Separable Space need not be First-Countable

Theorem

Let $T = \struct {S, \tau}$ be a topological space which is separable.

Then $T$ does not necessarily have to be first-countable.

Proof

Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.

We have that a Finite Complement Topology is Separable.

But we also have that an Uncountable Finite Complement Space is not First-Countable.

Hence the result, by Proof by Counterexample.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $4$