Double Pointed Finite Complement Topology is Compact

Theorem

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.

Let $T \times D$ be the double pointed topology on $T$.


Then $T \times D$ is compact.


Proof

The Finite Complement Topology is Compact.

Also, a Finite Topological Space is Compact.

So $D$, defined as the indiscrete space on a doubleton, is also compact.

The result follows directly from Tychonoff's Theorem.

$\blacksquare$


Sources

  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $7$
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