Singleton Set in Discrete Space is Compact
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Theorem
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Let $x \in S$.
Then $\set x$ is compact.
Proof 1
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure.
As $\set {\set x}$ is (trivially) an open cover of $\set x$, it follows by definition that $\set x$ is compact.
$\blacksquare$
Proof 2
Follows directly from Finite Topological Space is Compact.
$\blacksquare$