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 space 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.