Alexandroff Extension is Compact

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Theorem

Let $T = \left({S, \tau}\right)$ be a non-empty topological space.

Let $p$ be a new element not in $S$.

Let $S^* := S \cup \left\{{p}\right\}$.

Let $T^* = \left({S^*, \tau^*}\right)$ be the Alexandroff extension on $S$.


Then $T^*$ is a compact topological space.


Proof

Let $\mathcal U$ be an open cover of $T^*$.

At least one $V \in \mathcal U$ contains $p$.

Because $p \notin S$, $V$ is not an open set of $T$.

Therefore, by definition of the Alexandroff extension, $V$ must be the complement relative to $S^*$ of a closed, compact subset $\complement_{S^*} \left({V}\right)$ of $T$.

Because $\complement_{S^*} \left({V}\right)$ is compact, it can be covered by a finite number of sets in $\mathcal U$.

So:

$T^*$ can be covered by $V$

and:

that finite number of sets in $\mathcal U$ are a cover for $\complement_{S^*} \left({V}\right)$.

That is, $\mathcal U$ has a finite subcover.

Hence the result by definition of compact topological space.

$\blacksquare$


Sources