Particular Point Space is Separable
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Theorem
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is separable.
Proof
By definition, $T$ is separable if and only if there exists a countable subset of $S$ which is everywhere dense in $T$.
Consider $U := \set p \subseteq S$.
By definition, $U$ is open in $T$.
From Closure of Open Set of Particular Point Space we have that $U^- = S$, where $U^-$ is the closure of $U$.
By definition, $U$ is everywhere dense in $T$.
$U$ is (trivially) countable.
Hence the result, by definition of 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: $8 \text { - } 10$. Particular Point Topology: $6$