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单词 CompactnessAndAccumulationPointsOfNets
释义

compactness and accumulation points of nets


Theorem.

A topological spaceMathworldPlanetmath X is compactPlanetmathPlanetmath if and only if every net in X has an accumulation pointMathworldPlanetmath.

Proof.

Suppose X is compact and let (xα)αA be a net in X. For each αA, put Eα={xβ:βα}; the collectionMathworldPlanetmath {Eα¯:αA} of closed subsets of X has the finite intersection property, for given α1,,αnA, because A is directed, there exists βA satisfying βαi for each i{1,,n}, so that xβi=1nEαi¯. Therefore, by compactness, αAEα¯; let x be a point of this intersectionMathworldPlanetmath. If U is any open subset of X and αA, then because xEα¯, EαU, and thus there exists βαA for which xβU. It follows that x is an accumulation point of (xα). For the converseMathworldPlanetmath, assume that X fails to be compact, and let {Ui:iI} be an open cover of X with no finite subcover. If B is the set of finite subsets of I, then B is directed by inclusion. For each set SB, let xS be a point in the complement of iSUi. We contend that the net (xS)SB has no accumulation points; indeed, given xX, we may select i0I such that xUi0; if SB is such that i0S, that is, if S{i0}, then by construction, xSUi0, establishing our contention.∎

Corollary.

The following conditions on a topological space X are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    X is compact;

  2. 2.

    every net in X has an accumulation point;

  3. 3.

    every net in X has a convergentMathworldPlanetmathPlanetmath subnet;

Proof.

The preceding theorem establishes the equivalence of (1) and (2), while that of (2) and (3) is established in the entry on accumulation points and convergent subnets.∎

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