compactness and accumulation points of nets
Theorem.
A topological space is compact if and only if every net in has an accumulation point.
Proof.
Suppose is compact and let be a net in . For each , put ; the collection of closed subsets of has the finite intersection property, for given , because is directed, there exists satisfying for each , so that . Therefore, by compactness, ; let be a point of this intersection. If is any open subset of and , then because , , and thus there exists for which . It follows that is an accumulation point of . For the converse, assume that fails to be compact, and let be an open cover of with no finite subcover. If is the set of finite subsets of , then is directed by inclusion. For each set , let be a point in the complement of . We contend that the net has no accumulation points; indeed, given , we may select such that ; if is such that , that is, if , then by construction, , establishing our contention. ∎
Corollary.
The following conditions on a topological space are equivalent:
-
1.
is compact;
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2.
every net in has an accumulation point;
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3.
every net in has a convergent 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. ∎
Title | compactness and accumulation points of nets |
---|---|
Canonical name | CompactnessAndAccumulationPointsOfNets |
Date of creation | 2013-03-22 18:37:50 |
Last modified on | 2013-03-22 18:37:50 |
Owner | azdbacks4234 (14155) |
Last modified by | azdbacks4234 (14155) |
Numerical id | 7 |
Author | azdbacks4234 (14155) |
Entry type | Theorem |
Classification | msc 54A20 |
Related topic | Net |
Related topic | Compact |
Related topic | AccumulationPointsAndConvergentSubnets |