25
$\begingroup$

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.

$\endgroup$
6
  • 7
    $\begingroup$ 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 $\operatorname{cf}(2^{\aleph_0}) > \omega$. $\endgroup$ Sep 16, 2015 at 2:52
  • 4
    $\begingroup$ Let me restate François' last sentence: "$2^{\aleph_0}$ can be anything it ought to be", which is the title of Solovay's article. $\endgroup$
    – Burak
    Sep 16, 2015 at 4:02
  • 4
    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Sep 16, 2015 at 13:59
  • $\begingroup$ @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. $\endgroup$ Sep 23, 2015 at 2:26
  • $\begingroup$ @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". $\endgroup$
    – Burak
    Sep 24, 2015 at 2:27

1 Answer 1

30
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ How is this related to Easton's theorem?: en.wikipedia.org/wiki/Easton%27s_theorem $\endgroup$ Sep 16, 2015 at 2:53
  • 2
    $\begingroup$ How funny---I was writing about Easton's theorem just as you typed that. $\endgroup$ Sep 16, 2015 at 2:54
  • 5
    $\begingroup$ 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). $\endgroup$ Sep 16, 2015 at 3:51
  • 2
    $\begingroup$ @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. $\endgroup$ Sep 16, 2015 at 13:27
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 23, 2015 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.