Right Order Topology on Real Numbers is Topology

Theorem
Let $\tau$ be the right order topology on $\R$.

Then $\tau$ forms a topology on $\R$.

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

Proof
Write $\O = \openint \infty \infty$ and $\R = \openint {-\infty} \infty$.

Then $\tau$ can be written as $\set {\openint j \infty: j \in \overline \R}$.

First we note that:
 * $m \le n \implies \openint n \infty \subseteq \openint m \infty$

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

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

Let $\family {\openint j \infty}_{j \mathop \in S}$ be an indexed family of open sets of $T$, where $S \subseteq \overline \R$.

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

Let $m := \inf S \in \overline \R$.

We claim that:
 * $\ds \openint m \infty = \bigcup_{j \mathop \in S} \openint j \infty$

Indeed, for each $x \in \openint m \infty$ we have $x > m$.

By Characterizing Property of Infimum of Subset of Real Numbers:
 * $\exists y \in S: y < x$

and thus $\ds x \in \openint y \infty \subseteq \bigcup_{j \mathop \in S} \openint j \infty$.

This shows that $\ds \openint m \infty \subseteq \bigcup_{j \mathop \in S} \openint j \infty$.

For the other direction, note that $j \ge m$ for each $j \in S$.

Therefore $\openint j \infty \subseteq \openint m \infty$ for each $j \in S$.

By Union of Subsets is Subset, $\ds \bigcup_{j \mathop \in S} \openint j \infty \subseteq \openint m \infty$.

Thus our claim is true by definition of set equality.

Hence $V = \openint m \infty$ is open by definition.

Let $A = \openint m \infty$ and $B = \openint n \infty$, where $m, n \in \overline \R$.

let $m < n$.

Then $\openint n \infty \subseteq \openint m \infty$.

Hence by Intersection with Subset is Subset:
 * $\openint n \infty \cap \openint m \infty = \openint n \infty \in \tau$

Hence $\openint n \infty \cap \openint m \infty$ is open by definition.

Therefore the intersection of any two elements of $\tau$ is an element of $\tau$.

$\R \in \tau$ follows directly from definition.

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