Linearly Ordered Space is T1

Theorem
Let $T = \left({S, \preceq, \tau}\right)$ be a linearly ordered space.

Then $T$ is a $T_1$ (Fréchet) space.

Proof
Let $p \in S$.

By definition of linearly ordered space, the rays:
 * $R_1 := \left\{ {x \in S: x \prec p}\right\}$
 * $R_2 := \left\{ {x \in S: p \prec x}\right\}$

are open in $T$.

Thus their union:
 * $R_1 \cup R_2 = \left\{ {x \in S: x \prec p \lor p \prec x}\right\}$

is also open in $T$.

Consider the complement in $S$ of $R_1 \cup R_2$:
 * $\complement_S \left({R_1 \cup R_2}\right) = \left\{ {y \in S: y \notin R_1 \cup R_2}\right\}$

By definition of closed set, $\complement_S \left({R_1 \cup R_2}\right)$ is closed in $T$

But $\complement_S \left({R_1 \cup R_2}\right) = \left\{ {p}\right\}$.

Thus $p$ is a closed point of $T$.

Hence the result, by definition of $T_1$ (Fréchet) space.