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I'm wondering about the following (I've written a couple set theory papers but don't consider myself a set theorist, so please keep this in mind when answering):

Is it consistent w/ ZFC that there is a cardinal $\kappa$ such that $\aleph_0<\kappa<2^{\aleph_0}$, yet for every cardinal $\beta\geq2^{\aleph_0}$, there are no cardinals properly between $\beta$ and $2^\beta$? Any references would be most appreciated.

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    $\begingroup$ The consistency of the statement in your question can be easily obtained by adding $\aleph_2$ Cohen reals over $L$. $\endgroup$
    – Hanul Jeon
    Commented Dec 14, 2016 at 16:52

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Well, by Easton's theorem, we immediately get the following:

It is consistent that $2^{\aleph_0}=\aleph_{17}$ (say) and $2^\kappa=\kappa^+$ for every regular cardinal $\kappa\ge\aleph_{17}$.

In principle, singular cardinals could pose a problem, and indeed this comes up when trying to make the continuum function $\kappa\mapsto 2^\kappa$ do more complicated things. However, it's easy to see that they don't: it's not hard to check that adding $\aleph_{17}$-many Cohen reals to a model of GCH does the job.

In fact, this is actually much easier to prove than Easton's theorem; I mostly mentioned Easton above for context (and because it's cool).


Here are the details:

Let our ground model $V$ satisfy GCH, and let $\mathbb{P}$ be the forcing adding $\aleph_{17}$-many Cohen reals. The trick is nice names - if $x$ is a set in $V$, a nice name for a subset of $x$ is a family $(A_i)_{i\in x}$ of $x$-many maximal antichains in $\mathbb{P}$, together with maps $e_i: A_i\rightarrow 2$ for each $i\in x$.

It's not hard to see that every subset $\alpha$ of $x$ in the generic extension $V[G]$ has a corresponding nice name $\nu_\alpha=((A_i)_{i\in x}, (e_i)_{i\in x})$ - that is, such that $$i\in\alpha\iff \exists a\in A_i\cap G[e_i(a)=1].$$ So counting the subsets of $x$ in $V[G]$ amounts to counting the nice names. Since $\mathbb{P}$ has size $\aleph_{17}$, there are at most $2^{\aleph_{17}}=\aleph_{18}$-many maximal antichains; for any $\kappa\ge\aleph_{17}$, there are $\aleph_{18}^\kappa=2^{(\aleph_{17}})^\kappa=2^{\aleph_{17}\cdot \kappa}=2^\kappa=\kappa^+$-many $\kappa$-sequences of antichains; and similarly, only $\kappa^+$-many sequences of appropriate maps.

So we get, for $\kappa\ge\aleph_{17}$, that there are at most $\kappa^+$-many nice names for subsets of $\kappa$, and clearly there are at least $\kappa^+$-many nice names for subsets of $\kappa$; so $(2^\kappa)^{V[G]}=\kappa^+$.

(Note that we also need a nice name argument to say that $2^{\aleph_0}=\aleph_{17}$ in $V[G]$; all the forcing gives us trivially is that the continuum is at least $\aleph_{17}$, and weird things can happen (if you add $\aleph_\omega$-many Cohens to a model of GCH, you get $\aleph_{\omega+1}$-many reals). However, the argument is basically the same.)

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    $\begingroup$ I prefer $\aleph_{42}$. $\endgroup$ Commented Dec 14, 2016 at 16:57
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    $\begingroup$ @AndrejBauer In a talk I gave a while ago, I referred to the ground model of a set forcing extension as the "towel", so that when I presented the Woodin/Laver result on the definability of the ground model, I could phrase it as "$V[G]$ always knows where its towel is." $\endgroup$ Commented Dec 14, 2016 at 17:00
  • $\begingroup$ And I believe John Bell proves in "Boolean valued models and independance proofs" that when forcing the negation of the GCH, under some barely restrictive conditions, GCH has to hold above a certain cardinal, so that you can have $2^\aleph_0 > \kappa$ for any cardinal, but under these restrictions, there has to be some $\lambda$ such that for all $\mu>\lambda$, $2^\mu = \mu^+$ $\endgroup$ Commented Dec 14, 2016 at 17:16
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    $\begingroup$ @Max Yup, and that's a useful fact - but note that that doesn't immediately answer this question, which asked for GCH to hold immediately. $\endgroup$ Commented Dec 14, 2016 at 17:18
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    $\begingroup$ You didn't really mean "every regular cardinal $\kappa$" in the emphasized statement at the beginning. You probably meant "every regular cardinal $\geq \aleph_{17}$" which is what your proof shows. In fact, it's OK for $k=\aleph_{16}$ also, but not any lower. (Getting the result for $\aleph_{16}$ involves using the countable chain condition to bound the number of antichains by $\aleph_{17}^{\aleph_0}=\aleph_{17}$ instead of by $\aleph_{18}$.) $\endgroup$ Commented Dec 14, 2016 at 21:21

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