Double Pointed Finite Complement Topology is Compact
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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: $7$