Fortissimo Space is not Separable
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Theorem
Let $T = \struct {S, \tau_p}$ be the fortissimo space on an uncountable set $S$.
Then $T$ is not a separable space.
Proof
Let $U$ be a countable subset of $S$.
By the definition of the fortissimo space, $U$ is closed.
From Closed Set Equals its Closure, $U^- = U \ne S$.
Thus, by definition, $U$ is not everywhere dense in $T$.
Thus, there exists no countable subset of $S$ which is everywhere dense in $T$.
So, by definition, $T$ is not a separable space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $1$