It is claimed that if there are two weakly compact cardinals, then there is a generic extension in which $\aleph_2$ and $\aleph_4$ have the tree property. Assuming one knows Mitchell's forcing, what is the argument? If you just iterate Mitchell's forcing twice, then why does the second stage preserve the tree property at $\aleph_2$? The second stage is $\aleph_2$ directed closed, but having the tree property be indestructubile under such forcings seems to require larger cardinals.
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asked Sep 10, 2015 at 3:57
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$\begingroup$ See Theorem 5.35 of "THE TREE PROPERTY AT MORE CARDINALS" (search the name in google). It is proved using two measurable cardinals, but can be modified to obtain the same result using two weakly compact cardinals. $\endgroup$– Mohammad GolshaniCommented Sep 10, 2015 at 4:53
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$\begingroup$ You can see the paper here $\endgroup$– Mohammad GolshaniCommented Sep 10, 2015 at 4:55
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Starting from $\omega$-many weakly compact cardinals, you can for example force the tree property at all $\aleph_{2n}$'s, $0<n<\omega.$ See theorem 5.1 of "The tree property at the $\aleph_{2n}$'s and the failure of SCH at $\aleph_\omega$''
answered Sep 10, 2015 at 6:32