p / D E: Conversely, suppose E D H \ Y for some open set H in X. If p 2 E, then p 2 H, so B X.p; r/ H for some r > 0. It follows from (1) that B Y.p; r/ D B X.p; r/ \ Y E. Hence, E is open in Y.

11 1.2 Convergence 11 The claim on closed sets follows from this by taking the complements Convergence Recall that a sequence in a metric space X is a function p W N! X. We often denote the value p.n/ by p n and the entire sequence by fp n g n1 or simply fp n g. We say that fp n g is a sequence in E X if p n 2 E for all n. DEFINITION A sequence fp n g in X is said to converge to p 2 X if every ball centered at p contains p n for all large n. In other words, if for every " > 0 there is an integer N 1 such that d.p n ; p/ < " whenever n > N. In this case, we write lim n!1 p n D p or simply p n! p. Limits are unique: Suppose p n! p and p n! q and p q. The for every " > 0, the balls B.p; "/ and B.q; "/ must intersect since p n belongs to both for large n. This is impossible since by the triangle inequality B.p; "/ \ B.q; "/ D ; if 0 < " < 1 d.p; q/. 2 The next result justifies the terminology limit points for elements of E: THEOREM p 2 E if and only if there is a sequence fp n g in E such that p n! p. PROOF. The if part is trivial. For the only if part, suppose p 2 E. For each n 1, choose some p n 2 B.p; 1=n/ \ E. Evidently d.p n ; p/! 0 as n! 1, so p n! p. Notice that in the above statement, if p is an isolated point of E, then p n D p for all large n. COROLLARY A set E X is closed if and only if for every sequence fp n g in E, p n! p 2 X implies p 2 E. Convergence of sequences in R ties in nicely with the structure of the real line as an ordered field having the least upper bound property. DEFINITION The upper and lower limits of a sequence fp n g in R are defined, respectively, by lim sup p n D inf n!1 sup n1 kn p k and lim inf p n D sup inf p k : n!1 n1 kn

12 12 1 Topology of Metric Spaces I We often use the simpler notations p D lim sup p n and p D lim inf p n: n!1 n!1 Both p and p are well-defined quantities in Œ 1; C1, with p < 1 if and only if fp n g is bounded above and p > 1 if and only if fp n g is bounded below. Here is a more explicit description: For each n 1, consider the sequence tail L n D fp n ; p nc1 ; p nc2 ; : : :g and set x n D inf L n, y n D sup L n. Since L 1 L 2 L 3, the sequences fx n g and fy n g are increasing and decreasing, respectively. Hence p D sup x n D lim x n and p D inf y n D lim y n: n1 n!1 n1 n!1 Since x n y n for all n, the inequality p p holds. THEOREM For any sequence fp n g in R, lim n!1 p n p D p D p. D p if and only if PROOF. First suppose p n! p. Given ", find N 1 such that jp n pj < " whenever n > N. Then L n.p "; p C "/, hence L n Œp "; p C ". By Theorem 1.20, x n ; y n 2 L n, which shows jx n pj " and jy n pj " if n > N. Taking the limit as n! 1, we obtain jp pj " and jp pj ". Since " was arbitrary, we obtain p D p D p. Conversely, suppose p D p D p. Given " > 0 there is an N 1 such that jx n pj < " and jy n pj < " whenever n > N. This shows L n.p "; p C "/, or jp n pj < " whenever n > N. Thus, p n! p as n! Continuity Let us now take two metric spaces.x; d X / and.y; d Y / and a map f W X! Y between them. DEFINITION We say f W X! Y is continuous at p 2 X if for every " > 0 there is a ı > 0 such that d X.x; p/ < ı implies d Y.f.x/; f.p// < ". We say f is continuous on X if it is continuous at every point of X. The implication d X.x; p/ < ı H) d Y.f.x/; f.p// < " in the above definition can be expressed briefly as f.b X.p; ı// B Y.f.p/; "/. Thus, continuity of f at p means that sufficiently small balls centered at p map under f into arbitrarily small balls centered at f.p/. THEOREM The following conditions on f W X! Y are equivalent: (i) f is continuous at p 2 X.

13 1.3 Continuity 13 (ii) For every sequence fp n g in X, p n! p implies f.p n /! f.p/. (iii) For every set E X, p 2 E implies f.p/ 2 f.e/. (iv) For every set H Y, f.p/ 2 V H implies p 2.f 1.H //V. PROOF. (i) H) (ii): Take any " > 0 and find ı > 0 such that f.b X.p; ı// B Y.f.p/; "/. If p n! p in X, there is an N 1 such that p n 2 B X.p; ı/ whenever n > N. Hence, f.p n / 2 B Y.f.p/; "/ if n > N. This proves f.p n /! f.p/. (ii) H) (iii): If p 2 E, there is a sequence fp n g in E which converges to p. The sequence ff.p n /g in f.e/ then converges to f.p/, hence f.p/ 2 f.e/. (iii) H) (iv): Let E D f 1.H c /. If p is not an interior point of f 1.H /, every ball centered at p meets E, so p 2 E. It follows that f.p/ 2 f.e/ H c, so f.p/ HV by Theorem (iv) H) (i): Take any " > 0. The preimage of the set H D B Y.f.p/; "/ has p in its interior, so there is a ı > 0 such that B X.p; ı/ f 1.H /. This shows f.b X.p; ı// B Y.f.p/; "/. THEOREM The following conditions on f W X! Y are equivalent: (i) f is continuous on X. (ii) For every convergent sequence fp n g in X, lim n!1 f.p n/ D f. lim n!1 p n/. (iv) For every set E X, f.e/ f.e/. (iv) For every open set U Y, f 1.U / is open in X. (v) For every closed set F Y, f 1.F / is closed in X. PROOF. The equivalence of (i) through (iv) follows from applying Theorem 1.32 to every p 2 X. The equivalence of (iv) and (v) is a trivial consequence of the relation.f 1.F // c D f 1.F c /. COROLLARY If f W X! Y and g W Y! Z are continuous, so is the composition g ı f W X! Z. PROOF. We use the equivalence of (i) and (iv) in Theorem For every open set U Z, the preimage V D g 1.U / is open in Y. Hence the preimage f 1.V / D.g ı f / 1.U / is open in X. COROLLARY Suppose f; g W X! Y are continuous and f D g on a set E X. Then f D g on the closure E.

14 14 1 Topology of Metric Spaces I PROOF. If p 2 E, take a sequence fp n g in E converging to p. Then f.p/ D lim n!1 f.p n/ D lim n!1 g.p n/ D g.p/: As a special case, when E is dense in X in the sense that E D X, the relation f D g on E implies f D g everywhere on X. In other words, a continuous map is uniquely determined by its values on a dense subset of its domain. EXAMPLE Suppose f W R! R satisfies the Cauchy functional equation (2) f.x C y/ D f.x/ C f.y/ for all x; y 2 R: Let us explore some of the consequences of this equation: Setting x D y D 0 in (2) shows f.0/ D 0. Setting y D x in (2) and using f.0/ D 0 shows f. x/ D f.x/, so f is an odd function. Setting y D x and inductively applying (2) gives f.mx/ D mf.x/ for every positive integer m. Since f is odd, it follows that (3) f.mx/ D mf.x/ for all x 2 R and m 2 Z: Augustin-Louis Cauchy ( ) Setting x D 1=m in (3) gives f.1=m/ D f.1/ 1=m for every integer m 0. Setting x D 1=n back in (3) then gives f.m=n/ D mf.1=n/ D f.1/ m=n. Calling f.1/ D c, it follows that f.x/ D cx for all x 2 Q: Now suppose in addition that f is continuous. Since Q is dense in R, it follows from Corollary 1.35 that f.x/ D cx for all x 2 R. Thus, every continuous solution of the Cauchy functional equation is linear. This equation, however, has many discontinuous solutions as well. See?? 1.4. Completeness Once again, we assume.x; d/ is a given metric space. DEFINITION A sequence fp n g in X is a Cauchy sequence if for every " > 0 there is an integer N 1 such that d.p n ; p m / < " whenever n; m > N. Equivalently, if diamfp n ; p nc1 ; p nc2 ; : : :g! 0 as n! 1. Every convergent sequence is evidently a Cauchy sequence, but not vice versa. For example, f1=ng n1 is a Cauchy sequence in.0; 1/ R (equipped with the Euclidean distance), but it fails to converge to a point of.0; 1/. DEFINITION A set E X is complete if every Cauchy sequence in E converges to a point in E. THEOREM Every complete set is closed. Every closed subset of a complete set is complete.

15 1.4 Completeness 15 Thus, for subsets of a complete metric space, being closed is a necessary and sufficient condition for being complete. PROOF. Suppose E X is complete. If fp n g is a sequence in E which converges to some p 2 X, then fp n g is a Cauchy sequence in X, hence a Cauchy sequence in E. By completeness, p 2 E. Corollary 1.28 now proves that E is closed. Now let us assume F E is closed in E. Every Cauchy sequence fp n g in F is a Cauchy sequence in E, hence fp n g converges to some p 2 E. Another application of Corollary 1.28 then shows p 2 F. Thus, F is complete. THEOREM The real line R (equipped with the Euclidean distance) is complete. PROOF. Let fp n g be a Cauchy sequence in R. Set L n D fp n ; p nc1 ; p nc2 ; : : :g, so L 1 L 2 L 3 and diam L n! 0. For large n, diam L n < 1. Since the difference L 1 L n is finite, it follows that diam L 1 < C1, so fp n g is a bounded sequence. If, as before, we set x n D inf L n and y n D sup L n, it follows that 1 < x n y n < C1 for all n, hence the upper and lower limits p and p are real numbers. But p p D lim n!1.y n x n / D lim n!1 diam L n D 0; so fp n g converges to the common value p D p by Theorem REMARK In the above proof, completeness of R was deduced from the least upper bound axiom, which is also called completeness when R is viewed as an ordered field. This is an unfortunate double use of terminology since the least upper bound axiom is stronger than completeness of R viewed as a metric space. One can show that completeness of R together with the Archimedean property (for every " > 0 there is a natural number n such that 1=n < ") are equivalent to the least upper bound axiom. COROLLARY The Euclidean space R n is complete. PROOF. For each 1 i n, let i W R n! R denote the projection onto the i-th coordinate. Evidently, j i.p/ i.q/j kp qk for every p; q 2 R n. If fp k g is a Cauchy sequence in R n, it follows that f i.p k /g is a Cauchy sequence in R. Since R is complete, there is a q i 2 R such that i.p k /! q i as k! 1. It is then clear that p k!.q 1 ; : : : ; q n / 2 R n as k! 1. EXAMPLE It follows from Theorem 1.39 and Theorem 1.40 that a subset of R is complete if and only if it is closed. In particular, the set Q of rational numbers is not complete. Of course, this could be seen directly: The sequence f1; 1:4; 1:41; 1:414; 1:4142; 1:41421; 1:414213; : : :g of rational approximations to p 2 is a Cauchy sequence in Q which fails to converge to any point in Q.

