Is Vopenka's Principle + "ORD has the tree property" consistent? Vopenka's principle implies the existence of weakly compact cardinals (a proper class of them, I believe). My question is whether Vopenka's principle is consistent with the assertion that the universe itself is weakly compact. Alternatively, can a Vopenka cardinal be weakly compact?
There are several versions of this question, depending on how classes are treated in the formulation of the two statements. I'm hopeful that the answer is not too different for the different formulations.
I suspect that VP does not imply that ORD is weakly compact. After all, VP implies that ORD is Woodin, and I've read that the first Woodin cardinal is not weakly compact. So if consistent, the conjunction VP + ORD is weakly compact might have higher consistency strength than either of its conjuncts.
 A: Yes, a Vopenka cardinal can be weakly compact, at least assuming the consistency of a huge cardinal (though this is certainly a bit of an overkill).  A huge cardinal is a weakly compact (in fact, measurable) Vopenka cardinal.
EDIT:  Actually almost huge cardinals suffice to get measurable Vopenka cardinals.   Theorem 24.18 of Kanamori's "Higher Infinite" says that if $\kappa$ is almost huge, then there is a normal ultrafilter $U$ on $\kappa$ such that there are $U$-many $\alpha < \kappa$ such that $\alpha$ is a Vopenka cardinal.  The argument actually shows that there are $U$-many $\alpha$ such that $\alpha$ is Vopenka AND measurable.  To see why, note that part (b) of that theorem follows from the fact that if $j: V \to M$ witnesses the almost-hugeness of $\kappa$, then $M \models$ "$\kappa$ is Vopenka".  But $M$ also models "$\kappa$ is measurable" (in fact $U$ itself is in $M$), because

*

*$\kappa$ is measurable in $V$

*$j(\kappa)$ is inaccessible in $V$; so in particular all $\kappa$-complete ultrafilters on $\kappa$ are in $V_{j(\kappa)}$; and

*$V_{j(\kappa)} \subset M$ because $M$ is closed under $<j(\kappa)$ sequences.

EDIT 2 (regarding Tim's comment):  actually $\kappa$ itself is a Vopenka cardinal, because ``$\kappa$ is a Vopenka cardinal" is absolute between any transitive models that have the same powerset of $\kappa$ (and $V$ and $M$ in the above argument have the same powerset of $\kappa$, and $M$ believes that $\kappa$ is a Vopenka cardinal).
