What is the smallest cardinal $\beta$ such that it is provable in ${\sf (ZFC)}$ that $2^{\aleph_\beta} > 2^{\aleph_0}$?
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asked Jan 16, 2019 at 8:40
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4$\begingroup$ There is no such $\beta$. All the many theorems about the size of the continuum say exactly otherwise. $\endgroup$– Asaf Karagila ♦Jan 16, 2019 at 9:37
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7$\begingroup$ I guess $\beta=2^{\aleph_0}$ is one such answer. The question is a bit subtle though - ZFC can't directly talk about a statement for arbitrary ordinal, we need the ordinal to be defined by some formula. And the formula can essentially say "$\beta$ is the least such that $2^{\aleph_\beta}>2^{\aleph_0}$"... One statement Asaf (I presume) refers to is Easton's theorem which is a model-theoretic in nature, but should answer some reasonable version of this question. $\endgroup$– WojowuJan 16, 2019 at 9:52
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3$\begingroup$ The question is ill-posed. It asks about min of a certain "collection" of cardinals. But this "collection" is not definable in ZFC. I mean, the scheme of specification cannot be applied, because "such that it is provable in ZFC" is not a formula of ZFC. Maybe logicians would interpret the question by allowing some embedding between models, but it's not currently asked this way. $\endgroup$– YCorJan 16, 2019 at 10:48
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1$\begingroup$ Just an analogy: I recently read a paper (published in a serious journal) in which the authors apply Cohen's results on independence of GCH as "for every cardinal $\kappa$ there exists a model of ZFC in which $\kappa<2^{\aleph_0}$". Applying this to $\kappa=2^{\aleph_0}$ results in an obvious paradox, showing that some care is needed is formulating those results. $\endgroup$– YCorJan 16, 2019 at 10:56
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1$\begingroup$ @user57888 thanks, you're right. So the question sounds strictly equivalent to (in ZFC): let $\gamma$ be the min of the (nonempty) set of cardinals $\alpha\le 2^{\aleph_0}$ such that $2^\alpha>c$. What is $\gamma$ (or, writing $\gamma=\aleph_\delta$, what is $\delta$)? Under GCH, $\delta=1$, and there are models of ZFC for which $\delta>0$. It sounds like the intended question somewhat is "what is the sup of $\delta$ over all models of ZFC?" and I'm not sure this makes sense. $\endgroup$– YCorJan 16, 2019 at 17:07
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1 Answer
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Perhaps the following may clarify the comments: for any ordinal $\delta$, there is a Boolean-valued extension of the universe of sets where $2^{\aleph_0}>\aleph_\delta$ holds. If you rather talk of models than Boolean-valued extensions, what this says is that we can force while preserving all ordinals, and in fact all initial ordinals, and make the continuum larger than any given value.
(In particular, if $\delta=2^{\aleph_0}$, then in the resulting extension the continuum is larger than $\delta$, so of course the statement that $\delta=2^{\aleph_0}$ is no longer true there. Note that we preserve initial ordinals, that is, the ordinals $\alpha$ that are cardinals remain the same, but naturally other sets change, for instance, if we add new reals to the universe, then the set that used to be $\mathbb R$ is no longer the $\mathbb R$ of the extension. The equality $\delta=|$"old $\mathbb R$"$|$ is still true, though.)
Anyway, the point of the above is that there is no bound. This was established by Solovay and was one of the first results following Cohen's. Solovay's result shows that in fact there are no "hidden" restrictions: By a theorem of König, the cofinality of the continuum must be larger than $\aleph_0$. Solovay showed that, starting with a model of GCH, for any initial ordinal $\delta$ of uncountable cofinality, there is a forcing that preserves ordinals, initial ordinals, and cofinalities, and makes $2^{\aleph_0}=\delta$. Note that we may even have $\delta=\aleph_\delta$. Naturally, many large cardinal properties may be destroyed in the process, but, starting with appropriate assumptions, we may obtain models where the continuum is weakly inaccessible, weakly Mahlo, or more.
On the other hand, of course $2^{\aleph_0}<2^{2^{\aleph_0}}$, and there are models where $2^{\aleph_0}=2^\kappa$ for any $\kappa<2^{\aleph_0}$, so in a sense $\beta=2^{\aleph_0}$ could be seen as an answer to the question.)
answered Jan 16, 2019 at 18:27
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$\begingroup$ Thanks for your lovely answer @andresecaicedo! - So I take it that it is consistent that $\aleph_\beta \leq 2^{\aleph_0}$ for all $\beta\in 2^{\aleph_0}$? In that case, the last sentence of your post provides the perfect answer to my question. $\endgroup$ Jan 17, 2019 at 14:40
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1$\begingroup$ @Dominic Yes, indeed, we may have $2^{\aleph_0}$ to be a fixed point of the sequence of alephs, that is, a $\kappa$ such that $\kappa=\aleph_\kappa$ so that for any $\beta<2^{\aleph_0}$, also $\aleph_\beta<2^{\aleph_0}$, and simultaneously we may arrange that $2^\lambda= 2^{\aleph_0}$ for all infinite $\lambda<2^{\aleph_0}$. $\endgroup$ Jan 17, 2019 at 16:36