16 16 1 Topology of Metric Spaces I EXAMPLE For any set X, the space B.X/ of all bounded functions X! R with the uniform metric (see Example 1.7) is complete. Take a Cauchy sequence ff n g in B.X/, so for every " > 0 there is an N 1 such that (4) sup jf n.x/ f m.x/j D d 1.f n ; f m / < " if n; m N: x2x This shows that ff n.x/g is a Cauchy sequence in R for each x 2 X. It follows from completeness of R that f.x/ D lim n!1 f n.x/ exists for each x 2 X. Now take any " > 0 and find N 1 such that (4) holds. Letting m! 1 in (4) gives (5) sup jf n.x/ f.x/j " if n N: x2x In particular, sup jf.x/j sup jf N.x/j C " < C1; x2x x2x which shows f 2 B.X/. Hence, by (5), d 1.f n ; f / < " if n N. Since " > 0 was arbitrary, we conclude that f n! f in B.X/. We conclude this section with a useful characterization of complete metric spaces: THEOREM 1.45 (Cantor). For a metric space X to be complete, it is necessary and sufficient that the following condition holds: For every sequence fe n g n1 of non-empty closed sets in X, with E 1 E 2 E 3 and diam E n! 0, the intersection T n1 E n is a single point. If we drop the assumption diam E n! 0, the intersection T n1 E n may be empty, even when X is complete (see problem 18). PROOF. First suppose X is complete and take a sequence fe n g with the given properties. Choose some p n 2 E n for each n 1. Given " > 0 there is an N 1 such that diam E n < " whenever n > N. If n > m > N, both p n ; p m belong to E m, so d.p n ; p m / diam.e m / < ". This proves that fp n g is a Cauchy sequence, and therefore converges to some p 2 X. We show T n1 E n D fpg. To this end, fix any n 1. For every r > 0 we can find an m > n such that p m 2 B.p; r/. Since E m E n, it follows that B.p; r/ \ E n ;. Since r > 0 was arbitrary, this shows p 2 E n D E n. Since this holds for every n 1, we conclude that p 2 T n1 E n. Finally, letting n! 1 in the trivial inequalities 0 diam T n1 E n diam En gives diam T n1 E n D 0, which shows this intersection cannot contain any point other than p. Conversely, suppose X is a metric space in which the given condition holds. Let fp n g be a Cauchy sequence in X and for each n 1 define L n D fp n ; p nc1 ; p nc2 ; : : :g and E n D L n : Evidently each E n is non-empty and closed, and E 1 E 2 E 3. Moreover, the Cauchy condition on fp n g implies diam L n! 0. Since diam E n D diam L n for each n, it is also true that diam E n! 0. By our hypothesis, T n1 E n is a single point

17 1.5 Compactness 17 p. Given " > 0, take an N 1 such that diam E n < " if n > N. Since p; p n 2 E n, it follows that d.p n ; p/ diam E n < " if n > N. This means p n! p. Thus, X is complete Compactness We continue assuming that.x; d/ is a metric space. DEFINITION A set K X is compact if every open cover of K has a finite subcover. Explicitly, this means for every collection fu g of open sets in X with K S U there are finitely many indices 1; : : : ; n such that K U 1 [ [ U n. Here is an equivalent formulation: THEOREM K X is compact if and only if for every collection ff g of closed sets in K with T F D ; there are finitely many indices 1; : : : ; n such that F 1 \ \ F n D ;. PROOF. By Theorem 1.25, each F is of the form E \ K for some closed set E in X. Then T F D ; if and only if fec g is an open cover of K, and \ \ D ; if and only if fe c 1; : : : ; E c ng is a subcover. F 1 F n Notice that the condition in Theorem 1.47 is formulated in terms of closed subsets of K only, irrespective of what is happening outside K. This shows that compactness is an intrinsic property: If K Y X, then K is compact as a subset of Y if and only if K is compact as a subset of X. EXAMPLE In R, the interval.0; 1/ is not compact since 1 n ; 1 1 is an open cover n n3 of it with no finite subcover. Similarly, the interval Œ0; 1/ is not compact since f. 1; n/g n1 is an open cover of it with no finite subcover. The set E D f2 n W n D 1; 2; : : :g is not compact either: The open cover 2.nC1/ ; 2.n 1/ of E has no finite subcover since each element of this cover n1 contains exactly one point of E, and E itself is infinite. However, the set K D E [ f0g is compact: For every open cover fu g of K there is an element Uˇ which contains 0. Since Uˇ is open and lim n!1 2 n D 0, there must be an N 1 such that 2 n 2 Uˇ if n > N. For each 1 i N there is a such that 2 U i i 2. It follows that K U i Uˇ [ [ [. U 1 U N THEOREM Every compact set is bounded and closed. PROOF. Let K X be compact. Fix p 2 X and for each n 1 let U n D B.p; n/. Evidently S n1 U n D X, so fu n g is an open cover of K. As U 1 U 2 U 3, compactness of K implies that K U n for some n. Hence K is bounded. To see K is closed, pick any p 2 K c. For each q 2 K, take a ball U q centered at p and a ball V q centered at q, both of radius < d.p; q/=2, so as to guarantee U q \V q D ;.

18 18 1 Topology of Metric Spaces I Since K S q2k V q, compactness of K implies that K V q1 [ [ V qn for some q 1 ; : : : ; q n 2 K. It follows that the ball U q1 \ \ U qn centered at p is disjoint from K. This shows p is an interior point of K c. Thus, K c is open. THEOREM Suppose K X is compact and F X is closed. Then K \ F is compact. PROOF. Let fu g be an open cover of K \ F. Then fu g together with the open set F c is an open cover of K. Hence K [ [ U 1 U n [ F c for some 1; : : : ; n. It follows that K \ F [ [ U 1 U n. COROLLARY Closed subsets of compact sets are compact. THEOREM Suppose fk g is a collection of compact sets in X in which every finite intersection is non-empty. Then T K ;. PROOF. Suppose T K D ;. Fix some Kˇ in the collection and set F D K \ Kˇ. Then each F is closed in Kˇ and T F D ;. By Theorem 1.47, \ \ D ; for some 1; : : : ; n. This implies F 1 F n Kˇ \ \ \ D ;, K 1 K n which contradicts our assumption. The main application of the preceding result is the following THEOREM 1.53 (Cantor). Suppose fk n g n1 is a sequence of non-empty compact sets in X, with K 1 K 2 K 3. Then T n1 K n ;. If in addition diam K n! 0, then T n1 K n is a single point. Despite the similarity of this result with Theorem 1.45, neither of the two versions of Cantor s theorem implies the other. PROOF. That K D T n1 K n is non-empty follows from Theorem For the second assertion, simply note that 0 diam K diam K n for each n. Letting n! 1 gives diam K D 0, which shows K is a single point. The next result shows that compactness is preserved under continuous maps of metric spaces. THEOREM Suppose X is compact and f W X! Y is continuous. Then the image f.x/ Y is compact. PROOF. Let fu g be an open cover of f.x/. Continuity of f shows that each f 1.U / is open in X. Hence ff 1.U /g is an open cover of X. Since X is compact, there are indices 1; : : : ; n such that X f 1.U 1/ [ [ f 1.U n/. This shows f.x/ [ [ U 1 U n, which proves compactness of f.x/.

19 1.5 Compactness 19 f [ ) 0 2 [) (1,0) FIGURE 2. A continuous bijective map from Œ0; 2/ onto the unit circle whose inverse is discontinuous at the point.1; 0/. Applying the above theorem to real-valued functions yields the extreme value theorem of classical analysis: COROLLARY Suppose X is compact and f W X! R is continuous. Then f takes on a maximum and a minimum value on X. PROOF. The image f.x/ is a compact subset of R by Theorem 1.54, so it is bounded and closed by Theorem It follows from Theorem 1.20 that sup f.x/ and inf f.x/ are both finite and belong to f.x/. Recall that f W X! Y is a homeomorphism if it is bijective (=one-to-one and onto), with both f and f 1 continuous. Continuity of the inverse is an essential assumption here. For example, let T be the unit circle x 2 C y 2 D 1 in the plane and f W Œ0; 2/! T be defined by f.t/ D.cos t; sin t/. It is easy to see that f is bijective and continuous, but f 1 is discontinuous at the point.1; 0/ 2 T (Fig. 2). THEOREM Suppose f is a continuous one-to-one map from a compact metric space X onto a metric space Y. Then f is a homeomorphism. PROOF. f is a bijective map X! Y, so it suffices to prove that f 1 W Y! X is continuous. Let F X be closed. Then F is compact by Corollary 1.51, so f.f / Y is compact by Theorem 1.54, which implies f.f / is closed. Continuity of f 1 now follows from the equivalence (i) (v) in Theorem Another useful property of continuous maps on compact spaces is related to the following

20 20 1 Topology of Metric Spaces I DEFINITION A map f W X! Y between metric spaces is uniformly continuous on X if for every " > 0 there is a ı > 0 such that p; q 2 X and d X.p; q/ < ı imply d Y.f.p/; f.q// < ". The point of uniform continuity is that the choice of ı corresponding to each " can be made independent of the pair p; q. It is easy to see that uniform continuity of f W X! Y is equivalent to the following: For every " > 0 there is a ı > 0 such that E X and diam E < ı imply diam f.e/ < ". The following example shows that uniform continuity is indeed stronger than continuity. EXAMPLE The quadratic function f W R! R defined by f.x/ D x 2 is continuous on R but not uniformly continuous there: No matter what ı > 0 we pick, the distance jf.x C ı/ f.x/j D j2ıx C ı 2 j gets arbitrarily large if x is large enough. EXAMPLE A map f W X! Y between metric spaces is called Hölder of exponent > 0 if there is a constant C > 0 such that d Y.f.p/; f.q// C.d X.p; q// for all p; q 2 X: (A Hölder map of exponent D 1 is often called a Lipschitz map.) It is easy to see that every Hölder map is uniformly continuous: For a given " > 0, choosing ı D."=C / 1= will do the trick. THEOREM Suppose f W X! Y is continuous. If X is compact, then f is uniformly continuous on X. PROOF. Let " > 0 be given. By continuity, for each p 2 X there is a ı p > 0 such that f.b X.p; 2ı p // B Y.f.p/; "=2/. The collection fb X.p; ı p /g p2x is an open cover for X, so by compactness X B X.p 1 ; ı p1 / [ [ B X.p n ; ı pn / for some p 1 ; : : : ; p n 2 X: Set ı D min 1in ı pi > 0. Take any p; q 2 X with d X.p; q/ < ı. Find 1 i n such that d X.p; p i / < ı pi. Then d X.q; p i / d X.q; p/ C d X.p; p i / < ı C ı pi 2ı pi. Since p; q are both in B X.p i ; 2ı pi /, it follows that f.p/; f.q/ both belong to B Y.f.p i /; "=2/. Thus, by the triangle inequality, d Y.f.p/; f.q// < ". The rest of this section will discuss several properties that are equivalent to compactness, culminating in an important characterization of compact sets in Euclidean spaces.

21 1.5 Compactness 21 DEFINITION We say that a set K X has the Bolzano-Weierstrass property if every infinite subset of K has an accumulation point in K. is sequentially compact if every sequence in K has a subsequence which converges in K. is totally bounded if for every r > 0 one can cover K by finitely many r-balls. A totally bounded set is clearly bounded but not vice versa. EXAMPLE In the space BŒ0; 1 of all bounded functions Œ0; 1! R with the uniform metric, the closed unit ball E D ff 2 BŒ0; 1 W d 1.f; 0/ 1g is bounded but not totally bounded. To see this, consider the characteristic functions f n D Œ0;1=n for n D 1; 2; 3; : : : which all lie in E and have pairwise distance 1. As each 1=2-ball can contain no more than one f n, any collection of 1=2-balls covering E must be infinite. THEOREM The following conditions on a set K X are equivalent: (i) K is compact. (ii) K has the Bolzano-Weierstrass property. (iii) K is sequentially compact. (iv) K is complete and totally bounded. Bernhard Placidus Johann Nepomuk Bolzano ( ) PROOF. (i) H) (ii): Let E K be an infinite set with no accumulation point in K. Then every p 2 K is the center of a ball U p such that U p \ E D ; or fpg. The collection fu p g p2k is an open cover of K, so K U p1 [ [ U pn for some p 1 ; : : : ; p n 2 K. Taking the intersection with E now gives E fp 1 ; : : : ; p n g, contradicting that E is infinite. (ii) H) (iii): Suppose fp n g is a sequence in K. If the range of this sequence is finite, there is clearly a convergent subsequence. If infinite, the sequence must have an accumulation point p 2 K. For each i 1 choose a p ni such that d.p ni ; p/ < 1=i. Then p ni! p as i! 1. (iii) H) (iv): Take a Cauchy sequence fp n g in K. By sequential compactness, there is a subsequence fp ni g which converges to some p 2 K as i! 1. Given " > 0 find an N 1 such that d.p n ; p m / < " whenever n; m > N. Choose i so large that n i > N and d.p ni ; p/ < ". Then, if n > N, d.p n ; p/ d.p n ; p ni / C d.p ni ; p/ < 2": This shows p n! p. Thus, K is complete.

