Linearly Ordered Space is T1

Theorem

Let $T = \struct {S, \preceq, \tau}$ 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:

$R_1 := \set {x \in S: x \prec p}$

$R_2 := \set {x \in S: p \prec x}$

are open in $T$.

Thus their union:

$R_1 \cup R_2 = \set {x \in S: x \prec p \lor p \prec x}$

is also open in $T$.

Consider the complement in $S$ of $R_1 \cup R_2$:

$\relcomp S {R_1 \cup R_2} = \set {y \in S: y \notin R_1 \cup R_2}$

By definition of closed set, $\relcomp S {R_1 \cup R_2}$ is closed in $T$

But $\relcomp S {R_1 \cup R_2} = \set p$.

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

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

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $39$. Order Topology: $6$