User:Dfeuer/Convex Component is Closed

Theorem
Let $(S, \preceq, \tau)$ be a totally ordered space.

Let $(T, \tau')$ be a topological subspace of $S$.

Let $C$ be a convex component of $T$ in $S$.

Then $C$ is closed relative to $\tau'$.

Proof
Let $x \in T \setminus C$.

Let $a \in C$.

Assume WLOG that $a \prec x$.

By Definition 2 of convex component, there is a point $b \in S \setminus T$ such that $a \preceq b \preceq x$.

Since $a,x \in C \subseteq T$, we must have $a \prec b \prec x$.

Because $C$ is convex, $x \in T \cap {\uparrow} b \in T \setminus C$.

Since this holds for any such $x$, $C$ is closed relative to $\tau'$.