Singleton Set in Discrete Space is Compact

From ProofWiki
Jump to navigation Jump to search


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.


Proof 2

Follows directly from Finite Topological Space is Compact.