Order Topology on Natural Numbers is Discrete Topology

Theorem
Let $\le$ be the standard ordering on the natural numbers $\N$.

Then the order topology $\tau$ on $\N$ is the discrete topology.

Proof
By Topology is Discrete iff All Singletons are Open, it suffices to show that for all $n \in \N$, the singleton $\set n$ is an open of $\tau$.

Now observe that $\map {\mathop{\downarrow} } 1 = \set 0$, since for all $n \in \N$, $n < 1 \implies n = 0$.

It follows that $\set 0$ is an open set of $\tau$.

Suppose now that $n \in \N$ and $n \ne 0$.

Then it is known that for all $m \in \N$, $n - 1 < m < n + 1$ implies $m = n$.

Thus, $\map {\mathop{\uparrow} } {n - 1} \cap \map {\mathop{\downarrow} } {n + 1} = \set n$.

It follows that $\set n$ is an open set of $\tau$.

Hence the result, from Proof by Cases.