User:Dfeuer/Convex Component is Closed
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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 $a,x \in T$, 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 $\tau'$.