Theorem   A topological space $X$ is compact if and only if every net in $X$ has an accumulation point.

Proof. Suppose $X$ is compact and let $(x_\alpha)_{\alpha\in A}$ be a net in $X$ . For each $\alpha\in A$ , put $E_\alpha=\set{x_\beta:\beta\geq\alpha}$ ; the collection $\set{\overline{E_\alpha}:\alpha\in A}$ of closed subsets of $X$ has the finite intersection property, for given $\alpha_1,\ldots,\alpha_n\in A$ , because $A$ is directed, there exists $\beta\in A$ satisfying $\beta\geq\alpha_i$ for each $i\in\set{1,\ldots,n}$ , so that $x_\beta\in\bigcap_{i=1}^n\overline{E_{\alpha_i}}$ . Therefore, by compactness, $\bigcap_{\alpha\in A}\overline{E_\alpha}\neq\emptyset$ ; let $x$ be a point of this intersection. If $U$ is any open subset of $X$ and $\alpha\in A$ , then because $x\in\overline{E_\alpha}$ , $E_\alpha\cap U\neq\emptyset$ , and thus there exists $\beta\geq\alpha\in A$ for which $x_\beta\in U$ . It follows that $x$ is an accumulation point of $(x_\alpha)$ . For the converse, assume that $X$ fails to be compact, and let $\set{U_i:i\in I}$ 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 $S\in B$ , let $x_S$ be a point in the complement of $\bigcup_{i\in S}U_i$ . We contend that the net $(x_S)_{S\in B}$ has no accumulation points; indeed, given $x\in X$ , we may select $i_0\in I$ such that $x\in U_{i_0}$ ; if $S\in B$ is such that $i_0\in S$ , that is, if $S\geq\set{i_0}$ , then by construction, $x_S\notin U_{i_0}$ , establishing our contention.

Corollary   The following conditions on a topological space $X$ are equivalent:

1. $X$ is compact;
2. every net in $X$ has an accumulation point;
3. every net in $X$ 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.