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:

Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

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.

$\Box$

Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets

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

Without loss of generality 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 $\tau$.

$\Box$

Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

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

$\Box$

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

$\blacksquare$