If someone asks me how to prove that two order structures $\langle A,\leq \rangle$ and $\langle B,\preceq \rangle$ are isomorphic I would immediately suggest: try to find a function $f:A\to B$ such that $a \leq b$ iff $f(a) \preceq f(b)$, for all $a,b \in A$. Because given the definitions, this is the natural way the proof goes.
But if the same person asks me to show that $\langle A,\leq \rangle$ and $\langle B,\preceq \rangle$ aren't isomorphic, then I'm in trouble. I would start suggesting him: (1) pick up an arbitrary function $f:A\to B$ such that $a \leq b$ iff $f(a) \preceq f(b)$, for all $a,b \in A$, (2) try to find a contradiction. Then most of the times this is where I get stuck. Because I don't find any further information than this generic assumption, and I simply cannot check all possible functions.
This started with my (failed) attempt to show that $\langle \mathbb{N},< \rangle$ and $\langle \mathbb{Z},< \rangle$ aren't isomorphic.
Then I googled for some answers and I realize most people prove such "$X$ and $Y$ aren't isomorphic" statements by showing a property of $X$ (or $Y$) that is not satisfied by $Y$ (or $X$). Like:
$\langle \mathbb{N},< \rangle$ is well-ordered, $\langle \mathbb{Z},< \rangle$ isn't. QED.
Question: Why is this a legitimate approach? Why is it enough? As a logician, and looking at the definitions, my natural approach to show that two structures aren't isomorphic, is to assume a generic order-preserving mapping and then deriving a contradiction. How is this "property approach" in harmony with that?
Best Answer
The "property approach" does generate a contradiction. The full proof tends to follow these lines:
Suppose the structures $X$ and $Y$ are isomorphic. Then there exists an isomorphism $f : X \rightarrow Y$ between them. Property $P$ is preserved under isomorphism (due to some other proof), so because $X$ has property $P$, so must $Y$. But $Y$ does not have property $P$. Therefore, $X$ and $Y$ are not isomorphic.