4
$\begingroup$

Recall that a cardinal $\kappa$ is $(\lambda,\infty)$-almost-strongly-compact if every $\kappa$-complete filter can be refined to a $\lambda$-complete ultrafilter. A cardinal $\mu$ has the tree property if every $\mu$-sized tree with $\mu$-small levels has a branch of length $\mu$. (If in addition $\mu$ is inaccessible then $\mu$ is weakly compact.)

Question: Can the following constellation occur?

  • $\mu$ -- weakly inaccessible with the tree property

  • $\kappa$ -- a $(\mu^+,\infty)$-strongly-compact cardinal

  • every regular $\nu \in [\mu, \kappa)$ -- has the tree property.

I suspect this would be too good to be true. But I don't know much -- for all I know, maybe almost strong compactness implies inaccessibility, in which case of course the answer is no. But I'm having trouble tracking down even that information.

If $\kappa$ can be taken to be $(\mu,\infty)$-strongly-compact, that might be good enough for what I need. Also it should suffice for only the successor cardinals in $(\mu,\kappa)$ to have the tree property.

I apologize for the repeated changes to the question.

$\endgroup$
  • 2
    $\begingroup$ What fails when $\mu$ is strongly compact? $\endgroup$ – Asaf Karagila Aug 5 '18 at 7:57
  • 1
    $\begingroup$ @AsafKaragila Thanks, I realized that I was not asking what I intended to ask. I think you're right that clearly if $\mu$ is strongly compact then it has the tree property and $\mu^+$ is almost strongly compact. It turns out I had asked the wrong question! $\endgroup$ – Tim Campion Aug 5 '18 at 13:57
  • 1
    $\begingroup$ Do you know if a double successor cardinal can be almost strongly compact at all? I think that it would mean that there is a non-principle ultrafilter with completeness which is not a measurable cardinal, which is impossible. $\endgroup$ – Yair Hayut Aug 5 '18 at 14:32
  • 1
    $\begingroup$ @YairHayut I wouldn't be surprised if I am missing something very, very basic. At the risk of being a bit ridiculous, I might change my question again. $\endgroup$ – Tim Campion Aug 5 '18 at 14:32
  • 1
    $\begingroup$ I think that this is still impossible (assuming $\mu^+ < \kappa$). Since $\kappa$ is $(\mu,\infty)$-strongly compact, there is a measurable cardinal $\mu \leq \zeta \leq \kappa$. In particular, there is a strongly inaccessible cardinal, $\rho$, in the half-open interval $[\mu, \kappa)$. In particular, $\rho^{<\rho} = \rho$ and the tree property will fail at $\rho^+$. $\endgroup$ – Yair Hayut Aug 5 '18 at 20:06

Your Answer

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

Browse other questions tagged or ask your own question.