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 $T = \struct {S, \tau}$ is second-countable.

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

From Basis for Partition Topology, the set:
 * $\BB = \set {\openint {n - 1} n: n \in Z_{>0} }$

is a basis for $T$.

There is an obvious one-to-one correspondence $\phi: \Z_{> 0} \leftrightarrow \BB$ between $\Z_{> 0}$ and $\BB$:
 * $\forall x \in \Z_{>0}: \map \phi x = \openint {x - 1} x$

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

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

Thus $\BB$ is also countable by definition of countability.

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