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Of course, if you want your cardinals with the tree property to be strongly inaccessible, then you're asking about weakly compact cardinals. But what if you don't want them to be strongly inaccessible?

I see it's consistent that no successor cardinal has the tree property. I suspect this means it's consistent that no regular cardinal has the tree property?

As a non-set-theorist, I'm not sure whether this question is trivial, or hopeless!

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    $\begingroup$ In $L$, no successor cardinal has the tree property, so if you assume there are no weakly compact cardinals, then in $L$, no uncountable regular cardinal has the tree property. $\endgroup$
    – Mohammad Golshani
    Jan 23, 2018 at 6:08
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    $\begingroup$ The difficult question is to get tree property at successor regular cardinals $> \aleph_1.$ The famous question of Magidor asks if it is consistent that all such cardinals have the tree property, and it is widely open, despite many partial results. $\endgroup$
    – Mohammad Golshani
    Jan 23, 2018 at 6:12
  • $\begingroup$ Does that mean it's easy to get the tree property at regular limit cardinals which are not strong limit cardinals? $\endgroup$
    – Tim Campion ♦
    Jan 23, 2018 at 7:19

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Paul B. Larson mentions a partial result in A Brief History of Determinacy, page 48. Sadly he doesn't give a precise source.

Foreman, Magidor and Schindler showed that if there exist infinitely many cardinals $δ$ above the continuum such that the tree property holds at $δ$ and at $δ^+$, then [the axiom of projective determinacy] holds. The hypothesis of this statement had been shown consistent relative to the existence of infinitely many supercompact cardinals by James Cummings and Foreman. It is not known whether the conclusion can be strengthened to $\text{AD}^{L(\mathbb{R})}$.

So the tree property for all successors >$\omega_1$ has to be a very strong statement, at least as strong as the existence of infinitely many Woodin cardinals. This is the best partial/related result I know of.

EDIT: See this paper, which has the following abstract:

Starting from a strong cardinal and a measurable cardinal above it, we construct a model of ZFC, in which, for every singular cardinal $δ$, $δ$ is strong limit, $2^δ = δ^{+++}$, and the tree property holds at $δ^{++}$. It answers a question of Friedman, Honzik and Stejskalova. We also produce, relative to the existence of a strong cardinal and two measurable cardinals above it, a model of ZFC in which the tree property holds at all regular even cardinals. The result answers questions of Friedman, Halilovic and Honzik .

Which answers the question from the title, i.e. it is possible to have arbitrarily large cardinals with the tree property (relatively to very large cardinals).

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    $\begingroup$ As I think about it more, I think what I really wanted to know about is the assertion "There exists a cardinal $\kappa$ such that for all regular $\lambda \geq \kappa$, $\lambda$ has the tree property". You're saying the consistency strength of this statement (or rather, a variant just for successors) for $\kappa = \aleph_2$ is very high (and possibly inconsistent). I suspect it doesn't make much difference to allow $\kappa$ to vary... $\endgroup$
    – Tim Campion ♦
    Jan 24, 2018 at 22:34
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    $\begingroup$ If every regular cardinal $\geq\kappa$ has the tree property, then there will be infinitely many pairs of successive cardinals above the continuum with the tree property, whatever $\kappa$ you chose. So it always have high consistency strength by the result above. $\endgroup$
    – Julian Barathieu
    Jan 25, 2018 at 13:35
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    $\begingroup$ @JulianBarathieu The reference is The consistency strength of successive cardinals with the tree property $\endgroup$
    – Mohammad Golshani
    Jan 30, 2018 at 5:15
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    $\begingroup$ There's a different version of the history paper on my webpage which has references in it : users.miamioh.edu/larsonpb/Cabal_Determinacy.pdf $\endgroup$
    – Paul Larson
    Jan 31, 2018 at 16:28
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    $\begingroup$ I just changed my accepted answer to this one since the edit actually answers the title question. The irony is that the edit answers the title question via a paper of Mohammad Golshani.... whose answer was the one I had previously accepted! $\endgroup$
    – Tim Campion ♦
    Jul 21, 2018 at 14:10
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Let me address your question in the comments.

If one starts with a weakly compact cardinal $\kappa,$ then there exists a forcing extension in which all cardinals and cofinalities are preserved (in particular, $\kappa$ remains a regular limit cardinal) and in which the tree property holds at $\kappa$. In fact $\kappa$ can become the least weakly Mahlo cardinal.

This result is due to Boos, see Boolean extensions which efface the Mahlo property.

As far as I know the question of whether the least weakly inaccessible cardinal can have the tree property is open (it is asked in Boos paper).

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  • $\begingroup$ So, I understand now that it's consistent that no uncountable cardinal has the tree property : this holds in V=L with no weakly compact cardinals, because every regular limit cardinal in V=L is strongly inaccessible. I think this theory is equiconsistent with ZFC, right? On the other hand, it follows from the existence of arbitrarily large weakly compact cardinals that there are arbitrarily large cardinals with the tree property. To complete the circle -- is the existence of arbitrarily large cardinals with the tree property equiconsistent with arbitrarily large weakly compacts? $\endgroup$
    – Tim Campion ♦
    Jan 23, 2018 at 22:04
  • $\begingroup$ I think they are equiconsistent as follows from the work of Silver (I assume by large cardinals you mean regular limit cardinals). $\endgroup$
    – Mohammad Golshani
    Jan 24, 2018 at 9:24
  • $\begingroup$ Thanks. Sorry for the confusing terminology -- by "there are arbitrarily large cardinals such that ____", I just mean "there is a proper class of cardinals such that ______". $\endgroup$
    – Tim Campion ♦
    Jan 24, 2018 at 20:04

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