What is the smallest cardinal $\beta$ such that it is provable in ${\sf (ZFC)}$ that $2^{\aleph\_\beta} > 2^{\aleph\_0}$?



Smallest $\beta$ such that it is provable that $2^{\aleph_\beta} > 2^{\aleph_0}$
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.




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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.)