22 22 1 Topology of Metric Spaces I Heinrich Eduard Heine ( ) To show K is totally bounded, assume by way of contradiction that there is an r > 0 with the property that no finite collection of r-balls can cover K. Pick any p 1 2 K. Since B.p 1 ; r/ does not cover K, we can pick some p 2 2 K B.p 1 ; r/. In general, having chosen the points p 1 ; : : : ; p n, since the balls B.p 1 ; r/; : : : ; B.p n ; r/ do not cover K, we can pick some p nc1 2 K S 1in B.p i; r/. The sequence fp n g constructed inductively in this fashion has the property that d.p n ; p m / r whenever n m. As such, it can have no convergent subsequence. This contradicts sequential compactness of K. (iv) H) (i): Suppose there is an open cover fu g of K with no finite subcover. By total boundedness, we can cover K by finitely many 1=2-balls. Among them, there must be at least one ball B 1 such that B 1 \ K is not covered by finitely many of the U. Next, we cover K by finitely many 1=4-balls. Among those meeting B 1 \ K, there must be at least one ball B 2 such that B 2 \ K is not covered by finitely many of the U. This process continues inductively and gives a sequence fb n g of balls such that B n has radius 1=2 n, B n \ K meets B nc1 \ K, and B n \ K is not covered by finitely many of the U. For each n 1 take a point p n in B n \ K. It is easy to estimate d.p n ; p nc1 /: Take any x 2 B n \ B nc1 \ K and use diam.b n \ K/ 1=2 n 1 to obtain d.p n ; p nc1 / d.p n ; x/ C d.x; p nc1 / < 1 2 C 1 n 1 2 D 3 n 2 : n Thus, if n > m 1, Xn 1 Xn 1 1X d.p m ; p n / d.p i ; p ic1 / < 3 2 i < 3 2 i D 6 2 m : idm Given " > 0, choose N large enough so 2 N < ". The above estimate then shows that d.p m ; p n / < 6" whenever n; m > N. Thus, fp n g is a Cauchy sequence in K. Since K is complete, fp n g converges to some p 2 K. The limit p lies in U for some, and there is an r > 0 such that B.p; r/ U. If we choose n large enough to assure d.p n ; p/ < r=2 and diam.b n \ K/ < r=2 (the latter being possible since diam.b n \ K/! 0), it follows that B n \ K U. This is impossible by the choice of the B n. idm idm THEOREM 1.64 (Heine-Borel). A subset of bounded and closed. R n is compact if and only if it is PROOF. The only if part follows from Theorem For the if part, suppose K R n is bounded and closed. Choose a > 0 large enough so that K is contained in the n-dimensional box Πa; a n D.p 1 ; : : : ; p n / 2 R n W jp i j a for all 1 i n :

23 1.6 Connectedness 23 For each integer k > 0, express Œ a; a R as the union of k closed subintervals I 1 ; : : : ; I k, each of length ` D 2a=k. The box Œ a; a n will then be the union of k n smaller boxes I i1 I i2 I in for all choices 1 i 1 ; : : : ; i n k, each having diameter p n`. Given r > 0, choose k so large that p n` D 2a p n=k < r. Then K is contained in a finite union of boxes of diameter < r, and each such box is clearly contained in an r-ball. This shows K is totally bounded. Since K is a closed subset of the complete space R n, it must be complete (Theorems 1.39 and 1.40). It follows from Theorem 1.63 that K is compact. REMARK Bounded and closed sets may fail to be compact in others metric spaces. For example, any infinite set equipped with the discrete metric is bounded and closed, but not compact. Another example is the closed unit ball in BŒ0; 1 which fails to be compact since it is not totally bounded (see Example 1.62). COROLLARY 1.66 (Bolzano-Weierstrass). Every bounded infinite set in R n has an accumulation point. PROOF. Suppose E R n is bounded. Then E is closed and bounded since diam E D diam E. By Theorem 1.64, E is compact. By Theorem 1.63, E has the Bolzano-Weierstrass property. Hence, its infinite subset E must have an accumulation point Connectedness It will be convenient to call a subset A of a metric space X clopen if A is both closed and open. Equivalently, if both A and X A are open. The two subsets ; and X are trivially clopen, so we call any other clopen subset of X non-trivial. DEFINITION A metric space X is connected if ; and X are its only clopen subsets. X is called disconnected if it is not connected, i.e., if it has a non-trivial clopen subset. Before going further, let us comment on the definition of connectedness when applied to a subset E X. By definition, E is connected if it has no non-trivial clopen subsets. Being clopen in E means being both closed and open in E when we consider E as a metric space it its own right, i.e., with the metric space structure it inherits from X. This shows that, just like compactness, connectedness is an intrinsic property: If E Y X, then E is connected as a subset of Y if and only if E is connected as a subset of X. A non-trivial clopen set A E and its complement B D E A are separated in the sense that A \ B D A \ B D ;, a property that is much stronger than mere disjointness of A; B. To see this, simply note that A \ E D A since A is closed in E, so A \ B D ;. Similarly, B \ E D B since B is also closed in E, so A \ B D ;.

24 24 1 Topology of Metric Spaces I Conversely, suppose E is the union of two non-empty separated sets A; B. Since A\B D ;, we have A\E A. This, together with the reverse inclusion A\E A, which holds trivially, prove that A \ E D A. Thus, A is closed in E by Theorem The same argument applied to B proves that B is closed in E. This shows A (as well as B) is a non-trivial clopen set in E. We have shown COROLLARY A subset of a metric space is connected if and only if it is not the union of two non-empty separated sets. The formulation of connectedness in terms of separated sets may be easier to work with when recognizing clopen subsets of E seems tricky (for example if E itself is a complicated subset of X). EXAMPLE The punctured line R f0g is disconnected since it is the union of the separated sets. 1; 0/ and.0; 1/. It is not hard to characterize connected subsets of the real line: THEOREM A subset of R is connected if and only if it is an interval. PROOF. It will be convenient to use the following characterization of intervals: I R is an interval if and only if a < c < b and a; b 2 I imply c 2 I. If E R is not an interval, find a < c < b such that a; b 2 E but c E. Then A D E \. 1; c/ and B D E \.c; 1/ are separated and their union is E. Hence, E is disconnected. Conversely, suppose E R is disconnected, so it is the union of two non-empty separated sets A; B. Fix a 2 A and b 2 B. Assume without loss of generality that a < b. We prove E is not an interval by finding a point in.a; b/ which is not in E. Define c D sup.a \ Œa; b /. Then c 2 A, hence c B. In particular, a c < b. If c A, then a < c < b and c E. If c 2 A, then c B, so there is an " > 0 such that c C " < b and c C " B. Since c C " A by the definition of c, we conclude that c C " E. This completes the proof. THEOREM (i) If E is non-empty and connected and E L E, then L is connected. In particular, E is connected. (ii) If S fe g is a family of non-empty connected sets with E is connected. T E ;, then PROOF. (i) Assume L is disconnected, A is a non-trivial clopen set in L, and B D L A. The set A 1 D A \ E is clopen in E by Theorem If A 1 D ;,

25 1.6 Connectedness 25 then E B, so A L E B. Using the fact that B is closed in L, we obtain A D A \ L B \ L D B, which is impossible. It follows that A 1 ;, and a similar argument shows A 1 E. Hence A 1 is a non-trivial clopen set in E, contradicting connectedness of E. (ii) Fix some p 2 T E. Suppose S E is disconnected and A is a non-trivial clopen set in it. Without loss of generality, assume p 2 A (otherwise, we can replace A with its complement in T E ). By Theorem 1.25, for each the intersection A D A \ E is clopen in E. Moreover, A ; since p 2 A. Connectedness of E now shows that A D E, or E A for every. This implies S E D A, which is a contradiction. As to the intersection of connected sets, there is nothing that we can say in general: The two subsets E 1 D T f.1; 0/g and E 2 D T f. 1; 0/g of the unit circle T D f.x; y/ 2 R 2 W x 2 C y 2 D 1g are connected since they are the images of the intervals.0; 2/ and. ; / under the continuous map t 7!.cos t; sin t/. But E 1 \ E 2 is clearly disconnected. There is one situation, however, in which the intersection of connected sets turns out to be connected: THEOREM Suppose fk n g n1 is a sequence of non-empty connected compact sets in a metric space, with K 1 K 2 K 3. Then T n1 K n is connected. Compactness of each K n is an essential assumption here; see problem 34. PROOF. Let K D T n1 K n. By Theorem 1.53, K ;. Suppose, to reach a contradiction, that K is disconnected. Take a non-trivial clopen set A K and set B D K A. Since A; B are closed subsets of the compact set K, they are compact. We claim there is an " > 0 such that d.p; q/ " for every p 2 A and q 2 B. Otherwise, for each n 1 there are points p n 2 A and q n 2 B such that d.p n ; q n / < 1=n. By compactness of A, there is a subsequence fp ni g which converges to some p 2 A as i! 1. The corresponding subsequence fq ni g then tends to the same limit p. This gives p 2 A \ B D A \ B, which is impossible since A \ B D ;. Now with " > 0 as above, let U D S p2a B.p; "=2/ and V D S p2b B.p; "=2/. Then U and V are disjoint open sets containing A and B, respectively. The sets A n D K n V and B n D K n U are therefore non-empty closed subsets of K n. Since A n [ B n D K n, connectedness of K n shows that A n \ B n ;. For each n 1, choose some x n 2 A n \ B n. The sequence fx n g lies in the compact set K 1, so it has a subsequence fx ni g which converges to some x 2 K 1 as i! 1. Since x ni 2 K n if n i > n, we have x 2 K n D K n for every n 1. It follows that the limit x is in K, and therefore d.x n ; x/ " for all n since x n U [ V. This contradicts x ni! x.

26 26 1 Topology of Metric Spaces I DEFINITION Let E be a non-empty set in a metric space. For each p 2 E, the union C p of all connected subsets of E which contain p is called the connected component of E containing p. We say that E is totally disconnected if it has more than one point and C p D fpg for every p 2 E. Notice that C p itself is connected by Theorem 1.71(ii), so it can be described as the largest connected subset of E containing p. It follows in particular that E is connected if and only if it has a single connected component. THEOREM (i) For every p; q 2 E, either C p \ C q D ; or C p D C q. In particular, E is the disjoint union of its connected components. (ii) For every p 2 E, C p is closed in E. (iii) If E has finitely many connected components, for every p 2 E, C p is clopen in E. PROOF. If C p \ C q ;, then C p [ C q is connected by Theorem 1.71(ii), so it must be contained in both C p and C q. This shows C p D C q, which proves (i). For (ii), since C p C p \ E C p, Theorem 1.71(i) shows that C p \ E is connected. By maximality of C p, we have C p D C p \ E. It follows from Theorem 1.25 that C p is closed in E. Finally, suppose E has finitely many connected components. Pick one, say C p. The complement E C p is the union of the finitely many remaining connected components (possibly empty). As each connected component is closed in E by (ii), E C p is also closed in E. It follows that C p is open in E. This proves (iii). EXAMPLE Here are a few examples in R: The set R f0g has two connected components, namely. 1; 0/ and.0; 1/. The set E D S 1 n1 2n ; 1 2n 1 [ f0g has countably many connected components: Each interval in the union is a connected component of E which is clopen in E. The singleton f0g is a connected component of E which is closed but not open in E. The sets Q of rational numbers and I D R Q of irrational numbers are both totally disconnected. Beyond the real line, it is rather difficult to prove connectedness of a set using the definition. The following result provides an indirect method to achieve this: THEOREM Suppose X is connected and f W X! Y is continuous. Then f.x/ is connected. PROOF. Suppose f.x/ is disconnected, and take a non-trivial clopen set A in f.x/. Continuity of f W X! f.x/ together with Theorem 1.33 show that f 1.A/

