Accoding to wiki, weak compact cardinal is a very weak property in the large cardinal ladder. But, like ZFC+CH to ZFC, weak compact has some "useless part", so that even the first Woodin cardinal may be smaller then the first weak compact cardinal. I want to know the "pure part" of weak compact property.
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First of all, you're wrong.
The first Woodin is always larger than the first weakly compact. Namely becaues a Woodin cardinal is always much larger than a measurable cardinal---and in fact many many larger cardinals. However it is true that the first Woodin cardinal is itself not a weakly compact cardinal.
This is unlike how every measurable cardinal is a weakly compact cardinal.
If you want to talk about something which is equiconsistent with weakly compacts, then the tree property is exactly that. Namely, if every tree of height $\omega_2$, where every level is at most of size $\aleph_1$ has a branch, then $\omega_2$ is weakly compact in $L$. This can be extended to any other successor of regular (except $\omega_1$ itself).
You could look at other combinatorial properties for "small cardinals" are equiconsistent with weakly compact cardinals also.
answered Nov 23, 2016 at 11:49
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3$\begingroup$ +1. Re: "$\omega_1$ itself," Asaf is referring to Aronszajn trees, which provably exist in ZFC. It's worth observing that the converse of the statement Asaf mentions holds, too: if $\kappa$ is weakly compact, then there is a forcing extension in which $\omega_2$ has the tree property (by collapsing $\kappa$ to $\omega_2$ appropriately - see e.g. section 5.1 of this paper by Neeman for a neat proof of this fact, which is originally due to Mitchell). $\endgroup$ Nov 23, 2016 at 17:21