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On one hand due to Kunen's inconsistency theorem it is known that within $\sf ZF$, large cardinal axioms beyond Reinhardt cardinal are inconsistent with $\sf AC$.

Also some recent results of Bagaria, Koellner and Woodin (see here) suggest that very large cardinals beyond Reinhardt (e.g. Berkeley cardinals), could be inconsistent with weaker choice principles like $\sf DC$ under some "plausible assumptions" (borrowed from Koellner's words in his lecture slides).

Now consider the Konig's Infinity Lemma which implies tree property at $\aleph_0$, the statement that every $\aleph_0$ - tree has a cofinal branch. It is not hard to see that there is a similarity between such cofinal branches in $\kappa$ - trees and the $R$-chains that $\sf DC$ produces for a binary relation $R$ on a set $X$. In the other words tree property could be considered as a kind of Axiom of Dependent Choice. (For more information see here).

Now by replacing $\sf DC$ with tree property at Bagaria, Koellner and Woodin's observation, it is natural to ask:

Question. Within $\sf ZF$, is there any inconsistency between very large cardinal axioms beyond Reinhardt cardinals and tree property at one or more regular cardinals?

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    $\begingroup$ Would the downvoter like to explain? I'm not in this field but the question looks reasonable to me. $\endgroup$
    – Nate Eldredge
    Commented Jan 2, 2016 at 2:43
  • $\begingroup$ Usually large cardinals give tree property, say like Magidor-Shelah theorem. $\endgroup$
    – Mohammad Golshani
    Commented Jan 2, 2016 at 5:27
  • $\begingroup$ @MohammadGolshani I'm aware of such consistency results which all happen beneath the Reinhardt level and in fact beneath proper class many supercompact but I think this case is a bit different, particularly because the phenomena which happens for $\sf DC$ when we make our large cardinal axiom extraordinarily large. $\endgroup$
    – Morteza Azad
    Commented Jan 2, 2016 at 5:30

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Recall that a super Reinhardt cardinal $κ$, is a cardinal which is the critical point of elementary embeddings $j:V→V$, with $j(κ)$ as large as desired.

Claim. If $\kappa$ is super Reinhardt, then tree property holds for a class of cardinals.

First note that it is easily seen that if $\kappa$ is as above, then $\{\alpha < \kappa: \alpha$ is supercompact$ \}$ is unbounded in $\kappa.$ It follows that $\{ \alpha < \kappa: TP(\alpha) \}$ is unbounded in $\kappa,$ where $TP(\alpha)$ is the assertion ``Tree property holds at $\alpha$''. The point is the Magidor-Shelah theorem that tree property holds at successor of singular limits of supercompact cardinals (one has to check their proof works without the use of AC and using the definition of supercompact cardinals given in the non-AC case).

It is now evident that $\{ \alpha: TP(\alpha) \}$ should be a proper class: if not, let $\lambda$ be a bound, and let $j: V \to V$ be such that $crit(j)=\kappa$ and $j(\kappa) > \lambda^+.$ As $\{ \alpha < \kappa: TP(\alpha) \}$ is unbounded in $\kappa,$ by elementarity $\{ \alpha < j(\kappa): TP(\alpha) \}$ is unbounded in $j(\kappa),$ which is in contradiction with the choice of $\lambda.$

Remark. The same idea seems to work for extendible cardinals or even a storng limit of supercompact cardinals $\kappa.$

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