# Isomorphism of preferences

Suppose we have a denumerable set $S$ with a total order $\preceq$. Is there necessarily an isomorphism $\phi:S\to\mathbb{N}$ such that $x\preceq y \Leftrightarrow \phi(x)\leq\phi(y)$?

Specifically in the case of $S=\mathbb{Q}$ I can't find one, so I wonder about the general principle.

If we allow the image to be $\mathbb{Q}$ then it seems possible, but I'm not sure it will work if we restrict it to the naturals (or even the integers).

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No, of course not. For example, $\mathbb{N}$ has a smallest element, and hence so does any order isomorphic to it. But $\mathbb{Q}$ has no smallest element, so is not isomoprhic to $\mathbb{N}$.