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. ∎
| 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 |