27 1.6 Connectedness 27 p q p p 0 FIGURE 3. A convex set (left) vs. a star-like set (right). is clopen in X. Moreover, since A ; and A f.x/, it is clear that f 1.A/ is neither ; nor the whole X. This contradicts our assumption that X is connected. Applying the above theorem to real-valued functions yields the intermediate value theorem of classical analysis: COROLLARY Suppose X is connected and f W X! R is continuous. If p; q 2 X, and if f.p/ < c < f.q/, then f takes on the value c somewhere in X. PROOF. The image f.x/ is a connected subset of R by Theorem 1.76, hence it is an interval by Theorem Since f.p/; f.q/ 2 f.x/, it follows that c 2 f.x/. There is a stronger notion of connectedness which is often easier to verify than connectedness itself: DEFINITION Let E be a subset of a metric space. A path in E is a continuous map W Œ0; 1! E. We often say that is a path from.0/ 2 E to.1/ 2 E. We call E path-connected if for every pair of points p; q 2 E there is a path in E from p to q. EXAMPLE For p; q 2 R n, the segment Œp; q is the set of all points in R n of the form.1 t/p C tq for some 0 t 1. A set E R n is convex if p; q 2 E implies Œp; q E (Fig. 3). A convex set E is clearly path-connected: Given p; q 2 E, the affine map.t/ D.1 t/p C tq is a path in E from p to q. More generally, suppose E R n is star-like, that is, suppose there is a base point p 0 2 E such that Œp 0 ; p E for every p 2 E (Fig. 3). Then E is path-connected. In fact, given p; q 2 E, there is a path in E from p to q whose image is the segment Œp; p 0 followed by Œp 0 ; q. For example, W Œ0; 1! E defined by (.1 2t/p C 2tp 0 if 0 t 1=2.t/ D.2 2t/p 0 C.2t 1/q if 1=2 t 1

28 28 1 Topology of Metric Spaces I is such a path. THEOREM Every path-connected set is connected. The converse statement is not true; see problems 36 and 37. PROOF. Suppose E is path-connected and fix p 2 E. For each q 2 E, there is a path q in E from p to q. The image q.œ0; 1 / E is connected by Theorem The union of the collection f q.œ0; 1 /g q2e is E and its intersection is non-empty since it contains p. It follows from Theorem 1.71(ii) that E is connected. EXAMPLE The Euclidean space R n is connected because it is convex and therefore pathconnected. For each a 2 R, the slit-plane R 2. 1; a/ is connected because it is star-like (take the base point to be p 0 D.a C 1; 0/), hence path-connected. Problems (1) In every metric space, show that the set fx W d.x; p/ rg is closed and contains B.p; r/ (but this inclusion can be proper). Conclude that B.p; r/ B.p; s/ if r < s. (2) Give an example of a metric space in which there are balls B.p; r/ and B.q; s/ such that r > s but B.p; r/ B.q; s/. (3) Let X be a metric space and E X. For r > 0, define the r-neighborhood of E by N r.e/ D fp 2 X W d.p; q/ < r for some q 2 Eg: Show that N r.e/ is an open set containing E. What is T r>0 N r.e/? (4) Show that every closed set in a metric space is a G ı -set, i.e., a countable intersection of open sets. Conclude that every open set in a metric space is an F -set, i.e., a countable union of closed sets. (5) (i) Prove that for any collection fe g of sets in a metric space X, [ \ and E : E [ E E \ Give examples in which these inclusions are proper. (ii) Prove that for any finite collection fe 1 ; : : : ; E n g of sets in X, [ E i D [ E i : 1in (6) Let E be a set in a metric space X. 1in (i) The boundary of E, denoted is the set of p 2 X such that B.p; r/ meets both E and E c for every r > 0. Verify the D E \ E c and E D VE in is closed. (ii) Can the unit interval Œ0; 1 in R be the boundary of a set? What if one replaces Œ0; 1 with other closed sets in R? In R n? In more general metric spaces?

29 1.6 Problems 29 (7) Prove that for any set E in a metric space, the derived set E 0 is closed. Give an example of a closed set E R for which E 0 and E 00 D.E 0 / 0 are non-empty and the inclusions E 00 E 0 E are proper. (8) For two metrics d; d 0 on a set X, we write d d 0 if d 0.p; q/! 0 implies d.p; q/! 0; more precisely, if for every " > 0 there is a ı > 0 such that p; q 2 X and d 0.p; q/ < ı imply d.p; q/ < ". We say that d; d 0 are equivalent if d d 0 and d 0 d. (i) Prove that if d d 0, every open set in.x; d/ is also open in.x; d 0 /. Conclude that equivalent metrics define the same open (hence closed) sets. (ii) Give an example where d d 0, but some open set in.x; d 0 / is not open in.x; d/. (iii) If.X; d 0 / is compact and d d 0, prove that d; d 0 are equivalent. (Hint: The condition d d 0 means the identity map.x; d 0 /!.X; d/ is uniformly continuous. For (iii), use Theorem 1.56) (9) Let d be any metric on a set X. Define, for p; q 2 X, d 0.p; q/ D d.p; q/ 1 C d.p; q/ and d 00.p; q/ D min d.p; q/; 1 : Show that d 0 ; d 00 are metrics on X with respect to which all sets have diameter at most 1. Prove that d; d 0 ; d 00 are equivalent metrics, so.x; d/,.x; d 0 / and.x; d 00 / have the same open (hence closed) sets. (10) Suppose.X i ; d i /, i D 1; : : : ; n, are metric spaces. Verify that each of the following defines a metric on the Cartesian product X D Q n id1 X i: For p D.p 1 ; : : : ; p n / and q D.q 1 ; : : : ; q n / in X, d.p; q/ D max d i.p i ; q i /; 1in nx d 0.p; q/ D d i.p i ; q i /; d 00.p; q/ D id1! 1=2 nx d i.p i ; q i / 2 : id1 Show that these metrics are equivalent by proving the inequalities d.p; q/ d 00.p; q/ d 0.p; q/ nd.p; q/ for all p; q 2 X. As a special case, the metrics max jp i q i j; 1in nx jp i q i j; id1! 1=2 nx.p i q i / 2 on R n are equivalent, and the topological notions of convergence, continuity, compactness, and connectedness would be independent of which of these metrics is used. (11) Let fp n g be a bounded sequence in R and E be the set of all x 2 R for which there is a subsequence of fp n g converging to x. Verify that E is closed, and that sup E D lim sup n!1 p n and inf E D lim inf n!1 p n (see Definition 1.29). (12) Show that.x; y/ 7! x C y and.x; y/ 7! xy are continuous functions on R 2. Show that.x; y/ 7! x=y is continuous on R 2 f.x; y/ W y 0g. What do these results say about sums, products, and ratios of continuous real-valued functions on a metric space? id1

30 30 1 Topology of Metric Spaces I (13) Let C Œa; b be the space of all continuous functions Œa; b! R with the uniform metric. Is the map I W C Œa; b! R defined by I.f / D R b a f.x/ dx continuous? Uniformly continuous? (14) Let X be a metric space. (i) For each p 2 X, show that x 7! d.x; p/ is continuous on X. (ii) Show that.x; y/ 7! d.x; y/ is continuous on X X. Here you can put any of the equivalent metrics on X X as in problem 10. (15) In a metric space X, show that p n! p if and only if f.p n /! f.p/ for every continuous function f W X! R. (16) Suppose fp n g and fq n g are Cauchy sequences in a metric space. Show that the sequence fd.p n ; q n /g converges. (17) Let fp n g be a sequence in a metric space. If 1X d.p n ; p nc1 / < 1; nd1 show that fp n g is a Cauchy sequence. (Application: In a complete metric space, if d.p n ; p nc1 /! 0 geometrically fast, then fp n g converges.) (18) Give an example of a sequence E 1 E 2 E 3 of non-empty closed sets in a complete metric space for which T n1 E n D ;. This shows that the condition diam E n! 0 in Theorem 1.45 cannot be dispensed with (contrast this with Theorem 1.53). (19) Does there exist a metric d on R such that.r; d/ is an incomplete metric space? (20) A metric d on a set X is called an ultra-metric if it satisfies the stronger form of the triangle inequality d.x; y/ maxfd.x; z/; d.z; y/g for all x; y; z 2 X: Suppose d is an ultra-metric on X. (i) Show that if d.x; z/ d.z; y/, then d.x; y/ D maxfd.x; z/; d.z; y/g. This can be interpreted as saying that all triangles are isosceles. (ii) Show that fp n g is a Cauchy sequence in.x; d/ if and only if d.p n ; p nc1 /! 0. (21) Let X be the space of all sequences P D fp n g n1 in a given set. For P D fp n g and Q D fq n g in X, let n.p; Q/ be the smallest index n 2 N for which p n q n (as usual, we set n.p; Q/ D 1 if there is no such n). Define d.p; Q/ D 1 n.p; Q/ : Show that d is an ultra-metric on X and the resulting metric space.x; d/ is complete. (22) Give a direct proof for compactness of Œa; b R by completing the following sketch: Let fu g be an open cover of Œa; b and define E D fx 2 Œa; b W Œa; x is covered by finitely many of the U g: Show that sup E D b 2 E. (23) In a metric space, let K be a compact subset of an open set U. Show that N ".K/ U if " > 0 is sufficiently small. Here N ".K/ is the "-neighborhood of K defined in problem 3.

31 1.6 Problems 31 (24) The previous result has a useful generalization in which a single open set is replaced by an arbitrary open cover. Let K be a compact set in a metric space and fu g be an open cover of K. Show that there is an " > 0 such that if p 2 K, then B.p; "/ U for some. Such " is called a Lebesgue number for the cover fu g. Give an example of an open cover of a non-compact subset of R which does not have a Lebesgue number. (25) Give an alternative proof of Theorem 1.60 using the fact that X is sequentially compact. (Hint: Assuming the claim is false, there is an " > 0 and sequences fp n g, fq n g in X such that d X.p n ; q n / < 1=n but d Y.f.p n /; f.q n // " for every n 1.) (26) Let E R be bounded but not closed. Find a continuous function f W E! R which is (i) unbounded; (ii) bounded but without a maximum; (iii) not uniformly continuous. (27) Uniformly continuous functions on R grow at most linearly: Suppose f W R! R is uniformly continuous on R. Prove that there are constants a; b > 0 such that jf.x/j ajxj C b for all x 2 R: (28) Give an example of a bounded continuous function R! R which is not uniformly continuous. (29) Prove that either of the following assumptions on f W R! R implies that f is uniformly continuous on R: (i) If f is differentiable and sup x2r jf 0.x/j < C1. (ii) If f is continuous, monotonic, and bounded. (30) Prove that every non-empty proper subset of a connected metric space has non-empty boundary. (31) Suppose E is a countable subset of a metric space which has more than one point. Prove that E is disconnected. (Hint: Use a ball centered at a point of E, of appropriate radius, to find a pair A; B that separate E.) (32) Verify that balls in R n are connected. Give an example of a metric space in which there are disconnected balls. (33) Suppose fe n g n1 is a sequence of connected sets in a metric space such that E n \E nc1 ; for every n 1. Show that S n1 E n is connected. (34) Give an example of a sequence E 1 E 2 E 3 of connected subsets of R 2 for which T n1 E n is disconnected (Hint: Think of slit-planes). (35) Let f be a continuous map from a connected metric space X to a metric space Y. Show that the graph of f defined by.f / D f.x; y/ W y D f.x/g X Y is connected. Here you can put any of the equivalent metrics on X Y as in problem 10. (Hint:.f / is the image of X under the map x 7!.x; f.x//.) (36) Show that the set E D.x; y/ 2 R 2 W x 0 and y D sin.1=x/ [.0; y/ W 1 y 1 is connected but not path-connected. (Hint: E is the closure of a graph.)

