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 Discrete iff All Singletons Open, it suffices to show that for all $n \in \N$, the singleton $\left\{{n}\right\}$ is an open of $\tau$.


Now observe that $\mathop{\downarrow} \left({1}\right) = \left\{{0}\right\}$, since for all $n \in \N$, $n < 1 \implies n = 0$.

It follows that $\left\{{0}\right\}$ 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, $\mathop{\uparrow} \left({n - 1}\right) \cap \mathop{\downarrow} \left({n + 1}\right) = \left\{{n}\right\}$.

It follows that $\left\{{n}\right\}$ is an open set of $\tau$.


Hence the result, from Proof by Cases.

$\blacksquare$