Since the continuum has to be a regular/singular uncountable cardinal, and since by Easton Theorem it can be anything we want it to be, what is the forcing that makes the continuum a weakly compact cardinal?

If we can do that then $2^{\aleph_0}$ will have the tree property. But I have to find out first what is the forcing that can make $2^{\aleph_0}$ have the partition property $\kappa \to \(\kappa)^2$ i.e the forcing has to ensure that among the subsets that are added to $2^{\aleph_0}$ we have some $H \subset 2^{\aleph_0}$ of order type $2^{\aleph_0}$ such that $F: $[$2^{\aleph_0}$]$^2 \to 2$ is constant on $[H]^2$ for every such $F$. How do we get such an $H$?

  • $\begingroup$ I added the forcing tag. $\endgroup$ Jul 16, 2010 at 1:28
  • $\begingroup$ Did you mean in ZF without choice? $\endgroup$ Oct 15, 2018 at 18:42

2 Answers 2


The continuum can't be weak compact. The continuum can't be a strong limit (basically by definition of a strong limit) and weak compact cardinals are always strong limits.

You could start with a weakly compact cardinal in the ground model and make it the continuum by your favorite way of changing the continuum, but by the above you'll destroy the weak compactness. (Under the assumption that weak compacts are consistent, of course.)

If you are just interested in getting the continuum to have the tree property, this is done in Mitchell's Aronszajn trees and the independence of the transfer property by collapsing a weak compact to $\omega_2$.

  • $\begingroup$ It is true, I did not see it can't be a strong limit! $\endgroup$ Jul 15, 2010 at 23:14
  • $\begingroup$ So does that mean that the continuum can be any of the aleph of uncountable cofinality before the first inaccessible and that's all? is that right? $\endgroup$ Jul 15, 2010 at 23:17
  • 1
    $\begingroup$ That's essentially correct -Easton's theorem says that the only constraints are $cf(2^{\aleph_0}) > \aleph_0$ and $2^{\aleph_0} \leq 2^\kappa, \forall\kappa\in Card$. en.wikipedia.org/wiki/Easton%27s_theorem has your reference. $\endgroup$ Jul 16, 2010 at 1:03

There is a (weaker) version of weak compactness which does not imply strong inacessibility: just ask that the infinitary language allowing conjunctions and disjunctions of $< \kappa$ sentences is $( \kappa, \kappa)$-compact (this means that every set of $\kappa$-many sentences is satisfiable iff every subset of $< \kappa$ sentences is satisfiable).

The continuum cannot be weakly compact even in this weaker sense, by a characterizability argument: consider a model with a definable bijection which assigns to each element of $\kappa$ a function from $\omega$ to $\omega$. If this is a bijection, the $L _{\kappa, \omega}$-theory of this model implies that you cannot add new elements to $\omega$, hence no new element to $\kappa$ (this would give a new function from $\omega$ to $\omega$, which is impossible). This contradicts $( \kappa, \kappa)$-compactness.

However, what is interesting is that you can have $2^\omega > \kappa$ in the above situation, that is, you can have "weak compactness (in the weaker sense) without strong inaccessibility". This is the main result of an old paper by W. Boos: Boos, William, Infinitary compactness without strong inaccessibility. J. Symbolic Logic 41 (1976), no. 1, 33-38.

Even if the continuum cannot be weakly compact, there are interesting connected results.


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.