# A separable space that is not second-countable

In topology, a second-countable space (also called a completely separable space) is a topological space having a countable base.

It is well known that a second-countable space is separable. For the proof consider a second-countable space $$X$$ with countable basis $$\mathcal{B}=\{B_n; n \in \mathbb{N}\}$$. We can assume without loss of generality that all the $$B_n$$ are nonempty, as the empty ones can be discarded. Now, for each $$B_n$$, pick any element $$b_n$$. Let $$D=\{b_n;n \in \mathbb{N}\}$$. $$D$$ is countable. We claim that $$D$$ is dense in $$X$$. To see this let $$U$$ be any nonempty open subset of $$X$$. $$U$$ contains some $$B_p$$, hence $$b_p \in U$$. So $$D$$ intersects $$U$$ proving that $$D$$ is dense.

What about the converse? Is a separable space second-countable? The answer is negative and I present below a counterexample. Continue reading A separable space that is not second-countable