Alexandroff Extension is Compact

Theorem

Let $T = \struct {S, \tau}$ be a non-empty topological space.

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

Let $S^* := S \cup \set p$.

Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.


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


Proof

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

At least one $V \in \UU$ 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 $\relcomp {S^*} V$ of $T$.

Because $\relcomp {S^*} V$ is compact, it can be covered by a finite number of sets in $\UU$.

So:

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

and:

that finite number of sets in $\UU$ are a cover for $\relcomp {S^*} V$.

That is, $\UU$ has a finite subcover.

Hence the result by definition of compact topological space.

$\blacksquare$


Sources

  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $34$. One Point Compactification Topology: $1$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.