Proposition 2.7.14603E36ECC71

Proof. Suppose that $X$ is compact, and let $(x_ i)_{i\in I}$ be a net in $X$. The closures $T_{i_0}$ of the tails $\{ x_ i:i\geq i_0\} $ have the finite intersection property, since finitely many of them contain the tail of a common upper bound of their indices. As $X$ is compact, they have a common point $x$; otherwise their complements would form an open cover without a finite subcover. Thus every neighborhood of $x$ meets every tail of the net.

We now construct a subnet converging to $x$. Let $J$ be the set of triples $(i,U,k)$ with $i,k\in I$, $k\geq i$, $U$ an open neighborhood of $x$ and $x_ k\in U$, preordered by

It contains $(i,X,i)$ for every $i$, and it is directed: given two triples $(i_1,U_1,k_1)$ and $(i_2,U_2,k_2)$, choose $i\geq i_1,i_2,k_1,k_2$, put $U=U_1\cap U_2$, and choose $k\geq i$ with $x_ k\in U$, which is possible because $U$ meets the tail of $i$; then $(i,U,k)$ dominates both triples. The map $h(i,U,k)=k$ is increasing, and it is final because $(i,U,k)\geq (i_0,X,i_0)$ implies $k\geq i_0$. The subnet $(x_{h(j)})_{j\in J}$ converges to $x$: if $V$ is a neighborhood of $x$, choose an open neighborhood $U\subset V$ of $x$ and a triple $(i,U,k)\in J$; then $x_{k'}\in U'\subset V$ for every $(i',U',k')\geq (i,U,k)$.

Conversely, suppose that $X$ is not compact, and let $\mathcal U$ be an open cover without a finite subcover. The finite subfamilies $F\subset \mathcal U$, ordered by inclusion, form a directed set, and we may choose points $x_ F\notin \bigcup F$. If a subnet $(x_{h(j)})_{j\in J}$ converged to a point $x$, let $U\in \mathcal U$ contain $x$. Then $x_{h(j)}\in U$ for all large $j$, while finality gives $h(j)\supset \{ U\} $, and hence $x_{h(j)}\notin U$, for all large $j$. Since $J$ is directed, some $j$ satisfies both conditions, a contradiction. $\square$