Order Topology on Natural Numbers is Discrete Topology
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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 $\tau$.
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Hence the result, from Proof by Cases.
$\blacksquare$