# 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.

Before looking at the counterexample, we first notice that the converse is true for metric spaces.
Proof: consider a metric space $$(X,d)$$ and a dense subset $$D=\{b_n;n \in \mathbb{N}\}$$. We claim that the set of open balls $$\mathcal{B}=\{B(b_n,\frac{1}{m});(n,m) \in \mathbb{N} \times \mathbb{N}^*\}$$ is a countable base. $$\mathcal{B}$$ is countable as a product of two countable sets is countable. It remains to prove that $$\mathcal{B}$$ is a base. To do this let $$U$$ be any nonempty open subset of $$X$$. For $$x \in U$$ we can find an open ball $$B(x,\epsilon)$$ included in $$U$$ with $$\epsilon > 0$$. As $$D$$ is dense we can pickup $$b_{n_x} \in D$$ with $$b_{n_x} \in B(x,\epsilon/2)$$. For $$m_x \in \mathbb{N}$$ and $$d(x,b_{n_x}) < \frac{1}{m_x} < \epsilon/2$$ the open ball $$B(b_{n_x},\frac{1}{m_x})$$ is included in $$B(x,\epsilon)$$. Finally:
$U =\bigcup_{x \in U} B(b_{n_x},\frac{1}{m_x})$ leading to the desired conclusion.

We now shift our attention to the counterexample. Namely the Moore plane, also called sometimes Niemytzki plane.

Let $$X$$ be the closed upper half-plane. $$X = A \uplus B$$ is the disjoint union of the open upper half-plane $$A$$ and the X-axis $$B$$. We next define a topology from a neighborhood system:

• For points in $$A$$, the local basis consists in the open discs in the plane which are small enough to lie within $$A$$. Thus the subspace topology inherited by is the same as the subspace topology inherited from the standard topology of the Euclidean plane.
• Elements of the local basis at points $$p \in B$$ are the union sets $$\{p\} \cup D$$ where $$D$$ is any open disc in the upper half-plane which is tangent to the X-axis at $$p$$.

One can verify that the neighborhoods defined above verify the requirements of a neighborhood system. Hence, there is a unique topology $$\mathcal{T}$$ on $$X$$ inherited from above neighborhood system.

$$X$$ contains a countable dense subset, namely the subsets of points having rational coordinates. Hence $$X$$ is separable. If $$X$$ was second-countable, the X-axis subset equipped with the subspace topology would also be second-countable. However, one can notice that any singleton of the X-axis is an open subset of $$X$$ as for any $$a \in \mathbb{R}$$ we have $$\{(a,0)\}=V_{a,1} \cap B$$, where $$V_{a,1}$$ is the open subset of $$X$$ consisting of the union of the disc centered on $$(a,1)$$ of radius $$1$$ with the point $$(a,0)$$. But the X-axis has the cardinality of the continuum. Hence $$X$$ cannot admit a countable base.