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Compact and Totally Bounded Metric Spaces Appendix A Compact and Totally Bounded Metric Spaces Definition A.0.1 (Diameter of a set) The diameter of a subset E in a metric space (X, d) is the quantity diam(E) = sup d(x, y). x,y∈E ♦ Definition A.0.2 (Bounded and totally bounded sets)A subset E in a metric space (X, d) is said to be 1. bounded, if there exists a number M such that E ⊆ B(x, M) for every x ∈ E, where B(x, M) is the ball centred at x of radius M. Equivalently, d(x1, x2) ≤ M for every x1, x2 ∈ E and some M ∈ R; > { ,..., } 2. totally bounded, if for any 0 there exists a finite subset x1 xn of X ( , ) = ,..., = such that the (open or closed) balls B xi , i 1 n cover E, i.e. E n ( , ) i=1 B xi . ♦ Clearly a set is bounded if its diameter is finite. We will see later that a bounded set may not be totally bounded. Now we prove the converse is always true. Proposition A.0.1 (Total boundedness implies boundedness) In a metric space (X, d) any totally bounded set E is bounded. Proof Total boundedness means we may choose = 1 and find A ={x1,...,xn} in X such that for every x ∈ E © The Editor(s) (if applicable) and The Author(s), under exclusive 429 license to Springer Nature Switzerland AG 2020 S. Gentili, Measure, Integration and a Primer on Probability Theory, UNITEXT 125, https://doi.org/10.1007/978-3-030-54940-4 430 Appendix A: Compact and Totally Bounded Metric Spaces inf d(xi , x) ≤ 1. 1≤i≤n Set M = max1≤i≤n d(x1, xi ). We know there is an index i ∈{1,...,n} such that d(xi , x) ≤ 1, so the triangle inequality implies d(x1, x) ≤ d(x1, xi ) + d(xi , x) ≤ M + 1. Hence E ⊆ B(x1, M + 1), and E is bounded. Example A.0.1 (Bounded versus totally bounded sets) The notions of boundedness and total boundedness differ, because the former is weaker. The discrete distance of distinct points is always 1, and equals 0 for coinciding points. • The space X ={a, b, c} with discrete metric is surely bounded. It is totally bounded, too, since {a}, {b}, {c} have zero diameter (less than any > 0), there is a finite number of them, and their union is X. • The space [0, 1] equipped with the discrete distance, rather than the usual Euclidean metric, is bounded: its diameter is 1, the distance of any pair of distinct points. But it is not totally bounded. The subsets of diameter less than 1 are the singletons {x}, of which there is no finite number covering the whole space. • N with the discrete metric is bounded (distinct natural numbers are all 1 apart), but not totally bounded. We cannot cover N by finitely many balls of radius < 1 (a ball consists of a single point, its centre, but N is infinite). A Cauchy sequence of natural numbers is eventually constant, hence it converges. N is therefore a complete metric space. Proposition A.0.2 (Separability of totally bounded metric spaces) Totally bounded metric spaces (X, d) are separable. Proof As X is totally bounded, for every n ∈ N there exists An ={x1,...,xn}⊆X, finite, such that n 1 X = B x , . i n i=1 ∞ The union D = = An is countable (as countable union of finite sets). n¯ 1 We claim that D = X.Letx ∈ X be an arbitrary point. For any n ∈ N, there ∈ ∈ , / { } ⊆ exists some xin An such that x B xin 1 n . This gives a sequence xin n≥1 D Appendix A: Compact and Totally Bounded Metric Spaces 431 ( , )< / = ∈ with d x xin 1 n. Therefore lim xi x. This shows that every x X lies in the closure of D and so D¯ = X. Intuitively a separable space is one where any element is close to one from some (dense) countable subset. In a compact space, instead, any element is close to another taken from a finite number of subsets, so compact spaces can be considered kind of “pseudo-finite”. In analogy to the notion of compactness in R we have the following definition for generic metric spaces. Definition A.0.3 (Compactness) A subset E in a metric space (X, d) is said to be compact if any open cover of E admits a finite subcover (so-called Heine–Pincherle– Borel property). ♦ The next results generalises the Bolzano–Weierstraß theorem to compact metric spaces. Theorem A.0.1 (Bolzano–Weierstraß theorem for metric spaces) If (X, d) is a com- pact metric space, any infinite subset E ⊆ X has a limit point. Proof