Either-Or Topology is Compact

Theorem

Let $T = \struct {S, \tau}$ be the either-or space.

Then $T$ is a compact space.

Proof

Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$.

So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets.

So $\CC$ has a subcover containing at most three sets.

Hence $T$ is a compact space by definition.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $2$