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Let $X$ be a set and let $<_1,<_2$ be order relations on $X$.

Let $T_1,T_2$ be the topologies induced on $X$ respectively.

If $(X,T_1)$ is homeomorphic to $(X,T_2)$, does that imply that $(X,<_1)$ and $(X,<_2)$ are order isomorphic?

And a derived philisophical question: The other way around is easy to prove, so if this holds this means that, in some sense, homeomorphism between order topologies is equivalent to order isomorphism. What does that mean?

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up vote 5 down vote accepted

The answer is no. A trivial counterexample: if $\le$ is a linear order on $X$, $\langle X,\tau_\le\rangle$ and $\langle X,\tau_\ge\rangle$ are homeomorphic, but the orders are anti-isomorphic. Less trivially, the discrete topology on $X$ is generated by any discrete linear order on $X$, of which there are many. For example, if $\langle A,\le\rangle$ is any countably infinite linear order, the lexicographic order on $\langle A\times\Bbb Z\rangle$ induces the discrete topology on $A\times\Bbb Z$ and hence, via some bijection, on $\Bbb N$.

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