32 32 1 Topology of Metric Spaces I (37) Suppose U is an open and connected set in R n. Show that U is path-connected. (Hint: Fix a base point p 0 2 U and let A be the set of all p 2 U for which there is a path in U from p 0 to p. Show that A is clopen in U.) (38) Does there exist a discontinuous function R! R which maps every compact set to a compact set? Maps every connected set to a connected set? Has a connected graph? Has all three properties at the same time?

33 CHAPTER 2 Topology of Metric Spaces II 2.1. Some extension results Let X; Y be metric spaces, E X be non-empty, and f W E! Y be continuous. The general extension problem asks for conditions (on E, X, Y, or f ) that guarantee f can be extended to a continuous map X! Y. This section considers a few cases in which this problem can be solved. We begin with an extension result of the simplest type: THEOREM 2.1. Suppose X is the union of its closed subsets E; F. If f W E! Y and g W F! Y are continuous and f D g on E \ F, then the map h W X! Y defined by h D f on E and h D g on F is continuous. We informally refer to h as the result of gluing f on E to g on F. PROOF. Since f D g on E \ F, the map h is well defined on X. To prove continuity of h, take any closed set C Y. Continuity of f implies that f 1.C / is closed in E, and since E is closed in X, f 1.C / must be closed in X. Similarly, g 1.C / is closed in X. It follows that h 1.C / D f 1.C / [ g 1.C / is closed in X. EXAMPLE 2.2. The functions f W Œ0; C1/! R defined by f.x/ D x and g W. 1; 0! R defined by g.x/ D x are continuous and match along the intersection. 1; 0 \Œ0; C1/ D f0g. The glued function h W R! R (otherwise known as the absolute value function) is therefore continuous. REMARK 2.3. The assumption of E; F being closed in Theorem 2.1 is essential. For example, f.x/ D 1=x on.0; C1/ and g.x/ D x on. 1; 0 are continuous and there is no matching issue since their domains are disjoint. The glued function here is clearly discontinuous. However, Theorem 2.1 remains true if both E; F are assumed open in X. EXAMPLE 2.4. For any pair of continuous functions f; g W X! R, the functions maxff; gg and minff; gg are continuous. To see this, let E D fx 2 X W f.x/ g.x/g and F D fx 2 X W f.x/ g.x/g: If fp n g is a sequence in E such that p n! p, continuity of f shows that f.p/ D lim n!1 f.p n/ lim n!1 g.p n/ D g.p/;

34 34 2 Topology of Metric Spaces II so p 2 E. This proves E is closed, and we can check in the same way that F is closed. Now maxff; gg is the result of gluing f on E to g on F, so it is continuous by Theorem 2.1. Similarly, minff; gg is the result of gluing g on E to f on F, so it is continuous. THEOREM 2.5. Let X; Y be metric spaces, Y be complete and E X. Suppose f W E! Y is uniformly continuous. The f has a unique continuous extension fq W E! Y which is uniformly continuous. PROOF. Let p 2 E. For each n 1, define (6) Y n.p/ D f B X.p; 1=n/ \ E : Evidently, each Y n.p/ is a closed non-empty subset of Y, with Y 1.p/ Y 2.p/ Y 3.p/. Given " > 0, use uniform continuity of f on E to find a ı > 0 such that (7) A X and diam.a \ E/ < ı imply diam.f.a \ E// < ": Applying this to A D B X.p; 1=n/, and using the fact that a set and its closure have the same diameter, we see that diam.y n.p// < " if n > 2=ı. This shows diam Y n.p/! 0 as n! 1. By Theorem 1.45, T n1 Y n.p/ is a single point in Y. Define f Q.p/ to be this point. If p 2 E to begin with, then f.p/ 2 Y n.p/ for every n 1 by (6), so f Q.p/ D f.p/. Thus, fq is an extension of f. Now let " > 0 be given and find a ı > 0 such that (7) holds. Take any p; q 2 E with d X.p; q/ < ı=3. Choose an integer n > 3=ı so large that diam Y n.p/ < " and diam Y n.q/ < " (this is possible since both these diameters tend to 0 as n! 1). Find x; y 2 E such that d X.x; p/ < 1=n and d X.y; q/ < 1=n. Since d X.x; y/ d X.x; p/ C d X.p; q/ C d X.q; y/ < 1 n C ı 3 C 1 n < ı; (7) shows that d Y.f.x/; f.y// < ". It follows that d Y. Q f.p/; Q f.q// d Y. Q f.p/; f.x// C d Y.f.x/; f.y// C d Y.f.y/; Q f.q// diam Y n.p/ C " C diam Y n.q/ < 3": This proves uniform continuity of fq on E. Uniqueness of the extension f Q follows from Corollary EXAMPLE 2.6. Uniform continuity of f is essential in Theorem 2.5. For example, f.x/ D 1=x is continuous on E D R f0g but does not extend continuously to the closure E D R. Similarly, completeness of the target space is essential. For example, the identity map Q! Q does not extend to continuously to a map R! Q (although it extends continuously to a map R! R, i.e., if we take the completion of the target space.

35 2.2 More on complete spaces 35 The rest of this section focuses on the extension problem for real-valued functions. DEFINITION 2.7. Let.X; d/ be a metric space. The distance between a point x 2 X and a non-empty set E X is defined by d.x; E/ D inf d.x; p/: p2e LEMMA 2.8. (i) d.x; E/ D 0 if and only if x 2 E. (ii) The function x 7! d.x; E/ is uniformly continuous on X. PROOF. If x 2 E, there is a sequence fp n g in E which converges to x. Then d.x; p n /! 0, so clearly d.x; E/ D 0. If x E, there is an r > 0 such that B.x; r/ \ E D ;. Then d.x; p/ r for every p 2 E, so d.x; E/ r > 0. This proves (i). To see (ii), take a pair x; y 2 X and an arbitrary " > 0. By the definition of d.x; E/, we can find a p 2 E such that d.x; p/ < d.x; E/ C ". Then, d.y; E/ d.y; p/ d.y; x/ C d.x; p/ d.x; y/ C d.x; E/ C ": Letting "! 0, we obtain d.y; E/ d.x; E/ d.x; y/. By symmetry, d.x; E/ d.y; E/ d.x; y/. Together, these inequalities imply jd.x; E/ which proves the desired uniform continuity. d.y; E/j d.x; y/; THEOREM 2.9 (Urysohn). Suppose E; F are disjoint closed non-empty subsets of a metric space X. Then there is a continuous function f W X! Œ0; 1 such that E D f 1.1/ and F D f 1.0/. PROOF. The function f.x/ D d.x; F / d.x; E/ C d.x; F /.x 2 X/ is well defined since by Lemma 2.8, the condition E \ F D ; implies d.x; E/ C d.x; F / > 0. Other properties of f are easily verified. Pavel Samuilovich Urysohn ( ) 2.2. More on complete spaces DEFINITION A subset E of a metric space X is said to be dense in X if E D X. We say E is nowhere dense in X if E has empty interior.

36 36 2 Topology of Metric Spaces II Thus, E is dense E \ B ; for every ball B E is nowhere dense E 6 B for every ball B Note also that E is nowhere dense if and only if E is nowhere dense. EXAMPLE The set Q of rational numbers is dense in R, so is the set I D R Q of irrational numbers. The set Z of integers is nowhere dense in R, and its complement R Z is dense. There are uncountable subsets of R that are nowhere dense. An example is the Cantor set that we will describe in 2.6 As the above examples show, the complement of a dense set may or may not be nowhere dense. However, the following is true: LEMMA A set E X is nowhere dense if and only if X E is dense. In particular, a closed set E is nowhere dense if and only if X E is dense. PROOF. E is nowhere dense if and only if E contains no ball. This is equivalent to the condition that the complement X E meets every ball, which is just the density of X E. The intersection of two dense sets need not be dense and in fact could be empty (think of Q and I). But the intersection of two open dense sets is again dense: If U; V are open and dense in X and B is a ball, then V \ B ; since V is dense, and V \ B is open since both V and B are. The density of U then shows that U \ V \ B ;. It follows by induction that in any metric space the intersection of a finite collection of open dense sets is dense. The following example shows that this is no longer true for countably infinite collections. EXAMPLE Consider the metric space Q with the usual Euclidean distance. Let p 1 ; p 2 ; p 3 ; : : : be an enumeration of the points in Q. For each n, the set G n D Q fp n g is open and dense in Q, but T n1 G n D ;. Such a situation cannot occur in a complete metric space. This is the content of the following THEOREM 2.14 (Baire). Suppose X is a complete metric space and fg n g is a countable collection of open dense subsets of X. Then T n1 G n is dense in X. In particular, T n1 G n ;, which should be thought of as the main thrust of the theorem.

37 2.2 More on complete spaces 37 PROOF. We need to show that T n1 G n intersects every ball B.p 0 ; r 0 / in X. By the density of G 1, there is a point p 1 2 G 1 \ B.p 0 ; r 0 /. Since this intersection is open, we can choose 0 < r 1 < 1 such that B.p 1 ; r 1 / G 1 \ B.p 0 ; r 0 /. By the density of G 2, there is a point p 2 2 G 2 \ B.p 1 ; r 1 /, and since this intersection is open, we can choose 0 < r 2 < 1=2 such that B.p 2 ; r 2 / G 2 \ B.p 1 ; r 1 /. This process can be continued inductively to yield sequences fp n g in X and fr n g in R such that B.p n ; r n / G n \ B.p n 1 ; r n 1 / and 0 < r n < 1=n for every n 1. The closed sets B.p n ; r n / are nested with diameter tending to 0. Since X is complete, Cantor s Theorem 1.45 implies that T n0 B.p n; r n / is a single point p. Evidently, p 2 B.p 0 ; r 0 /. Also p 2 B.p n ; r n / G n for every n 1, so p 2 T n1 G n. Here is an equivalent formulation of Baire s theorem: COROLLARY Suppose X is a complete metric space and ff n g is a countable collection of closed nowhere dense subsets of X. Then S n1 F n has empty interior. René-Louis Baire ( ) In particular, S n1 F n X. PROOF. For each n, the set G n D X F n is open and dense by Lemma By Theorem 2.14, T n1 G n is dense, hence meets every ball. It follows that its complement S n1 F n does not contain any ball, hence has empty interior. Yet another formulation of Baire s theorem, which is especially useful in applications, is the following: COROLLARY Suppose X is a complete metric space and ff n g is a countable collection of closed subsets of X such that S n1 F n D X. Then at least one F n has non-empty interior. Before discussing some applications of Baire s theorem, let us establish some terminology. DEFINITION A subset of a metric space X is called residual if it contains a countable intersection of open dense sets in X. It is called meager if its complement is residual, i.e., if it is contained in a countable union of closed nowhere dense sets in X. Thus, according to Baire, in a complete metric space, residual sets are dense and meager sets have empty interior. Being residual is in fact much stronger than density and being meager is much stronger than having empty interior. Intuitively, residual sets are topologically thick while meager sets are topologically thin. These notions are the topological analogs of full measure and measure zero sets that we will discuss later. Compare Fig. 1.

