# Order Topology on Convex Subset is Subspace Topology

## Theorem

Let $\struct {S, \preceq,\tau}$ 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

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By the definition of the order topology, the sets of open rays in $\struct {S, \preceq}$ and $\struct {A, \preceq}$ 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 $\struct {S, \preceq}$ 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 the duality principle we need show this only for the upward-pointing rays.

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

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

Let $p \in A$.

Then by Strict Upper Closure in Restricted Ordering:

- $p^{\succ A} = A \cap p^{\succ S}$

Let $q \in S$.

If $q \in A$, then $A \cap q^{\succ S} = q^{\succ A}$, so $A \cap q^{\succ S}$ is an open ray in $A$.

Suppose instead that $q \notin A$.

If $A \cap q^{\succ S} = \O$, the proof is complete.

Otherwise, $A \cap q^{\succ S}$ must have some element $x$.

Aiming for a contradiction, suppose there is a $y \in A \setminus \paren {A \cap q^{\succ S} } = A \cap q^{\preceq S}$.

Then since $q \notin A$, it follows that $y \prec q$.

Thus $y \prec q \prec x$, $y, x \in A$, and $q \notin A$, contradicting the fact that $A$ is convex.

Thus $A \cap q^{\succ S} = A$.

$\blacksquare$