2
$\begingroup$

Was looking at the Cantor's attic page for weakly compact cardinals and thought it was kind of odd that they only stipulate $\kappa^{<\kappa}=\kappa$ when many of the equivalent conditions obviously imply inaccessibility, some even explicitly including it (I'm sure some of the ones I'm unfamiliar with don't imply it, and this particular condition probably has some significance I'm unaware of, but anyway).

That got me thinking about how much weaker the condition $\kappa^{<\kappa}=\kappa$ is than inaccessibility. First off, it holds for successor cardinals if and only if the predecessor satisfies GCH, but let's add the additional condition that $\kappa$ is a limit. It's fairly easy to see that it cannot hold for $\kappa$ singular, so we're left with the case where $\kappa$ is weakly inaccessible but not strongly.

The simplest example I can come up with is that it is consistent relative to some very mild assumptions (I think the consistency of one inaccessible suffices) that $\mathfrak c$ is weakly inaccessible and that $2^\alpha=\mathfrak c$ for all $\alpha < \mathfrak c.$ In this case $\kappa^{<\kappa}=\kappa$ holds for $\kappa=\mathfrak c$ and clearly it is not strongly inaccessible.

(I know this to be true from various results I've seen, but I think this even holds in the situation when we force $\kappa$ Cohen reals into a model of GCH where $\kappa$ is inaccessible... but please correct me if I'm wrong.)

  1. Is there a simpler example (either conceptually or in terms of proof difficulty), or a particularly interesting example of a different character?
  2. Is this an optimal example in terms of consistency strength? (It would seem to be provided my remark above about only needing one inaccessible is correct.)
$\endgroup$
1
$\begingroup$

So if I understand your question correctly, you are looking for examples in which $\kappa$ is weakly inaccessible, but not strongly, yet satisfies $\kappa^{<\kappa}=\kappa$. Conceptually, your example is the easiest. First note that to find such an example you need to start with a situiation which has consistency strength at least a (strongly) inaccessible, as ZFC+ a weakly inaccessible has the same strength as ZFC + a (strongly) inaccessible (this is as any weakly inaccessible is strongly inaccessible in Gödels constructible universe).

Now in this situation, there must be some cardinal $\lambda<\kappa$ so that $2^\lambda=\kappa$. Moreover, after $\lambda$ up to $\kappa$, the continuum function must constantly evaluate to $\kappa$. That is, there is some minimal $\lambda_0<\kappa$ so that for all $\lambda_0\leq\lambda<\kappa$, $2^\lambda=\kappa$. So the only thing that really is variable is the value of $\lambda_0$. The minimal possible value is, of course, $\omega$, which is exactly your example. Starting with an inaccessible $\kappa$ + GCH, for any regular $\delta<\kappa$ we can find a forcing extension in which $\lambda_0$ is exatly $\delta$. I am quite sure that it is even consistent that $\lambda_0$ is singular, though this would require much higher consistency strength than just an inaccessible (but dont quote me on that...) [edit: upon further reflection, you may quote me on that. It is however quite interesting how such a singular $\lambda_0$ can look. A prominent result of Shelah gives $\lambda_0\neq\aleph_\omega$. In fact his PCF theory proves that a singular $\lambda_0$ must be a fixpoint of the $\aleph$-function, if I am not mistaken.]. So this might be an interesting example.

Here is another example: In the usual construction of a model in which $2^\omega$ is real-valued measurable (or equivalently, in which there is a countably additive measure on the reals, that measures every subset of $\mathbb R$ and which extends the Lebesgue measure) one starts with a measurable cardinal $\kappa$ (so in particular, $\kappa$ is (strongly) inaccessible) and adds with a specific forcing $\kappa$ new reals. In the extension, $\kappa=2^\omega$ satisfies $\kappa^{<\kappa}=\kappa$, is weakly inaccessible and real-valued measurable. So this is a construction where such a situation naturally pops up.

$\endgroup$
  • $\begingroup$ Starting with a c.t.m. $M$ of ZFCI which satisfies "$k$ is strongly inaccessible", we can get a forcing extension M' that satisfies "$2^{\omega}=k$ and $k$ is weakly inaccessible" and force over M' to get an extension M'' that still satisfies "$2^{\omega}=k$ and $k$ is weakly inaccessible" but also satisfies MA (Martin's Axiom). And MA implies that if $2^{\omega}=k$ then $k^{<k}=k.$ $\endgroup$ – DanielWainfleet Jun 25 at 7:15
  • 1
    $\begingroup$ Did you mean to say, in the last paragraph, "... a $countably$ additive measure on the reals..." ? $\endgroup$ – DanielWainfleet Jun 25 at 7:21
  • $\begingroup$ Yes, you are right, it should have been countably additive. Regarding your first comment, if you force with the finite support product of $\kappa$-many Cohen forcings, then you will automatically get $\kappa^{<\kappa}=\kappa$, no need to additionally force MA. $\endgroup$ – Andreas Lietz Jun 25 at 8:43
  • $\begingroup$ Good point about MA.... BTW (but you likely know) that if we assume there is a least real-measurable cardinal $k$ then by elementary means there is a countably additive measure on $P(\Bbb R)$ that extends Lebesgue measure. $\endgroup$ – DanielWainfleet Jun 28 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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