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I hear it's consistent with ZFC to have

$$ 2^{\aleph_0} = \aleph_n $$

for any $n = 1, 2, 3, \dots $. How much worse can it get?

More precisely: are there models of ZFC with $2^{\aleph_0} \gt \aleph_n$ for all $n$? What's the 'world record' when it comes to finding models where $2^{\aleph_0}$ is 'very large' in some sense? And on the other hand, are there theorems putting interesting bounds on how large $2^{\aleph_0}$ can be?

(I count $2^{\aleph_0} \lt 2^{2^{\aleph_0}}$ as an uninteresting bound.)

According to Wikipedia:

A recent result of Carmi Merimovich shows that, for each $n \ge 1$, it is consistent with ZFC that for each $\kappa$, $2^\kappa$ is the $n$th successor of κ. On the other hand, László Patai (1930) proved, that if $\gamma$ is an ordinal and for each infinite cardinal $\kappa$, $2^\kappa$ is the $\gamma$th successor of $\kappa$, then $\gamma$ is finite.

However, I'm interested in failures of the Continuum Hypothesis, not the Generalized Continuum Hypothesis.

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John Baez
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    I think this is a serious gap in Wikipedia. Systematic failures of GCH are interesting (and require extremely high-level machinery) but, as far as $2^{\aleph_0}$ is concerned, the answer has been known since soon after Cohen! Basically, the only restriction is provided by König's Theorem that $\operatorname{cf}(2^{\aleph_0}) > \omega$. – François G. Dorais Sep 16 '15 at 02:52
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    Let me restate François' last sentence: "$2^{\aleph_0}$ can be anything it ought to be", which is the title of Solovay's article. – Burak Sep 16 '15 at 04:02
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    This question seems to be quite popular on Math.SE, as it turns out. Here are a few threads (My answers there are not on par, in terms of clarity, as Joel's, but others answered as well!): one, two, three; and if you wondered how the axiom of choice comes into play, then four, five and six will be interesting to you. – Asaf Karagila Sep 16 '15 at 13:59
  • @Burak: Would it be possible for you to provide a cite for that Solovay article? I looked for a downloadable copy on Solovay's homepage but couldn't find any. – Thomas Benjamin Sep 23 '15 at 02:26
  • @ThomasBenjamin: Unfortunately, I do not have an electronic copy either (My knowledge on this article is secondhand). According to Google, that article should have appeared in "The Theory of Models, Proceedings of the 1963 International Symposium at Berkeley". – Burak Sep 24 '15 at 02:27
  • @Burak: Thanks, I'll check that out. – Thomas Benjamin Sep 24 '15 at 02:57

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Solovay proved shortly after Cohen's result on the independence of CH that in any model of set theory $V$, if $\kappa^\omega=\kappa$, then there is a forcing extension in which $2^\omega=\kappa$. The forcing is simply $\text{Add}(\omega,\kappa)$, the forcing to add $\kappa$ many Cohen reals. Thus, the continuum $2^\omega$ can be $\aleph_{\omega+1}$ or $\aleph_{\omega_1}$ or $\aleph_{\omega_1+\omega^2+17}$ or what have you, even weakly inaccessible, if you had such a cardinal available in the ground model. But it cannot be $\aleph_\omega$ or $\aleph_{\omega_1+\omega}$ or any cardinal with cofinality $\omega$, because $(2^\omega)^\omega=2^\omega$ but König's theorem shows $\kappa^{\text{cof}(\kappa)}>\kappa$ for any infinite cardinal $\kappa$.

In particular, if you start in a model of the GCH, then the continuum can be made to be equal to $\delta^+$ for any infinite successor cardinal in the ground model, and this is cardinal-preserving forcing, so the meaning of the successor function is the same in the ground model as in the extension.

Easton vastly generalized this result to show that if you start in a model of the GCH, and if $E$ is any function defined on the infinite regular cardinals and obeying the following requirements (now called an Easton function)

  • $E(\kappa)>\kappa$
  • Further, $\text{cof}(E(\kappa))>\kappa$
  • $\kappa\leq\lambda\to E(\kappa)\leq E(\lambda)$

then $E$ can become the continuum function $\kappa\mapsto 2^\kappa=E(\kappa)$ as computed in a cardinal-preserving forcing extension of the universe. See Easton's theorem.

The Easton requirements amount to the trivial requirements on the continuum function imposed by the facts that $\kappa\leq\lambda\to 2^\kappa\leq 2^\lambda$ and König's theorem $\text{cof}(2^\kappa)>\kappa$.

As a result, we can have $2^{\aleph_n}=\aleph_{\omega^2+\omega+p_n}$ where $p_n$ is the $n^{th}$ prime, if we like. There is infinite flexibility, and basically anything is allowed subject to the Easton requirements, which are trivial limitations.

Easton's theorem does not apply at all to singular cardinals, however, and controlling the continuum function at singular cardinals is an extremely subtle issue.

Joel David Hamkins
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  • How is this related to Easton's theorem?: https://en.wikipedia.org/wiki/Easton%27s_theorem – Sam Hopkins Sep 16 '15 at 02:53
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    How funny---I was writing about Easton's theorem just as you typed that. – Joel David Hamkins Sep 16 '15 at 02:54
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    Very nice answer. The only thing to add is that (should there be the appropriate large cardinals in the ground model), in the extension, the continuum can be a significantly large weakly inaccessible cardinal (meaning, we can ensure it retains a good part of the combinatorics of the original large cardinal). – Andrés E. Caicedo Sep 16 '15 at 03:51
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    @AndresCaicedo Yes, I had mentioned that $2^\omega$ can be made weakly inaccessible, if such a cardinal is available in the ground model. Solovay has advocated that the continuum should turn out to be real-valued measurable, which would be the kind of case you are talking about (and this is consistent with ZFC assuming a measurable cardinal), and one can of course arrange even stronger properties, if there are larger cardinals available in the ground model; just pump up the continuum to that level. – Joel David Hamkins Sep 16 '15 at 13:27
  • @JoelDavidHamkins: Regarding your statement, "...one can of course arrange even stronger properties, if there are larger cardinals available in the ground model; just pump up the continuum to that level": could you provide an example (or a reference, if you prefer) of such a larger cardinal and how you would "pump up the continuum to that level"? – Thomas Benjamin Sep 23 '15 at 03:02
  • I just meant that if $\kappa$ is some kind of large cardinal in $V$, then there are various forcing extensions $V[G]$ in which $2^\omega=\kappa$, in which $\kappa$ will retain a residue of the large cardinal property it once had. For example, if $\kappa$ is measurable and one performs the random real forcing, then $\kappa=2^\omega$ will be real-valued measurable in $V[G]$. I think this is in Kanamori's book. – Joel David Hamkins Sep 23 '15 at 10:50
  • @JoelDavidHamkins: Thanks. But of course if $V$ contained $\kappa$ measurable, say, $\mathfrak c$ could not be 0-1-measurable, or supercompact, Vopenka, or $I1$-$I0$ if $V$ contained any of those cardinals, right? – Thomas Benjamin Sep 23 '15 at 10:58
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    No, $\frak{c}$ will not have those large cardinal properties, since it is not inaccessible, but it will retain some of the properties that those large cardinals have, such as being weakly inaccessible, weakly Mahlo and so on, or having various kinds of nice ideals and so on. – Joel David Hamkins Sep 23 '15 at 11:05
  • @JoelDavidHamkins: I happened to notice that in 'cantor's attic' there is no page for the term "real-valued measurable". Why is that? Since it is known that $\mathfrak c$ can be real-valued measurable, is that as large as $\mathfrak c$ can get? Where in cantor's attic could one expect to find real-valued measurable cardinals? – Thomas Benjamin Sep 24 '15 at 01:40
  • It will eventually be part of the page on measurable cardinals (see http://cantorsattic.info/Measurable), since these large cardinals have the same consistency strength. Every measurable cardinal is also real-valued measurable, but meanwhile, the real-valued measurable cardinals that are not measurable are at most continuum. – Joel David Hamkins Sep 24 '15 at 02:13