Order Topology on Convex Subset is Subspace Topology

Theorem
Let $(S,\le,\tau)$ be a linearly ordered space.

Let $A \subseteq S$ be a convex set in $S$.

Let $\upsilon$ be the order topology on $A$.

Let $\tau'$ be the $\tau$-relative subspace topology on $A$.

Then $\upsilon = \tau'$.

Proof
By the definition of the order topology, the sets of open rays in $(S,\le)$ and $(A,\le)$ form sub-bases for $\tau$ and $\upsilon$, respectively.

By Sub-Basis for Topological Subspace, we need only show that the set of open rays in $(S,\le)$ induces a subbase for $\upsilon$. Specifically, we will show that each open ray in $A$ is the intersection of $A$ with an open ray in $S$, and we will show that the intersection of any open ray in $S$ with $A$ is either an open ray in $A$, the empty set, or $A$.

By Dual Pairs (Order Theory), strict upper closure is dual to strict lower closure, so by ______ we need show this only for the upward-pointing rays.

For each $p \in S$, let ${\uparrow}_S p$ be the strict upper closure of $p$ in $S$.

For each $p \in A$, let ${\uparrow}_A p$ be the strict upper closure of $p$ in $A$.

Let $p \in A$.

Then by the definition of strict upper closure:

Let $q \in S$.

If $q \in A$, then $A \cap {\uparrow_S}q = {\uparrow_A}q$, so $A \cap {\uparrow_S}q \in \upsilon$.

Suppose instead that $q \notin A$.

If $A \cap {\uparrow_S}q = \varnothing$, we are done.

Otherwise, let $x \in A \cap {\uparrow_S}q$

Since $A$ is convex in $S$, no element of $A$ precedes $q$.

Thus $A \cap {\uparrow_S}q = A$.