Suppose $\kappa$ is a weakly inaccessible cardinal with the tree property. What can we say about the height of $\kappa$? Is it a weakly-hyper-Mahlo of some sort? Does it enjoy some kind of indescribability property? Of course it is weakly compact in $L$, but I am interested in what height properties we can say it has in a universe where $\kappa$ is not strongly inaccessible.
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asked Dec 21, 2020 at 16:04
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4$\begingroup$ By the work of Boos, the least weakly Mahlo cardinal can have the tree property. $\endgroup$– Mohammad GolshaniCommented Dec 21, 2020 at 17:26
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4$\begingroup$ I think it is open if the least weakly inaccessible can have the tree property. $\endgroup$– Mohammad GolshaniCommented Dec 21, 2020 at 17:28
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$\begingroup$ What is the height of $\kappa$? $\endgroup$– Asaf Karagila ♦Commented Dec 21, 2020 at 18:18
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$\begingroup$ @MohammadGolshani Great! Can you give a reference? $\endgroup$– Monroe EskewCommented Dec 21, 2020 at 22:21
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$\begingroup$ @AsafKaragila I just mean some description of “how many” cardinals are below $\kappa$, like not only $\kappa$-many but stationary many regulars, and moreover a stationary many with that property, blah blah. $\endgroup$– Monroe EskewCommented Dec 21, 2020 at 22:22
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1 Answer
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In his paper Boolean extensions which efface the Mahlo property William Boos proves the following consistency result:
Theorem. Assume GCH holds and $\kappa$ is weakly compact. Then there exists a cardinal preserving generic extension of the universe in which $\kappa$ is the least weakly Mahlo cardinal and the tree property holds at $\kappa$.
The proof of the theorem is very similar to that of Mitchell, the main difference is that at Mahlo cardinals below $\kappa$ he adds a club of singular cardinals into that cardinal.
The following is asked in the paper and is still open:
Question. Is it consistent that the tree property hold at the least weakly inaccessible cardinal?
answered Dec 22, 2020 at 3:46
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$\begingroup$ In my last visit of Vienna, Yair and I discussed this question and Yair had some ideas, but we could not find any other time to discuss it again. He may give some more information about the question. $\endgroup$ Commented Dec 22, 2020 at 3:47