Deleted Integer Topology is Second-Countable

Theorem
Let $S = \R_{\ge 0} \setminus \Z$.

Let $\tau$ be the deleted integer topology on $S$.

Then the topological space $T = \left({S, \tau}\right)$ is second-countable.

Proof
Let $\Z_{> 0}$ be understood as the set of strictly positive integers:
 * $\Z_{> 0} = \left\{{x \in \Z: x > 0}\right\} = \left\{{1, 2, 3, \ldots}\right\}$

From Basis for Partition Topology, the set:
 * $\mathcal B = \left\{{\left({n - 1 \,.\,.\, n}\right): n \in Z_{> 0}}\right\}$

is a basis for $T$.

There is an obvious one-to-one correspondence $\phi: \Z_{> 0} \leftrightarrow \mathcal B$ between $\Z_{> 0}$ and $\mathcal B$:
 * $\forall x \in \Z_{> 0}: \phi \left({x}\right) = \left({x - 1 \,.\,.\, x}\right)$

But $\Z_{> 0} \subseteq \Z$ and Integers are Countably Infinite.

So from Subset of Countably Infinite Set is Countable, $\Z_{> 0}$ is countable.

Thus $\mathcal B$ is also countable by definition of countability.

So we have that $T$ has a countable basis, and so is second-countable by definition.