order preserving implies isomorphism

Question

Let $M = (M, <)$ and $N = (N, <)$ be two dense totally ordered sets without endpoints.

Let $F$ be the collection of order preserving maps between finite subsets of $M,N$ respectively.

Then $F$ is a non-empty family of partial isomorphisms between $M$ and $N$ with the back-and-forth property.

It's clear that $F$ is non-empty and each member of $F$ is an injective map, but how can I show it's onto?

Answer 1

The crucial notion to define here is:

map between finite subsets of $M,N$, respectively.

It would seem that you interpreted this as:

$f: M_0\to N_0, \quad M_0\in \mathcal P^{<\omega}(M), ~N_0\in\mathcal P^{<\omega}(N)$

where $\mathcal P^{<\omega}(X)$ is the set of finite subsets of $X$ -- which is a very natural thing to do. Now, given such an $f:M_0\to N_0$, we can "extend" it to $f':M_0\to N_1$ with $N_0\subsetneq N_1$, and $f'$ wouldn't be onto.

So it seems that "between" in "map between [...]" includes a hidden/omitted surjectivity assumption -- i.e., the intended definition is:

$f: M_0\to N_0, \quad M_0\in \mathcal P^{<\omega}(M),~ N_0\in\mathcal P^{<\omega}(N),~ f[M_0]= N_0$

Now if we add the "order-preserving" assumption on $f$ (that is to say, $f \in F$), it readily follows that $f$ is an order isomorphism.

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Alternatively, one could drop the surjectivity condition for a "partial isomorphism from $M$ to $N$", but to me this definition means "an isomorphism between subsets of $M$ and $N$".