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;
- 2.
every net in has an accumulation point;
- 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.∎