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Can someone say something about my version of "order topology implies Hausdorff"

(WLOG) Let $a <b$, and let $U_1,U_2$ be a neighborhood of $a,b$ respectively. Denote $U_1 = (a - \epsilon, a + \epsilon)$ where $\epsilon = (b - a)/2$. So clearly $U_1 \cap U_2 = \emptyset$.

I am correct right? I just have to one pair of disjoint open sets

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An order topology on a set does not presupose a natural addition operator, it only presupposes a total order on the underlying set.

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I think you're assuming that every ordered space is a metric space, which might not be the case.

Try to think of examples of ordered spaces $X$ in which $d(a,b)$ is not defined for $a,b\in X$.

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