I'm currently studying topology, and in one of the lectures we were presented with a theorem that went something like this (rephrasing since I don't have the theorem in front of me):
Let $(X, <)$ be an ordered toplogy, and let $Y$ be a subset of $X$. Then $Y$ is convex iff $(a,b) \subseteq Y$ for all $a,b \in Y$.
Theorem. If $(X, <)$ is an ordered toplogical space and $Y \subseteq X$ is convex, then the subspace topology $Y$ inherts from $X$ is the same as the order topology with respect to $<$ restricted to $Y$.
I understand this theorem, but I don't understand why the same thing can't be said in general if $Y$ is open? The textbook lists some counterexamples of the theorem with non-convex sets, but in all cases these are closed. I haven't managed to find an counterexample with a non-convex open set either.
Could someone then please help me understand why openness apparently isn't sufficient for this theorem?
Consider the space $X=\{-1,0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the linear order topology; the subset $U=X\setminus\{0\}$ is open. $U$ is discrete in the subspace topology, but in the inherited order it is not discrete: $-1$ is a limit point.