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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.

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

1 Answer 1

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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.

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  • $\begingroup$ I see, but the problem is we were not yet introduced to the concept of isomorphism, so I doubt I can state it in regards to mapping endpoints to endpoints. Is there any way without it? $\endgroup$ Commented Feb 2, 2014 at 2:21
  • $\begingroup$ Then what is "alike" that you mention? (I am asking because I have never seen that term being used, and you did mention "order type" which is the equivalence class under isomorphism...) $\endgroup$
    – Asaf Karagila
    Commented Feb 2, 2014 at 2:26
  • $\begingroup$ It is isomorphism, however we were not introduced to the fact it must map endpoints to endpoints, and it was not regarded as isomorphism (yet, at least). $\endgroup$ Commented Feb 2, 2014 at 2:30
  • $\begingroup$ Then prove that endpoints must go to endpoints! It's not a difficult thing to prove! $\endgroup$
    – Asaf Karagila
    Commented Feb 2, 2014 at 2:31
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    $\begingroup$ Huh, didn't think about that. Cheers! $\endgroup$ Commented Feb 2, 2014 at 2:32

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