I've doubt about Dedekind-infinite sets, sets which are in bijection with a proper part, in the ZF axiomatic framework, without Axiom of Choice.

Assume a Dedekind-infinite set X exists.

Then it can be proved X contains a Dedekind-infinite N set which satisfy the Peano Axioms.

This set N can be well ordered using traditional arguments from Peano Axioms.

It can also be proved its initial chains, in this well order, I_n = { m < n }, are Dedekind-finite.

Can be proved they are also finite using the ZF definition of "finite set", without any further assumption?

  • $\begingroup$ You may find useful my blog posting on defining finite or infinite sets at dcproof.com/Infinity.html There, I start by developing a non-numeric definition of a finite set based on a walk through a finite village. This in contrast to the usual approach using Hilbert's Infinite Hotel to develop the notion of an infinite set. $\endgroup$ – Dan Christensen Mar 17 at 16:31

Yes. The definition of finite is being equipotent with a proper initial segment of $\omega$, that is the ordinal corresponding to $\Bbb N$.

Once you have established a bijection, and in fact an order isomorphism, between your "copy of $\Bbb N$" and $\omega$, and you have by virtue of it being a model of $\sf PA_2$, then the initial segments correspond exactly to the actual finite cardinals.

The axiom of choice is needed when you want to prove that a Dedekind-finite set is finite. This is not what you're doing here. You're not arguing that these are Dedekind-finite, you're quite literally defining them to be finite.

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  • $\begingroup$ Hence, taking your words, assuming a Dedekind-infinite set exists is enough to build a model of PA2 and allow the existence of a countable well ordered set, a well order isomorphic with the one of the first infinite ordinal, omega. But once we built the first infinite ordinal, each countable set, a set in bijection with a model of natural integers N, can be well ordered, using the bijection to induce a well order on the set. Does this mean assuming one Dedekind-infinite set exists imply the Countable Axiom of Choice, in the ZF axiomatic framework? $\endgroup$ – Giuseppe Vitillaro Mar 18 at 9:33
  • $\begingroup$ No, ZF proves the existence of infinite ordinals, all of which, almost trivially, are Dedekind infinite. Again, choice is only necessary when you try to prove the equivalence between infinite and Dedekind infinite. $\endgroup$ – Asaf Karagila Mar 18 at 9:45
  • $\begingroup$ I asked a different question. Assuming at least one Dedekind-infinite set exists, in the ZF axiomatic framework, imply Countable Axiom of Choice? Not the general Axiom of Choice, just the countable instance of the axiom. $\endgroup$ – Giuseppe Vitillaro Mar 18 at 9:54
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    $\begingroup$ And I answered. No. ZF proves there are Dedekind infinite sets, but it does not prove countable choice. $\endgroup$ – Asaf Karagila Mar 18 at 9:55
  • $\begingroup$ Right. Then something has to be wrong in the logic of my main question. If you read what I wrote, adding what you wrote, we may prove just this statement: "The existence of at least one Dedekind-infinite set exist IMPLY Countable Axiom of Choice". Which is FALSE. Something has to be wrong in my assumptions above. Isn't it? $\endgroup$ – Giuseppe Vitillaro Mar 18 at 10:06

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