Linearly Ordered Space is T5

Theorem
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.

Then $T$ is a $T_5$ space.

Proof
Let $A$ and $B$ be separated sets of $T$.

Let $A^*$ and $B^*$ be defined as:
 * $A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
 * $B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$

where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.

Let $A^*$, $B^*$ and $\complement_S \left({A^* \cup B^*}\right)$ be expressed as the union of convex components of $S$:
 * $\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$

where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.

Let $M$ be the linearly ordered set:


 * $M = \set {A_\alpha, B_\beta, C_\gamma}$

as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.

Let:
 * $S_\alpha$ be the set of strict upper bounds for $A_\alpha$.
 * $S_\beta$ be the set of strict upper bounds for $B_\beta$.

From Successor Sets of Linearly Ordered Set Induced by Convex Component Partition:
 * each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$


 * each of the sets $B_\beta \in M$ has an immediate successor in $M$ if $B_\beta$ intersects the closure of $S_\beta$


 * that immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.

For each $\gamma$, let $k_\gamma \in C_\gamma$ be selected and fixed.

Let $A_\alpha \cap {S_\alpha}^- \ne \O$.

Then there exists a unique ${k_\alpha}^+ \in {C_\alpha}^+$.

In such cases, let $I_\alpha = \hointr p { {k_\alpha}^+}$ where $p \in A \cap {S_\alpha}^-$.

The other case is when $A_\alpha \cap {S_\alpha}^- = \O$.

In this case, let $I_\alpha = \O$.

Let $J_\alpha$ be defined similarly for the strict lower bounds of $A_\alpha$, using the same set of points $k_\gamma \in C_\gamma$.

Then:
 * For each $\alpha$, let $U_\alpha = J_\alpha \cup A_\alpha \cup I_\alpha$.
 * For each $\beta$, let $V_\beta = J_\beta \cup A_\beta \cup I_\beta$.

Each $U_\alpha$ and $V_\beta$ is a convex open set in $S$ containing $A_\alpha$ and $B_\beta$ respectively.

Thus $\ds U = \bigcup U_\alpha$ and $\ds V = \bigcup V_\beta$ are open sets in $T$ containing $A^*$ and $B^*$ respectively.

We have that:
 * no $A_\alpha$ intersects no $B_\beta$
 * the same $K_\gamma$ being used throughout implies that no $J_\beta$ or $I_\beta$ can intersect any $J_\alpha$ or $I_\alpha$.

Thus no $U_\alpha$ can intersect any $V_\beta$.

Thus $U \cap V = \O$.

So we have two disjoint open sets $U, V$ in $S$ containing the two separated sets $A$ and $B$ respectively.

As $A$ and $B$ are arbitrary, it follows that such $U$ and $V$ can be found whatever $A$ and $B$ are, provided they are separated.

Hence the result by definition of $T_5$ space.