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If we assume $AC$ we can construct an uncountable well-ordered chain of subsets of $\mathbb{N}$ by well-ordering the reals and then using the same Dedekind's cut construction as in this question.

edit: as pointed out in the comments and Asaf's answer the construction above doesn't work

What if we don't assume $AC$? Can we show the existence of an uncountable, well-ordered chain of subsets of $\mathbb{N}$ in $ZF$ alone?

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Alessandro Codenotti
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    No, we can't, because it is consistent with $\mathsf{ZF}$ that $\omega_1$ does not inject into $\mathfrak c$. – Andrés E. Caicedo Feb 05 '16 at 19:53
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    Note that there's a slight ambiguity here: when you say "well-ordered chain," do you mean "a chain which can be well-ordered" or "a chain whose $\subseteq$-ordertype is well-ordered"? Note that uncountable instances of the latter can't happen, even with AC: if $A_\eta\subseteq \mathbb{N}$ for $\eta<\omega_1$ with $\alpha<\beta$ implies $A_\alpha\subsetneq A_\beta$, then for each successor $\alpha<\omega_1$ there is some $n_\alpha\in A_\alpha\setminus\bigcup_{\beta<\alpha} A_\beta$. Since there are uncountably many successor ordianls but only countably many naturals, this can't happen. – Noah Schweber Feb 05 '16 at 20:23

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You're confusing well-ordered with well-orderable.

There is no uncountable well-ordered chain in $\mathcal P(\Bbb N)$, because if there was such $\{A_\alpha\mid\alpha<\omega_1\}$, map $\alpha$ to the least $n$ in $A_{\alpha+1}\setminus A_\alpha$, and you got yourself an injection from $\omega_1$ into $\omega$. Something which you cannot get with or without choice.

But it is also consistent without the axiom of choice that there is no uncountable well-orderable chain in $\mathcal P(\Bbb N)$, as it is consistent that there is no injection from $\omega_1$ into $\mathcal P(\Bbb N)$ which means every well-orderable subset is countable. So we cannot prove in $\sf ZF$ the existence of an uncountable well-orderable subset of $\mathcal P(\Bbb N)$, chain or otherwise.

Asaf Karagila
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  • Note that this argument only works for successor $\alpha<\omega_1$ (since we could have $A_\lambda=\bigcup_{\beta<\lambda} A_\beta$) - but that's still enough. – Noah Schweber Feb 05 '16 at 20:25
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    I'm fairly sure that it also works for the chain ${{0,\ldots,n-1}\mid n\in\Bbb N}$ which has order type $\omega$. :-) Are you claiming $\omega$ is a successor ordinal? :-P – Asaf Karagila Feb 05 '16 at 20:26
  • Whoops, I read the argument as ". . . the least natural number in $A_\alpha$ which . . ." Silly me! – Noah Schweber Feb 05 '16 at 20:31
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    This is interesting. We had been discussing in chat that one should get such a chain by the same construction as in the linked question by first picking a well-ordering of the reals. But what fails is that we are no longer guaranteed that any open interval contains rationals (in fact, this shows that in any well-ordering of the reals there will be an open interval containing no rationals, and same if we replace the rationals by any countable subset. Quite interesting). – Tobias Kildetoft Feb 05 '16 at 20:38
  • Aha! So that's why this construction fails, interesting. @Asaf When you say that there are no well-ordered chains in $\mathcal{P}(\mathbb{N})$ you're speaking about uncountable chains, right? – Alessandro Codenotti Feb 05 '16 at 20:43
  • @Alessandro: Yes, thanks! – Asaf Karagila Feb 05 '16 at 20:46
  • Actually, I think this argument still needs a bit of work: we could for instance have $0\not\in\bigcup A_\alpha$, so that $A_\alpha$ always gets mapped to $0$. This is easily fixed (WLOG assume $\bigcup A_\alpha=\mathbb{N}$) but more problematically the map could still be constant for long periods of time. So all this really gets is a cofinal map from $\omega$ to $\omega_1$, which could exist in ZF. I think we do want to look at the least natural in $A_\alpha$ not in any prior $A_\beta$, and to restrict to successor $\alpha$s so that this works. Thoughts? – Noah Schweber Feb 05 '16 at 20:52
  • @Noah: You're right about the first part. I'm not sure about the second. If you map $\alpha$ to the least $n$ which wasn't used thus far, and does not appear in any previous set, you should get an injection into $\omega$. You're right about successors being the easy way out, though. Map $\alpha$ to the least element in $A_{\alpha+1}\setminus A_\alpha$, and you definitely get an injection into $\omega$. – Asaf Karagila Feb 05 '16 at 20:57
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There is none since the set of integers is countable. The dedekind cut produces an uncountable chain which isn't well ordered.

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    Countability of integers doesn't imply that there are no such chains. – Wojowu Feb 05 '16 at 20:19
  • The OP is not asking for the chain to be well-ordered according to $\subseteq$, just that it be well-orderable in some way; see my comment to the question. – Noah Schweber Feb 05 '16 at 20:23