# Likeness of strictly well-ordered sets

The question itself:

Given a strictly ordered set $\langle A, < \rangle$ (< merely symbolises some kind of an order in this case) which is dense, with a least and greatest elemnts that are not the same. Prove that if A is countable, then $\langle A, < \rangle$ is alike to $\langle Q \cap[0,1], < \rangle$ (in the latter case, < is the 'regular' kind of order>.

Two sets are alike to each other if they have the same order type, or if you can prove there is a one-to-one and unto function that also keeps the order.

Now, I tried to approach the question from few directions yet I have failed to reach anything meaningful. I considered trying to show that both are alike to the same order type, yet there is no order type they are both alike to. They are both close to $\eta$ but both have a first and last element, so they are not of that order type, and I can't think of a way to prove that since they are both different to that order type in the same way, they are alike. I also tried to look for a fitting function, with no success as I do not know what kind of an order there is in A.

Buttom point, I am stuck. Any hints, suggestions or thoughts will be extremely appreciated.

• Note, by the way, that neither $A$ nor $\Bbb Q\cap[0,1]$ are well-ordered. They just happened to have endpoints. Commented Feb 2, 2014 at 2:15

If you already know about the fact that every countable dense order without endpoints has order type $\eta$ then it's easy.
Observe that an isomorphism must map endpoints to endpoints. Therefore $A\setminus\{\min A,\max A\}$ is isomorphic to $\Bbb Q\cap(0,1)$ and both are isomorphic to $\Bbb Q$. Now since the isomorphism must map $\min A$ to $0$ and $\max A$ to $1$ we're done.