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It's now well known that Continuum Hypothesis (CH) is independent from standard axioms of set theory: one can assume that $c=\aleph_1$ or assume that $c \neq \aleph_1$. Let us assume the second case-then the natural queation rise:

What 'values' $c$ can take? I know that it's impossible that $c=\aleph_{\omega}$? Is it possible to give simple criterium for a cardinal $\alpha$ to have a property: $c=\alpha$ in some model of set theory? In particular, is there a cardinal $\alpha$ defined not using the term $c$ for which we have $c<\alpha$? And the last question, is it possible for $c$ to be weakly inaccessible?

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To add to Andreas's answer below, you may be interested in Easton's Theorem (en.wikipedia.org/wiki/Easton%27s_theorem) as a further generalization of Solovay's result. –  Trevor Wilson Aug 11 '12 at 21:29
    
You mentioned the impossibility of $\mathfrak{c} = \aleph_\omega$, but you put a question mark, so I thought I'd mention that the reason is $\mathfrak{c}^\omega = (2^\omega)^\omega = 2^{\omega \times \omega} = 2^\omega = \mathfrak{c}$, but if $\kappa$ has countably cardinality, e.g., $\aleph_\omega$, then $\kappa^\omega > \kappa$ by Koenig's theorem. –  Trevor Wilson Aug 11 '12 at 21:34
    
The notion of "a cardinal $\alpha$ defined not using the term $c$" (in the penultimate question) is too vulnerable to cheating. You can avoid the term $c$ by formulating things in the primitive language of ZFC, with only $\in$. A less absurd $\alpha$ would be the smallest cardinal such that, if you partition the set $[\alpha]^2$ of its 2-element subsets into two pieces, there is an uncountable $H\subseteq\alpha$ whose 2-element subsets all lie in the same piece. A theorem of Erdös and Rado says that this $\alpha$ is $c^+$. –  Andreas Blass Aug 11 '12 at 21:59
    
You could also characterize $\mathfrak{c}^+$ as the smallest cardinal $\alpha$ such that any countable family of partitions of $\alpha$ has a nontrivial common refinement. This might be cheating, but I think it's hard to define cheating. –  Trevor Wilson Aug 11 '12 at 22:27

1 Answer 1

Any cardinal of uncountable cofinality can consistently be the cardinal of the continuum. In particular, it is consistent that the cardinal of the continuum is weakly inaccessible (assuming, of course, that inaccessible cardinals are consistent.) The relevant paper is Solovay's "$2^{\aleph_0}$ can be anything it ought to be" but nowadays the result is also in standard textbooks.

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In more detail: If you start with a universe satisfying GCH and an uncountable cardinal $\kappa$ (of that universe) and adjoin $\kappa$ Cohen reals in the usual way, the resulting model will have $\mathfrak c$ equal to $\kappa$ or $\kappa^+$ according as the cofinality of $\kappa$ was uncountable or countable. Furthermore, the forcing won't change cardinals or cofinalities (because it satisfies the countable chain condition). If you prefer, you can use random reals instead of Cohen reals; or if $\kappa$ has uncountable cofinality, you can force Martin's axiom and $\mathfrak c=\kappa$. –  Andreas Blass Aug 11 '12 at 21:27
    
I'm confused by this since $2^{\aleph_0}$ cannot be strongly inaccessible. Why is there an asymmetry here between weakly and strongly inaccessibles, given that both have uncountable cofinality? Also, do you know if it's consistent that $2^{\aleph_\alpha}$ be the $\alpha$-th weakly inaccessible cardinal for all ordinals $\alpha$, assuming the consistency of a proper class of inaccessibles? –  Jesse Elliott Dec 11 at 23:44
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@JesseElliott After the forcing, $\kappa$ is no longer strongly inaccessible, but it remains weakly inaccessible, because weak inaccessibility is defined using only the notions of cardinal and cofinality, neither of which changes when the forcing notion satisfies the countable chain condition. –  Andreas Blass Dec 11 at 23:48
    
@JesseElliott Easton's results let you make $2^{\aleph_\alpha}$ the $\alpha$-th weakly inaccessible for all $\alpha$ such that $\aleph_\alpha$ is regular, but not for singular $\aleph_\alpha$. Getting anything like this for singular cardinals will require considerable large-cardinal strength, if it isn't outright inconsistent. –  Andreas Blass Dec 11 at 23:53

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