38 38 2 Topology of Metric Spaces II subsets of a complete metric space empty interior meager dense residual nowhere dense open and dense FIGURE 1. Schematic relation between various types of sets in a complete metric space. Caution: Non-meager doesn t mean residual. Non-residual doesn t mean meager. EXAMPLE Every countable set in R is meager. Thus Q is meager and I is residual. On the other hand, Q is dense in R but it is not residual. To see this, suppose Q T G n, where each G n is open and dense in R. Let p 1 ; p 2 ; p 3 ; : : : be an enumeration of Q and define V n D G n fp n g. Then V n is still open and dense, but T V n D ;, contradicting Baire s theorem. THEOREM Every complete metric space with no isolated point is uncountable. PROOF. Assume by way of contradiction that X is complete and without isolated points, but countable. Enumerate points of X as p 1 ; p 2 ; p 3 ; : : :. Each singleton fp n g is evidently closed. It is also nowhere dense since p n is not an isolated point of X. This implies X D S n1 fp ng being meager, which is impossible by Corollary COROLLARY R n is uncountable. EXAMPLE Suppose X is a complete metric space and F is a family of continuous functions X! R. Assume F is a pointwise bounded family, that is, for each x 2 X there is a constant M x > 0 such that jf.x/j M x for every f 2 F. We claim there is a non-empty open set U X and a number M > 0 such that jf.x/j M for every f 2 F and every x 2 U. To see this, for each n 1 consider the set F n D fx 2 X W jf.x/j n for every f 2 F g D \ f 1.Πn; n /: Then F n is closed in X since each preimage f 1.Πn; n / is closed by continuity of f. The assumption of pointwise boundedness tells us that X D S n1 F n. By Corollary 2.16, some F n must have nonempty interior U, which proves jf.x/j n for every f 2 F and x 2 U. This result is often called the principle of uniform boundedness. f 2F

39 2.2 More on complete spaces 39 EXAMPLE Take a continuous function f W Œ0; 1! R. Let f 1 be an anti-derivative of f (so f1 0 D f ), f 2 be an anti-derivative of f 1 and so on. Thus, we have a sequence ff n g of successive anti-derivatives of f such that f n.n/ D f. Suppose for each x 2 Œ0; 1 there is an integer n D n.x/ 1 so that f n.x/ D 0. We will show that f D 0 everywhere on Œ0; 1. To this end, take any interval Œa; b Œ0; 1 and for n 1 define F n D fx 2 Œa; b W f n.x/ D 0g: Each F n is closed since f n is continuous, and our assumption implies Œa; b D S n1 F n. As a closed subset of a complete space, Œa; b is complete. Hence, by Corollary 2.16, some F n must have non-empty interior. In other words, there is an open interval J Œa; b on which f n D 0. Differentiating n times, we see that f D 0 on J. Since Œa; b was chosen arbitrarily, this shows the set of zeros of f are dense in Œ0; 1. Since f is continuous, we conclude that f D 0 everywhere on Œ0; 1. We end this section by proving an important result which, loosely speaking, tells us that any metric space can be embedded in a smallest possible complete metric space in an essentially unique way. DEFINITION A map f W X! Y is said to be distance-preserving if d Y.f.p/; f.q// D d X.p; q/ for all p; q 2 X: A distance-preserving map from X onto Y is called an isometry between X and Y. When such isometry exists, we say that X and Y are isometric. Observe that every distance-preserving map f W X! Y is automatically oneto-one and (uniformly) continuous. It follows that every isometry f W X! Y is a homeomorphism, and the inverse f 1 W Y! X is also an isometry. EXAMPLE Every isometry f W R! R is of the form f.x/ D xcb for some constant b 2 R. In fact, if g.x/ D f.x/ f.0/ is an isometry with g.0/ D 0, so jg.x/j D jg.x/ g.0/j D jx 0j D jxj for all x. As easy calculus exercise shows that there are only four continuous maps R! R with this property, namely g.x/ D x, g.x/ D x, g.x/ D jxj, and g.x/ D jxj. Of these four, only the first two are isometries (the other two are not one-to-one). It follows that f.x/ D x C f.0/. For a classification of isometries R n! R n, see problem 9. DEFINITION A metric space Y is said to be a completion of X if Y is complete, and if X is isometric to a dense subset of Y. For example, R is a completion of Q. LEMMA Any two completions of a metric space are isometric. PROOF. Suppose Y; Z are both completions of X, so there are dense subsets Y 1 Y and Z 1 Z and isometries f W X! Y 1 and g W X! Z 1. The map

40 40 2 Topology of Metric Spaces II h D g ı f 1 W Y 1! Z is clearly distance-preserving, so it is uniformly continuous. Since Y 1 is dense in Y and Z is complete, Theorem 2.5 shows that h extends to a continuous map h Q W Y! Z. We will prove that h Q is an isometry between Y and Z. Let us first verify that Q h is distance-preserving. If p; q 2 Y, take sequences fp n g and fq n g in Y 1 with p n! p and q n! q. By continuity, Q h.p n /! Q h.p/ and Qh.q n /! Q h.q/. Using this and continuity of d Y W Y Y! R and d Z W Z Z! R (see problem 14 in chapter 1), we obtain d Z. Q h.p/; Q h.q// D lim n!1 d Z. Q h.p n /; Q h.q n // D lim n!1 d Z.h.p n /; h.q n // D lim n!1 d Y.p n ; q n / D d Y.p; q/: It remains to show that h.y Q / D Z. Since h.y Q / contains the dense subset Z 1, it suffices to prove that h.y Q / is closed in Z. Suppose fp n g is a sequence in Y with Qh.p n /! z 2 Z. Then fh.p Q n /g is a Cauchy sequence in Z, so fp n g is a Cauchy sequence in Y since h Q is distance-preserving. Since Y is complete, fp n g converges to some p 2 Y. Continuity of h Q then shows that h.p/ D z, so z 2 h.y Q /. THEOREM Every metric space is isometric to a subspace of a complete metric space. PROOF. Let X be an arbitrary metric space. Consider the space B.X/ of all bounded real-valued functions on X with the uniform metric d 1, which is complete (see Example 1.44). Fix some base point p 0 2 X. For p 2 X, consider the function ' p W X! R defined by ' p.x/ D d.x; p/ d.x; p 0 /. The triangle inequality shows that j' p j is bounded by d.p; p 0 /, so ' p 2 B.X/. The map f W X! B.X/ defined by f.p/ D ' p is distance-preserving. To see this note that for p; q 2 X, Since j' p.x/ j' p.x/ ' q.x/j D jd.x; p/ d.x; q/j d.p; q/: ' q.x/j takes the value d.p; q/ when x D p, it follows that d 1.' p ; ' q / D sup j' p.x/ x2x Thus, f is an isometry between X and f.x/ B.X/. ' q.x/j D d.p; q/: REMARK The above proof is based on the completeness of B.X/ which in turn depends on the completeness of R. Since the map f W X! f.x/ B.X/ in the proof of Theorem 2.27 is an isometry, the image f.x/ can be thought of as a copy of X sitting inside B.X/. The closure f.x/ is complete (being a closed subset of a complete space) and contains f.x/ as a dense subset. It follows that f.x/ is a completion of X.

41 2.3 Separability and local compactness 41 COROLLARY Every metric space has a completion which is unique up to isometry. Because of the uniqueness, we often speak of the completion of a metric space X and denote it by X. O For a lengthier but more intuitive construction of the completion X, O similar to the way R is constructed from Q, see problem Separability and local compactness This section will discuss a few conditions which make a metric space topologically tame. They will turn out to be useful in constructing sets and functions with nice properties. DEFINITION A metric space is called separable if it has a countable dense subset. EXAMPLE The Euclidean space R n is separable: Q n provides an example of a countable dense subset. The space C Œa; b with the uniform metric is also separable, but the proof is harder. In fact, one can show that the set of all polynomials with rational coefficients is dense in C Œa; b (see Corollary??). The space BŒa; b is not separable. To see this, consider the uncountable set ff t D Œa;t W t 2 Œa; b g of characteristic functions in BŒa; b with pairwise distance 1. If D BŒa; b is any dense set, for each t 2 Œa; b we can choose a g t 2 D with d 1.f t ; g t / < 1=2. If g t D g s for some t; s 2 Œa; b, the triangle inequality gives d 1.f t ; f s / d 1.f t ; g t / C d 1.g s ; f s / < 1 2 C 1 2 D 1; which implies t D s. This shows that the map Œa; b! D defined by t 7! g t is one-to-one, so D must be uncountable. THEOREM Every compact metric space is separable. PROOF. Let X be a compact metric space. By Theorem 1.63, X is totally bounded. For each n 1, cover X by finitely many 1=n-balls, and let D n denote the finite collection of the centers of these balls. The countable set D D S n1 D n is dense in X since for each n, every point of X is within the distance 1=n of some point of D. DEFINITION A collection fv g of open sets in a metric space X is a base if every open set in X is the union of a sub-collection of fv g. Equivalently, fv g is a base if for every open set U and every p 2 U there is a V such that p 2 V U.

42 42 2 Topology of Metric Spaces II EXAMPLE In R n, the collection of all balls B.p; r/ with p 2 R n and r > 0 is clearly a base. But so is the smaller collection of all balls B.p; r/ with p 2 Q n and r 2 Q. THEOREM A metric space is separable if and only if it has a countable base. PROOF. First suppose X is a separable metric space, having a countable dense subset D. Let V be the countable collection of all balls of the form B.p; r/ with p 2 D and r 2 Q. Suppose U is an open set in X and p 2 U. Choose a rational number r > 0 such that B.p; 2r/ U. Since D is dense, we can find a q 2 B.p; r/ \ D. The ball B.q; r/ belongs to V and p 2 B.q; r/ B.p; 2r/ U. This shows V is a base. Conversely, suppose X has a countable base fv n g n1. Form a countable set D X by selecting one point from each V n. Then D is dense since every open set in X, being a union of the V n, intersects D. Each point of a closed set is either an isolated point or an accumulation point. Closed sets without isolated points occur so frequently that are given a special name. DEFINITION A non-empty closed set in a metric space is called perfect if it has no isolated point. Equivalently, a non-empty set E is perfect if it coincides with its derived set, so E D E 0. EXAMPLE The interval Œ0; 1 R is perfect, while the countable set f1=n W n 2 Ng [ f0g is not. In fact, since R is complete, all perfect subsets of the real line are uncountable by Theorem Note, however, that there exist countable perfect subsets of incomplete spaces. For example, the set Œ0; 1 \ Q is perfect as a subset of Q. Our goal is to prove a basic structure theorem for closed sets in separable spaces. First, we need the following DEFINITION A point p is called a condensation point of a set E X if B.p; r/ \ E is uncountable for every r > 0. The set of all condensation points of E is denoted by E. Here is an alternative characterization of E in separable spaces: LEMMA Let X be a separable metric space with a countable base fv n g n1. Then, for any set E X, X E D [ Vn W V n \ E is at most countable :

43 2.3 Separability and local compactness 43 In particular, E is closed and E E is at most countable. PROOF. Let U denote the union on the right. Suppose p 2 X E, so B.p; r/\e is at most countable for some r > 0. Choose V n such that p 2 V n B.p; r/. Then V n \ E is at most countable, so p 2 U. Conversely, suppose p 2 U so there is a V n containing p such that V n \ E at most countable. Choose r > 0 small enough so B.p; r/ V n. Then B.p; r/ \ E is at most countable, so p 2 X E. Finally, E is closed since U D X E is open, and E E D U \ E D [ Vn \ E W V n \ E is at most countable At most countable means countably infinite, or finite, or empty. is clearly at most countable. THEOREM Let E be an uncountable subset of a separable metric space X. Then E is a perfect set. PROOF. By Lemma 2.39, E is closed. It is also non-empty since E is uncountable while E E is at most countable. Suppose p 2 X is an isolated point of E, so B.p; r/ \ E D fpg for some r > 0. Let fv n g n1 be a countable base for X. Then for every q 2 B.p; r/ fpg there is a V n such that q 2 V n B.p; r/ fpg and V n \E is at most countable. Since the union of all such V n is B.p; r/ fpg, it follows that B.p; r/ fpg intersects E in at most countably many points. This contradicts the assumption p 2 E, and proves that E has no isolated point. THEOREM 2.41 (Cantor-Bendixson). Let E be an uncountable closed subset of a separable metric space X. (i) E is the disjoint union of a perfect set P and an at most countable set A. (ii) If, in addition, X is complete, then the decomposition E D P [ A in (i) is unique. Ivar Otto Bendixson ( ) PROOF. Set P D E and A D E E. Then P is perfect by Theorem 2.40, P E since E is closed, and A is at most countable by Lemma This proves the existence of the decomposition E D P [ A claimed in (i). Now assume X is also complete. Then every perfect subset of X is uncountable by Theorem Let E D P [ A be any decomposition of E into a perfect set P and an at most countable set A. We first show that P D P. The inclusion P P is clear since P is closed. If p 2 P is not a condensation point of P, there is an r > 0 for which B.p; r/ \ P is at most countable. We may choose 0 < s < r such that fx 2 X W d.x; p/ D sg \ P D ;. Since B.p; s/ \ P is non-empty and closed, it