Let (X, d) be compact and E ⊆ X infinite. Assume, by contradiction, E has no limit points. Then for every x ∈ X there is an open ball B(x, rx ) such that B(x, rx ) ∩ E ⊆{x}. The family {B(x, rx )}x∈X is an open cover of X, so there exists ( , ) = ,..., a finite subcover B xi rxi , i 1 n. But then n = ∩ = ∩ ( , ) ⊆{ ,..., }, E E X E B xi rxi x1 xn i=1 i.e. E is finite. The contradiction proves the claim. Definition A.0.4 (Sequentially compact metric spaces) A metric space (X, d) is sequentially compact if every sequence in X admits a convergent subsequence. ♦ We will see straightaway that sequential compactness implies total boundedness. Proposition A.0.3 (Sequential compactness versus total boundedness) A sequen- tially compact metric space (X, d) is totally bounded. 432 Appendix A: Compact and Totally Bounded Metric Spaces Proof Suppose (X, d) is not totally bounded. Then there is an 0 such that X cannot be covered by balls centred at the points of any finite set A0 ⊆ X. Taking x1 ∈ X, then, we can find x2 ∈ X with d(x1, x2)>0, for otherwise X = B(x1, 0) and A0 ={x1}. Next, take x1, x2 ∈ X, and there is an x3 ∈ X with d(x3, x1)>0 and d(x3, x2)> 0, because otherwise X = B(x1, 0) ∪ B(x2, 0) and A0 ={x1, x2}. Inductively, at step k + 1wehave{x1,...,xk }⊆X and there is an xk+1 ∈ X with d(xk+1, xi )>0 for every i ∈{1,...,k}, because if not, once again, k X = B(xi , 0) i=1 and A0 ={x1,...,xk }. Overall we have a whole sequence {xk }k≥1 such that d(xk , x j )>0, k = j.It must diverge, hence it certainly cannot have a convergent subsequence. But then X would not be sequentially compact, contradicting the hypothesis. Definition A.0.5 (Countably compact metric spaces) A metric space (X, d) is countably compact if any countable open cover of X has a finite subcover. ♦ Proposition A.0.4 (Sequentially compact metric spaces and countable compact- ness) A sequentially compact metric space (X, d) is countably compact. Proof Let {Gn}n≥1 be a countable open cover of X ∞ X = Gn. n=1 Suppose X not countably compact, so {Gn}n≥1 does not have a finite subcover, meaning for every m ≥ 1 m Gn ⊆ X. n=1 ∈ ∈/ m { } The proof is by induction. Take xm1 X such that xm1 n=1 Gn, and since Gn n≥1 ∈ ∈ ∈/ covers X, there is an m1 such that xm1 Gm1 . Choose xm2 X such that xm2 m1 ∈ ∈ n=1 Gn. Then there is an m2 such that xm2 Gm2 . Then pick xm3 X such that ∈/ m2 ∈ xm3 n=1 Gn,soxm3 Gm3 for some index m3. At step k Appendix A: Compact and Totally Bounded Metric Spaces 433 mk−1 ∈ ∈/ . xmk Gmk and xmk Gn n=1 { } We end up with a sequence xmk k≥1 in the sequentially compact space X, so there { } ∈ { } is a subsequence xkn n≥1 with some limit point x X.But Gn n≥1 is a cover, so x belongs in some Gn0 . Furthermore, lim xk = x n→∞ n ∈ ≥ forces xkn Gn0 for all n n0. But because of how the sequence was constructed, − > ∈ ∈/ ⊆ mkn 1 there exists an mkn n0 such that xkn Gmkn , and xkn Gn0 n=1 Gn. Hence { } no subsequence of xmk k≥1 can converge to x. But this means X is not sequentially compact which is a contradiction. Therefore X is countably compact. In metric spaces compactness can be characterised by means of sequences. Namely, Theorem A.0.2 (Characterisation of compact metric spaces in terms of sequential compactness) A metric space (X, d) is compact if and only if it is sequentially compact. Proof (⇒)Let(X, d) be compact, and let us prove it is sequentially compact. Given a compact subset E ⊆ X, suppose there is a sequence {xn}n≥1 none of whose subsequences converge in E. Any such must have infinite range, otherwise it would become eventually constant and hence converge in E. Nor can it accumulate at a point in E, otherwise this point would be the limit of a subsequence. Take x ∈ E. This is not a limit point for {xn}n≥1,sothereisanx > 0 such that the ball B(x, x ) contains at most one term of the sequence. ( , ) Clearly B x x x∈E is an open cover of E, and by compactness we extract a finite subcover such that n ⊆ ( , ). E B xi xi i=1 { } ⊆ n ( , ) But xn n≥1 E, and since in i=1 B xi xi there are at most n distinct points, we deduce the sequence can only attain finitely many values. Hence it must have a convergent subsequence, contradicting the hypothesis.
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