# Every order topology is regular (proof check)

My proof:

Let $X$ be an space with the order topology, $x \in X$ and $F$ a closed set that does not contain $x$. Then, the set $X-F$ is an open set that contains $x$, hence there is an open set (basic) $(a,b)$ such that $x \in (a,b)\subseteq X-F$. Then $(a,b)$ and $(-\infty,a) \cup (b,\infty)$ are open disjoint sets that separate $x$ and $F$.

I am not sure because I’ve seen other proofs and they are much more complicated, like this one (source):

Besides, in that proof I don’t understand when do they use the fact that a point is closed in a Hausdorff space. I also found that the last union is not disjoint, I double check that, but I may be missing something.

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Your proof wouldn’t work as while your (a,b) could exist, there is nothing to say that a and b aren’t in F. So your open set wouldn’t necessarily contain F.