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单词 ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded
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proof that a metric space is compact if and only if it is complete and totally bounded


TheoremMathworldPlanetmath: A metric space is compactPlanetmathPlanetmath if and only if it is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and totally boundedPlanetmathPlanetmath.

Proof.  Let X be a metric space with metric d. If X is compact, then it is sequentially compact and thus complete. Since X is compact, the covering of X by all ϵ-balls must have a finite subcover, so that X is totally bounded.

Now assume that X is complete and totally bounded. For metric spaces, compact and sequentially compact are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath; we prove that X is sequentially compact. Choose a sequenceMathworldPlanetmath pnX; we will find a Cauchy subsequenceMathworldPlanetmath (and hence a convergentMathworldPlanetmath subsequence, since X is complete).

Cover X by finitely many balls of radius 1 (since X is totally bounded). At least one of those balls must contain an infiniteMathworldPlanetmath number of the pi. Call that ball B1, and let S1 be the set of integers i for which piB1.

Proceeding inductively, it is clear that we can define, for each positive integer k>1, a ball Bk of radius 1/k containing an infinite number of the pi for which iSk-1; define Sk to be the set of such i.

Each of the Sk is infinite, so we can choose a sequence nkSk with nk<nk+1 for all k. Since the Sk are nested, we have that whenever i,jk, then ni,njSk. Thus for all i,jk, pni and pnj are both contained in a ball of radius 1/k. Hence the sequence pnk is Cauchy.

References

  • 1 J. Munkres, TopologyMathworldPlanetmathPlanetmath , Prentice Hall, 1975.
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