44 44 2 Topology of Metric Spaces II Complete separable metric spaces are commonly known as Polish spaces. must have an isolated point q, for otherwise it would be an at most countable perfect set in X. We have q 2 B.p; s/ by the choice of the radius s. But this implies q being an isolated point of P, contradicting the assumption that P is perfect. This proves the reverse inclusion P P. Thus, P D P. Now, P E so P D P E. On the other hand, if p 2 E, every ball centered at p has uncountably many points of E, hence uncountably many points of P since A D E P is at most countable. Hence p 2 P D P. This shows P D E, and completes the proof of uniqueness. DEFINITION A metric space X is called locally compact if for every p 2 X there is an r > 0 such that the closed ball B.p; r/ is compact. For example, the Euclidean space R n is not compact, but it is locally compact. EXAMPLE The space C Œ0; 1 of all continuous functions Œ0; 1! R with the uniform metric d 1 is not locally compact. To see this, suppose there is a ball of some radius r > 0 centered at the constant function 0 whose closure is compact. Take 0 < s < r and consider g n 2 C Œ0; 1 defined by ( snx if 0 x 1 g n.x/ D n s if 1 n < x 1: Then d 1.g n ; 0/ D s, so g n 2 B.0; r/ for all n. Since B.0; r/ is assumed compact, there is a subsequence fg ni g which converges to some g 2 B.0; r/. In particular, g ni.x/! g.x/ for every x 2 Œ0; 1 as i! 1. This implies ( 0 if x D 0 g.x/ D s if 0 < x 1; which is a contradiction since this function is not continuous. The following is a useful property of locally compact spaces. It says that in such spaces, every compact set has arbitrarily small neighborhoods with compact closure. LEMMA Suppose X is a locally compact metric space and K U X, where K is compact and U is open. Then there is an open set V with compact closure such that K V V U. PROOF. For each p 2 K there is a ball V p centered at p such that V p is compact and contained in U. By compactness, K V p1 [ [ V pn for some p 1 ; ; p n 2 K. The union V D V p1 [ [ V pn is then an open set with the desired property. Recall that the support of a continuous function f W X! R is the closure of the set fx 2 X W f.x/ 0g.

45 2.4 Contraction maps 45 COROLLARY Suppose X is a locally compact metric space and K U X, where K is compact and U is open. Then there exists a continuous function f W X! Œ0; 1 with f j K D 1, whose support is a compact subset of U. PROOF. Find an open set V as in Lemma Apply Urysohn s Theorem 2.9 to the disjoint closed sets K and X V to find a continuous function f W X! Œ0; 1 such that K D f 1.1/ and X V D f 1.0/. The support of f is clearly V, which is a compact subset of U. Here is another useful statement about locally compact spaces; it will be invoked in the next chapter: THEOREM Let X be a locally compact and separable metric space. Then there is a sequence fk n g n1 of compact sets in X such that K n KV nc1 for every n 1, and X D S n1 K n. Such a collection fk n g is often called an exhaustion of X. Of course when X itself is compact, the sequence K n D X is an exhaustion. The most useful property here is that every compact set K X is contained in some K n. In fact, K is covered by the union of the interiors KV n, and since these interiors are nested, one of them should already cover K. PROOF. For each p 2 X take a ball U p centered at p such that U p is compact. The collection fu p g p2x is an open cover of X, so it has a countable subcover since X is separable. It follows that X is covered by a countable collection fe n g n1 of compact sets (each E n is of the form U p for some p 2 X). Set K 1 D E 1. By Lemma 2.44, there is an open set V 2 containing the compact set K 1 [ E 2 whose closure is compact. Set K 2 D V 2. Again by Lemma 2.44, there is an open set V 3 containing the compact set K 2 [ E 3 whose closure is compact. Set K 3 D V 3, and so on. Evidently, the compact sets K n constructed this way satisfy K n KV nc1 for every n 1. Their union is X since E n K n for all n Contraction maps DEFINITION Let X be a metric space. A self-map f W X! X is called a contraction if there is a number 0 < 1 such that (8) d.f.p/; f.q// d.p; q/ for all p; q 2 X: Notice that (8) is equivalent to diam f.e/ diam E for every E X:

46 46 2 Topology of Metric Spaces II Note also that every contraction is automatically (uniformly) continuous. What we call a contraction is sometimes called a uniform contraction or strong contraction to distinguish it from a weak contraction which merely satisfies d.f.p/; f.q// < d.p; q/ for all distinct p; q 2 X: EXAMPLE Every affine function f W R! R of the form f.x/ D ax C b, with jaj < 1, is a contraction. The function g W.1; C1/!.1; C1/ defined by g.x/ D x C 1=x is a weak contraction: If y > x 1, then 0 < g.y/ g.x/ D.y x/ C 1 y 1 D.y x/ 1 x 1 < y x: xy However, g is not a contraction since by the above calculation, the ratio.g.y/ g.x//=.y x/ tends to 1 as y! C1. EXAMPLE Suppose f W Œa; b! Œa; b is continuous, and differentiable in.a; b/. Then, f is a contraction sup jf 0.x/j < 1: x2.a;b/ This simply follows from the mean value theorem: Suppose D sup x2.a;b/ jf 0.x/j < 1. For every x; y 2 Œa; b, there is a z between x and y such that f.x/ f.y/ D f 0.z/.x y/. Then, jf.x/ f.y/j jx yj. Conversely, suppose f is a contraction by a factor of < 1. Then, for every x 2.a; b/, jf 0.x/j D lim f.y/ f.x/ y!x ˇ y x ˇ : EXAMPLE Let f W R n! R n be the affine map f.p/ D Ap C b, where A D Œa ij is an n n real matrix and b 2 R n is a fixed vector. Using the standard Euclidean distance on R n, we see that d.f.p/; f.q// D ka.p q/k nx h nx i 2 1=2 D a ij.p j q j / id1 j D1 nx h nx h D id1 nx i;j D1 j D1 a 2 ij a 2 ij ih nx.p j q j / 2i 1=2 j D1 i 1=2kp qk D h nx i;j D1 (by the Cauchy-Schwarz inequality) a 2 ij i 1=2d.p; q/: This shows f is a contraction with respect to the standard Euclidean distance if P n i;j D1 a2 ij < 1. Similarly, using the metrics nx d.p; q/ D jp i q i j or d.p; q/ D max i 1in q i j: id1 on R n, as in Example 1.4, it is easy to verify that f is a contraction if A satisfies the conditions nx nx max ja ij j < 1 or max ja ij j < 1; 1j n id1 respectively (see problem 16). 1in j D1

47 2.4 Contraction maps 47 Recall that a fixed point of a self-map f f.p/ D p. W X! X is a point p such that THEOREM 2.51 (Banach). Suppose X is a complete metric space and f W X! X is a contraction. Then, f has a unique fixed point. PROOF. Uniqueness is clear: If f.p/ D p and f.q/ D q, then d.p; q/ D d.f.p/; f.q// d.p; q/, where 0 < 1 is the contraction factor of f. This can happen only if d.p; q/ D 0 or p D q. To prove existence, take any base point p 0 2 X and set ı D d.p 0 ; f.p 0 //. Choose a radius R such that R > ı=.1 /. If d.x; p 0 / R, then d.f.x/; p 0 / d.f.x/; f.p 0 // C d.f.p 0 /; p 0 / R C ı < R: This shows f.b.p 0 ; R// B.p 0 ; R/. Define the sequence fe n g of closed subsets of X by E 0 D B.p 0 ; R/ and E nc1 D f.e n / for n 0: Then E 1 E 0 and inductively, E nc1 E n for all n 0. Moreover, since f is a contraction of factor, diam E nc1 diam E n, so diam E n n diam E 0, which shows diam E n! 0. It follows from Cantor s Theorem 1.45 that T n0 E n is a single point p. The condition E nc1 D f.e n / shows that f.p/ 2 T n0 E n, so f.p/ D p. Here is an important byproduct of the above proof: COROLLARY Let f be a contraction of a complete metric space X and p be its unique fixed point. Then, for any x 2 X, the iterate x n D f ın.x/ converges to p geometrically fast as n! 1. PROOF. Choose R > 0 large enough in the above proof to assure x 2 E 0, so x n 2 E n for all n 1. Then d.x n ; p/ diam E n n diam E 0 ; which proves x n! p geometrically fast. EXAMPLE For a given n n real matrix A and a fixed vector b 2 R n, the linear equation Ax C b D x has a unique solution in x 2 R n if A satisfies one of the conditions in Example EXAMPLE A photo of young Stefan Banach is placed somewhere over a print of the same photo. Then there is a unique point that corresponds to the same spot in both photos (see Fig. 2). This simply follows from the fact that the similarity R 2! R 2 that takes the large photo onto the smaller one is a contraction of factor 1=3, so it has a unique fixed point by Theorem 2.51.

48 48 2 Topology of Metric Spaces II F IGURE 2. A photo of young Stefan Banach randomly placed over a identical photo. There is a unique point that corresponds to the same spot in both photos. EXAMPLE Take any x0 > 0 and define xnc1 D 1=.xn C 2/ for n 0. Then fxn g converges. In fact, the function f W Œ0; C1/! Œ0; C1/ defined by f.x/ D 1=.x C 2/ is a contraction since jf 0.x/j D 1 1.x C 2/2 4 for all x 0: By Corollary 2.52, the sequence fxn g, the orbit of x0 under iterations of f, tends to the unique fixed point p of f in Œ0; C1/, which can be computed easily by solving an algebraic equation: 1 D x H) x 2 C 2x xc2 1 D 0 H) x D 1 p 2 H) p D p 2 1: Further applications of the Theorem 2.51 will be discussed in the next section, and problems at the end of this chapter The Hausdorff distance DEFINITION Let FX denote the space of all non-empty bounded and closed subsets of a metric space X. We define the Hausdorff distance of E; F 2 FX by ı.e; F / D max sup d.x; F /; sup d.x; E/ : x2e x2f

49 2.5 The Hausdorff distance 49 N (E) ε N (F) ε F E ε ε FIGURE 3. The Hausdorff distance between the closed disk E and the 10-pronged bouquet F is ". Observe that both sup x2e d.x; F / and sup x2f d.x; E/ are finite since E; F are assumed bounded. The condition ı.e; F / < " implies that every point of E is within the distance " of some point of F and every point of F is within the distance " of some point of E. Using this, it is easily verified that Compare Fig. 3. ı.e; F / D inf " > 0 W E N ".F / and F N ".E/ : THEOREM The Hausdorff distance defines a metric on F X. PROOF. If ı.e; F / D 0, then d.x; F / D 0 for every x 2 E, so E F, and d.x; E/ D 0 for every x 2 F, so F E. Hence E D F. The symmetry ı.e; F / D ı.f; E/ is trivial. It remains to check the triangle inequality. Take E; F; H 2 F X. Let x 2 E and " > 0. Find z 2 H such that d.x; z/ < d.x; H / C " and then find y 2 F such that d.z; y/ < d.z; F / C ". Then d.x; F / d.x; y/ d.x; z/ C d.z; y/ < d.x; H / C d.z; F / C 2" ı.e; H / C ı.h; F / C 2": Letting "! 0 and then taking the supremum over all x 2 E gives sup d.x; F / ı.e; H / C ı.h; F /: x2e The same reasoning shows sup x2f d.x; E/ ı.e; H / C ı.h; F /, which proves ı.e; F / ı.e; H / C ı.h; F /. We will need the following property of the Hausdorff metric:

