14
$\begingroup$

I have stumbled across a related question asking which large cardinal properties can hold for $\aleph_1$. This question is probably also related, asking in what ways $\aleph_0$ is a "large" cardinal.

To state my question:

For which large cardinal properties is it consistent with ZFC that $\frak{c}$, the cardinality of continuum, has this property? How does the answer change if we abandon choice?

Here are results I'm aware of:

  • $\frak{c}$ can be real-valued measurable (relative to existence of a measurable cardinal), which apparently implies it's weakly Mahlo. I believe all of these hold in ZFC.

  • Possibility of $\frak{c}$ being weakly inaccessible and weakly Mahlo (or, I believe, weakly 1-inaccessible or whatever intermediate condition we put on it) is also consistent relative to existence of respective cardinals, which can be established by forcing which doesn't disrupt in any way structure of ordinals, only the size of continuum.

  • Clearly $\frak{c}$ can't be strongly inaccessible or strongly Mahlo, because these directly require a cardinal to be strongly limit.

  • In ZFC, every measurable is strongly limit, so $\frak{c}$ can't be measurable. But it is also true in ZF, because we have $\aleph_0<\frak{c}$ and $2^{\aleph_0}\geq\frak{c}$, and standard proof that this can't happen for measurable goes through.

  • By same means, $\frak{c}$ can't be the critical point of any elementary embedding, so it can't land in any of the higher entries of large cardinals list.

  • Continuum can't be weakly compact since weak compactness implies strong limitness. I'm sure this is true in ZFC, but not sure about ZF.

What other properties can or can't continuum have? I believe axiom of choice disallows it to be in anywhere above weakly compact (correct me if I'm wrong), but I have hopes for ZF itself giving $\frak{c}$ more possibilites to be large.

Thank you in advance.

$\endgroup$
12
  • $\begingroup$ I find this question to be a bit too broad, especially in the choiceless context where there is too much to say, and not nearly enough known, as to "how do you formulate largeness" of certain types (just "inaccessible" has a handful of different inequivalent formulations in ZF). $\endgroup$
    – Asaf Karagila
    Commented Jan 1, 2016 at 23:08
  • 1
    $\begingroup$ I think this an excellent question, because it would shed light, for the rest of us, on how under-determined c is from standard axioms, even large cardinals. And c being a large cardinal is kinda mind-blowing, for me $\endgroup$
    – David Roberts
    Commented Jan 2, 2016 at 0:18
  • 3
    $\begingroup$ @Asaf: But I'm pretty sure Wojowu didn't know how well-researched the question was before asking it. $\endgroup$
    – Deedlit
    Commented Jan 2, 2016 at 1:26
  • 1
    $\begingroup$ If you start with $\kappa$ large, say for example supercompact and add $\kappa$ many Cohen reals, $\kappa$ becomes generically supercompact in the extension. $\endgroup$ Commented Jan 2, 2016 at 5:25
  • 3
    $\begingroup$ Also it is possible for the continuum to be weakly inaccessible and satisfies the tree property. $\endgroup$ Commented Jan 2, 2016 at 5:26

1 Answer 1

7
$\begingroup$

Let me add a few examples:

(1) If we start with a supercompact cardinal $\kappa$, and force with $Add(\omega, \kappa)$, then in the extension the cardinal $\kappa=2^\omega$ becomes generically supercompact. The same holds for many other large cardinals.

(2) If we start with a weakly compact cardinal, we can find a generic extension in which $2^\omega=\kappa$ is (the least) weakly Mahlo, and tree property holds at $\kappa.$ This result is due to Boos ``Boolean extensions which efface the Mahlo property''.

(3) The consistency of the theory $ZFC$+ "there is a supercompact cardinal'' implies the consistency of the theory $ZFC$ + "there exist a uniform measure $μ$ on the cardinal $2^ω$ and a set $X⊆2^ω$ of positive $μ$-measure such that for every $y∈X$ there is a uniform measure on $y$ which is $|y|$-additive.''

(4) The consistency of the theory $ZFC$+ "there is a measure concentrating on compact cardinals'' implies the consistency of the theory $ZFC$ + "there exist a uniform measure $μ$ on the cardinal $2^ω$ and a set $X⊆2^ω$ of positive $μ$-measure such that for every $y∈X$ there is a uniform measure on $2^ω$ which is $|y|$-additive.''

For (3) and (4) see Some combinatorial properties of measures

(5) Under $PFA, 2^\omega=\aleph_2$ has some large cardinal properties.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .