User:Dfeuer/Convex Component of Open Set in GO-Space is Open

Theorem

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

Let $A$ be a $\tau$-open subset of $S$.

Let $C$ be a convex component of $A$ in $S$

Then $C$ is open relative to $\tau$.

Note: the term convex will be used here to refer to a set that is convex in $S$.

Since $(S, \preceq, \tau)$ is a generalized ordered space, $\tau$ has a basis $\mathcal B$ consisting of open convex sets.

Let $x \in C$.

Then $x \in A$, and $A$ is open in $S$.

Thus by the definition of a basis, there is a $U \in \mathcal B$ such that $x \subseteq U \subseteq A$.

Since $U$ is convex and contains $x$, $U \subseteq C$.

Thus $C$ contains an open neighborhood of each of its points, so $C$ is open.

$\blacksquare$