Topology induced by Usual Metric on Positive Integers is Discrete

Theorem
Let $\Z_{>0}$ be the set of (strictly) positive integers.

Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.

Then the metric topology for $d$ is a discrete topology.

Proof
Let $\tau_d$ denote the metric topology for $d$.

Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.

Let $a \in \Z_{>0}$.

Recall the definition of the open $\epsilon$-ball of $a$ in $\left({\Z_{>0}, d}\right)$:
 * $B_\epsilon \left({a}\right) := \left\{{x \in A: d \left({x, a}\right) < \epsilon}\right\}$

But we have:
 * $\forall x \in \Z_{>0}, x \ne a: d \left({x, a}\right) \ge 1$

and so:
 * $\forall x \in \Z_{>0}, x \ne a: x \notin B_\epsilon \left({a}\right)$

It follows that:
 * $B_\epsilon \left({a}\right) := \left\{{a}\right\}$

Thus by definition of $\tau_d$:
 * $\forall a \in \Z_{>0}: \left\{{a}\right\} \in \tau_d$

It follows from Basis for Discrete Topology that $\left({\Z_{>0}, \tau_d}\right)$ is a discrete topological space.