Right Order Topology on Strictly Positive Integers is Topology

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

Let $\tau$ be the right order topology on $\Z_{>0}$.

Then $\tau$ forms a topology on $\Z_{>0}$.

That is:
 * $T = \struct {\Z_{>0}, \tau}$ is a topological space.

Proof
Let $S := \Z_{>0}$ to ease notational clutter.

First we note that:
 * $m \le n \implies O_n \subseteq O_m$

where $O_n := \set {x \in \Z_{>0}: x \ge n}$.

By definition we have that:
 * $\O \in \tau$

Then each of the open set axioms is examined in turn:

Let $\family {O_j}_{j \mathop \in S}$ be an indexed family of open sets of $T$.

Let $\ds V = \bigcup_{j \mathop \in S} O_j$ be the union of $\family {O_j}_{j \mathop \in S}$.

Let $m \in S$ be defined as:


 * $\ds m = \inf \set {n \in S: O_n \in \bigcup_{j \mathop \in S} O_j}$

Then $O_m = \bigcup_{j \mathop \in S} O_j$

Hence $V = O_m$ is open by definition.

Let $O_m$ and $O_n$ be open sets of $T$.

let $m < n$.

Then $O_n \subseteq O_m$.

Hence by Intersection with Subset is Subset
 * $O_n \cap O_m = O_n$

Hence $O_n \cap O_m$ is open by definition.

Then we note that:
 * $O_1 = S$

Hence $S \in \tau$ as required.

All the open set axioms are fulfilled, and the result follows.