50 50 2 Topology of Metric Spaces II LEMMA For any collections fe g and ff g in F X, ı. [ E ; [ F / sup ı.e ; F /: PROOF. Let E D S E, F D S F, and let s 2 Œ0; C1 denote the supremum on the right side of the above inequality. Take any x 2 E, so x 2 E for some. Then d.x; F / d.x; F / ı.e ; F / s; so sup x2e d.x; F / s. Similarly, sup x2f d.x; E/ s. Combining the two inequalities, we obtain ı.e; F / s. THEOREM If X is a complete metric space, so is F X. PROOF. Take a Cauchy sequence ff n g in F X. Let Y n D [ kn F k and F D \ n1 Y n : Felix Hausdorff ( ) (This should remind you of the definition of the upper limit of a sequence of real numbers in Definition 1.29). We will show that F 2 F X and ı.f n ; F /! 0. Evidently F, the intersection of the closed sets Y n, is closed. Since ff n g is a Cauchy sequence, it is bounded in F X. Hence there is an R > 0 such that ı.f n ; F 1 / < R for all n. This shows F n N R.F 1 / for all n, so Y 1 N R.F 1 /, so F N R.F 1 /. Since F 1 is a bounded subset of X, the same must be true of F. It remains to check that F ;. Use the Cauchy condition on the sequence ff n g to find an increasing sequence fn i g of positive integers such that ı.f ni ; F nic1 / < 2 i for all i. Starting with any x 1 2 F n1, we can inductively find x 2 ; x 3 ; x 4 ; : : : such that x i 2 F ni and d.x i ; x ic1 / < 2 i. The fx i g thus constructed clearly form a Cauchy sequence. Since X is complete, fx i g converges to some p 2 X. We have x i 2 Y n as soon as n i n. Letting i! 1, and using the fact that Y n is closed, we obtain p 2 Y n. Since this holds for every n, we must have p 2 F. This shows F ; and completes the proof of F 2 F X. Since ff n g is a Cauchy sequence, to prove F n! F in F X, it suffices to show that the subsequence ff ni g constructed above converges to F. Given " > 0, choose an integer M 1 such that 2 M < ". Take any integer j > M. Starting with any x j 2 F nj, we can construct a sequence fx i g ij as above, such that x i 2 F ni and d.x i ; x ic1 / < 2 i for all i j. Just as before, it follows that fx i g ij converges to some p 2 F. Now, for i > j, d.x j ; x i / X d.x k ; x kc1 / X 2 k D 2 j C1 : j ki 1 j ki 1 2 k < X kj

51 2.5 The Hausdorff distance 51 Letting i! 1, we obtain d.x j ; p/ 2 j C1 2 M < ". Since this holds for every x j 2 F nj, we conclude that (9) F nj N ".F / if j > M: On the other hand, given " > 0, there is an M 0 1 such that ı.f n ; F m / < "=2 if n; m > M 0. Choose M 00 1 so large that j > M 00 implies n j > M 0. Then, if j > M 00 and n > n j, we have F n N "=2.F nj /, so Y n N "=2.F nj / N ".F nj /. This proves (10) F N ".F nj / if j > M 00 : Combining (9) and (10), we conclude that ı.f nj ; F / " if j > maxfm; M 00 g. For our purposes in this section, we are more interested in the space K X of all non-empty compact subsets of X. Since K X is a subset of F X by Theorem 1.49, the Hausdorff distance makes it into a metric space. Of course, K X D F X when X itself is compact. LEMMA Let E be a closed subset of a complete metric space X. Suppose that for each " > 0 there is a compact set K X such that E N ".K/. Then E is compact. PROOF. As a closed subset of a complete space, E is complete (Theorem 1.39). To prove compactness of E, we must therefore show that E is totally bounded (Theorem 1.63). Let " > 0 and find a compact set K such that E N "=2.K/. Cover K by finitely many balls fb.p i ; "=2/g 1in. If x 2 E, there is a p 2 K with d.x; p/ < "=2 and there is a p i such that d.p; p i / < "=2. It follows from the triangle inequality that d.x; p i / < ". This shows E is covered by the balls fb.p i ; "/g 1in. COROLLARY If X is a complete metric space, then K X is closed in F X. PROOF. Take a sequence fk n g in K X which converges to E 2 F X. For each " > 0 there is a large enough n such that ı.k n ; E/ < ", so E N ".K n /. It follows from Lemma 2.60 that E 2 K X. We are now ready to prove the analog of Theorem 2.59 for K X. THEOREM If X is a complete metric space, so is K X. If X is a compact metric space, so is K X. PROOF. First suppose X is complete. Then, by Theorem 2.59, F X is complete. It follows from Theorem 1.39 and Corollary 2.61 that K X is complete. Now suppose X is compact. Then X is complete, so by the first case above we already now that K X is complete. To prove compactness of K X, we show that it is

52 52 2 Topology of Metric Spaces II totally bounded. Given " > 0, cover X by finitely many balls fb.p i ; "=2/g 1in. For each K 2 K X, let S D f1 i n W B.p i ; "=2/ \ K ;g: Clearly, K S i2s B.p i; "=2/ N ".K/, so the finite set F D fp i W i 2 Sg satisfies ı.k; F / < ". This shows that the "-balls in the Hausdorff metric centered at the 2 n subsets of fp 1 ; p 2 ; : : : ; p n g cover K X. DEFINITION Let X be a complete metric space and f i W X! X be contractions for 1 i n. The map f W K X! K X defined by f.k/ D f 1.K/ [ [ f n.k/ is called the iterated function system (IFS) generated by ff 1 ; : : : ; f n g. THEOREM Every IFS is a contraction on K X. PROOF. Suppose f is the IFS generated by ff 1 ; : : : ; f n g. Let i 2 Œ0; 1/ be the contraction factor of f i and set D max 1in i. Take distinct sets K; H 2 K X with " D ı.k; H / > 0. For each 1 i n, f i.h / f i.n ".K// N i ".f i.k//; and similarly f i.k/ N i ".f i.h //. It follows that Applying Lemma 2.58, we obtain ı.f.k/; f.h // D ı. [ ı.f i.k/; f i.h // i " ": 1in which proves f is a contraction of factor. f i.k/; [ 1in f i.h // "; COROLLARY Suppose f is the IFS on K X generated by f i W X! X, 1 i n. Then f has a unique fixed point which can be described as the smallest non-empty compact subset of X which is invariant under f i for every 1 i n. PROOF. K X is a complete metric space by Theorem 2.62 and f acts as a contraction on it by Theorem It follows from Theorem 2.51 that f has a unique fixed point K 2 K X. Since f.k/ D K, we have f i.k/ K for every 1 i n. Conversely, if H 2 K X satisfies f i.h / H for every 1 i n, the definition of f shows that f.h / H. We can then consider f as an IFS on K H by restricting all the f i to H, and repeating the above argument to conclude that f has a fixed point in K H. By uniqueness, this fixed point must be K, proving that K H.

53 2.6 The Cantor set The Cantor set We begin with a standard construction on the real line. Consider the following sequence of compact subsets of R C 0 D Œ0; 1 C 1 D Œ0; 1 3 [ Œ 2 3 ; 1 C 2 D Œ0; 1 9 [ Œ 2 9 ; 1 3 [ Œ 2 3 ; 7 9 [ Œ 8 9 ; 1 in which every C n is obtained by removing the open middle-third of each interval in C n 1. Evidently C n is the disjoint union of 2 n closed intervals, each of length 1=3 n. The intersection C D \ n0 C n is called the (middle-thirds) Cantor set; compare Fig. 4. For future references, let us establish some terminology. Each of the 2 n closed intervals that make up C n is called a level-n interval. Every level-n interval contains precisely two level-.n C 1/ intervals (think of them as a parent and two children). Each connected component of C n 1 C n is called a level-n gap. Thus, a level-n gap is an open interval of length 1=3 n that gets removed from the middle of a level-.n 1/ interval (there is no gap of level 0). The union of all gaps of various levels is Œ0; 1 C. By an endpoint of C we mean an endpoint of a level-n interval for some n 0. These are certainly points of C since an endpoint of a level-n interval is an endpoint of a level-k interval for all k n. THEOREM The cantor set is non-empty, compact, perfect, and totally disconnected. PROOF. That C is non-empty and compact follows from Cantor s Theorem 1.53 (alternatively, C ; because C contains all the endpoints at various levels). Given p 2 C and r > 0, choose n 1 such that 1=3 n < r. Then, the level-n interval I containing p is contained in B.p; r/ D.p r; p C r/. Both endpoints of I are in C and at least one of them is not p, so C meets B.p; r/ somewhere other than p. Since r > 0 was arbitrary, p must be an accumulation point of C. Thus, C is perfect. To see C is totally disconnected, let p 2 C and r > 0 and choose n and I as above. The level-.n C 1/ gap that appears in the middle third of I is contained in B.p; r/, so Œ0; 1 C meets B.p; r/. This proves Œ0; 1 C is open and dense in Œ0; 1, so C is nowhere dense in Œ0; 1 (Lemma 2.12). In particular, C contains no open intervals. It follows from Theorem 1.70 that every connected subset of C is a singleton. One byproduct of the above proof is worth mentioning:

54 54 2 Topology of Metric Spaces II C C C C C C 5 FIGURE 4. First five levels in the construction of the middle-thirds Cantor set C. At level n there are 2 n intervals and 2 n 1 gaps of length 1=3 n, and 2 nc1 endpoints all belonging to C. COROLLARY C is nowhere dense in Œ0; 1. Here is another corollary: COROLLARY C is uncountable. PROOF. As a closed subset of the real line, C is complete. The result now follows from Theorem 2.19 since C has no isolated point. The last corollary shows that C contains many more points than the endpoints, since the latter form a countable set. According to the Continuum Hypothesis, an uncountable subset of the real line has the same cardinality as the real line. If we accept this hypothesis, it follows that C has the same cardinality as R. But this can also be verified directly as follows. Similar to the usual decimal representation of real numbers, every x 2 Œ0; 1 has a ternary expansion x D 1X nd1 x n 3 n ; where the digits x n belong to f0; 1; 2g. We write the above relation as x D 0:x 1 x 2 x 3

55 2.6 The Cantor set 55 For example, if we use bars to denote repeating digits, we have the following ternary expansions: p 2 0 D 0:0 1 D 0:2 2 5 D 0: D 0:10 D 0: D 0:220 D 0: D 0: Notice that 1=3 and 8=9 have two different ternary expansions, one ending in a string of 2 s, the other ending in a string of 0 s. It is easy to check that this is the case for every rational in.0; 1/ of the form m=3 n with m; n positive integers. All other numbers in Œ0; 1 have a unique ternary expansion. Georg Ferdinand Ludwig Philipp Cantor ( ) LEMMA C is the set of all numbers in Œ0; 1 which have a ternary expansion in digits 0 and 2 only. PROOF. Looking back at the construction of C, we see that for n 1, the 2 n 1 level-n gaps consists precisely of those x 2 Œ0; 1 whose ternary expansion(s) have the first occurrence of 1 at the n-th digit. Each of the 2 n intervals that make up C n corresponds to one of the 2 n possible initial segments x 1 x n of the ternary expansion, with x i 2 f0; 2g (see Fig. 5). The details are easily supplied and will be left as an exercise. Now take any x 2 C, take the unique ternary expansion of x containing digits 0 and 2 only, convert every digit 2 to a 1 and interpret the resulting string of 0 s and 1 s as the binary expansion of a number in Œ0; 1. This gives a well-defined map C! Œ0; 1 which is surjective, but not quite injective since the endpoints of each gap in Œ0; 1 C map to the same point. For example, 1 3 D 0:02 and 2 3 D 0:20 map to 0:01 D 0:10 D 1 2 ; or 7 9 D 0:202 and 8 9 D 0:220 map to 0:101 D 0:110 D 3 4 : (one can verify that this map C! Œ0; 1 is in fact continuous). This, in particular, proves COROLLARY The Cantor set has the same cardinality as the real line.