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It is consistent with $\mathrm{ZF}$ that a countable union of countable sets may be uncountable. As far as I understand it, this is because in absence of $\mathrm{AC}$ we cannot necessarily choose a bijection between each set in our family and $\omega$. (As this entails infinitely many choices.) I imagine the situation might be somewhat different with countable sets of sufficiently small countable ordinals, since for these such bijections may be described by some kind of formula.

Since there are still many pretty big creatures among countable-in-$\mathrm{ZFC}$ ordinals (while casually reading about ordinals on the internet, $\omega_1^{CK}$ in particular caught my attention, because of its definition in terms of computability) I am wondering:

What is the smallest possible ordinal $\alpha$, for which it is consistent with $\mathrm{ZF}$ that $\alpha$ is uncountable?

In other words, what is the smallest possible ordinal $\alpha$ (describable (I guess) in some elementary terms other than "$\alpha$ is the smallest uncountable ordinal" and possibly countable in $\mathrm{ZFC}$) such that it is consistent with $\mathrm{ZF}$ that $\alpha=\omega_1$. I hope this is not too vague ...

For example: is it consistent with $\mathrm{ZF}$ that $\varepsilon_0=\omega_1$ or perhaps $\omega_1^{CK}=\omega_1$? What is the best result known along these lines?

Just curious. Thanks.

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My guess is that Shoenfield's absoluteness should absolve all "definable" ordinals, and therefore their limit. So probably the answer is no for both ordinals specified. –  Asaf Karagila May 3 '12 at 14:12
    
One can ask a similar question in the presence of $0^\#$, where $\omega_1^V$ is a large cardinal in $L$. What ordinal could be $\omega_1^L$ and whether or not we can describe the "large cardinals" of $L$ below $\omega_1^V$. –  Asaf Karagila May 3 '12 at 14:17
    
@AsafKaragila: Thanks! Shoenfield's absoluteness looks pretty fascinating. –  Dejan Govc May 3 '12 at 14:24
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1 Answer

up vote 3 down vote accepted

I have no idea how to make sense of your question other than as follows: If there is a transitive model of ZF, then there is a countable transitive model of ZF+V=L, and then there is such a model of smallest height. Its $\omega_1$, call it $\omega_1^m$, is the ordinal I would identify as the answer to your question.

The thing is, if you talk about consistency with a formal theory, you need to be able to identify the ordinal within the theory, in a way that its value is unambiguous. This is not so easy. "The ordinal $\omega_1$" may change as you move between transitive models, so your unambiguous identification needs to be more robust. It is even more serious, as it needs to be something that you can identify even if you are looking at a non-standard (perhaps non-$\omega$-model). But then, what does it even mean that it is unambiguous?

So I imagine that you need to restrict yourself to transitive models, and then $\omega_1^m$ is the answer to you question.

By the way, $\omega_1^{CK}$ is explicitly countable. No ordinal you'd be able to name that is a potential candidate is going to work for similar reasons. Asaf cites in comments Shoenfield absoluteness, but invoking it is an overkill. If you so much as restrict yourself to $\omega$-models, the class of Turing machines and their outcomes is completely determined, and that uniquely identifies $\omega_1^{CK}$. Of course, much larger countable ordinals are also unambiguously identified.

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Andres, is there an answer to my $0^\#$ question in the comments? –  Asaf Karagila May 3 '12 at 14:29
    
@AsafKaragila It is the same thing. If you want a "minimal $\omega_1^L$ in the presence of $0^\sharp$", then look at the smallest transitive model of "ZFC+$0^\sharp$ exists", which is countable, and its $\omega_1$, call it $\omega_1^M$ would be the answer to your question. Of course, $\omega_1^M>\omega_1^m$, as the new theory proves the existence of the latter (but not of the former, by incompleteness). –  Andres Caicedo May 3 '12 at 14:33
    
@AsafKaragila As of "describing" the large cardinals in $L$ below $\omega_1^V$ if $0^\sharp$ exists, I again wouldn't know what this means precisely. There is a club subset of $\omega_1^V$ consisting of $L$-inaccessibles, and I don't know if there is much more to say beyond this. (There may be more to say, of course, depending on how one formalizes the question.) –  Andres Caicedo May 3 '12 at 14:37
    
@AndresCaicedo: Thanks, this will certainly give me something to think about. –  Dejan Govc May 3 '12 at 14:47
    
@Andres: I meant to ask this question internally, that is from the view of $V$ there are a lot of countable ordinals which are inaccessible in $L$. Can we say something about them? –  Asaf Karagila May 3 '12 at 16:55
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