Particular Point Space is Separable

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: $6$
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