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In ZFC, Dedekind finite set and finite set are same things. So I have a set say A(which is equal to N in ZFC) all Dedekind finite set are equivalent to proper subsets of A and A is well ordered set.

But I just get to know, and there is a model of ZF theory in which there exist an infinite but Dedekind finite set. So I have one question. Can we find a set A such that all Dedekind finite sets are equivalent to its proper subset and A is well ordered.(Of course N will not work)

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No. First we note that if $A$ is Dedekind-finite then all its subsets are Dedekind-finite. If $A$ is infinite, then for every $n$ there is some $A_n$ such that $|A|=|A_n|+n$. Therefore the cardinals $\{|A_n|\mid n<\omega\}$ make an infinite decreasing sequence of cardinals below that of $A$.

You might want to ask if the cardinals are even linearly ordered, because they don't have to be. But it is also consistent. If $A$ is an infinite set which cannot be partitioned into two infinite sets we call $A$ amorphous. It is not hard to see that the cardinals below $|A|$ are exactly $\omega\cup\{|A_n|\mid n<\omega\}$ which is a linear order.

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  • $\begingroup$ Hi Asaf. Can we actually pick a sequence $(A_n)_n$ with sizes as stated? (Naturally, we can pick the sequence $(|A_n|)_n$.) $\endgroup$ – Andrés E. Caicedo Sep 25 '14 at 18:11
  • $\begingroup$ Andres, no, since $2^A$ might be Dedekind-finite. For example if $A$ is amorphous. $\endgroup$ – Asaf Karagila Sep 25 '14 at 18:36
  • $\begingroup$ Yes, but the $A_n$ don't need to be subsets of $A$, so why does this matter? $\endgroup$ – Andrés E. Caicedo Sep 25 '14 at 20:14
  • $\begingroup$ Right, we are interested in arbitrary sets. I'll think about it some more. $\endgroup$ – Asaf Karagila Sep 25 '14 at 20:29

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