Countable Complement Space Satisfies Countable Chain Condition
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Theorem
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ satisfies the countable chain condition.
Proof
We have that a Countable Complement Space is Irreducible.
Therefore there is no disjoint set of open sets of $T$.
So the result holds vacuously.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $3$