Timeline for What is the status of the assertion "There are arbitrarily large cardinals with the tree property"?
Current License: CC BY-SA 3.0
9 events
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| Jul 21, 2018 at 14:10 | comment | added | Tim Campion | I just changed my accepted answer to this one since the edit actually answers the title question. The irony is that the edit answers the title question via a paper of Mohammad Golshani.... whose answer was the one I had previously accepted! | |
| Jul 21, 2018 at 14:06 | vote | accept | Tim Campion | ||
| Feb 28, 2018 at 20:48 | history | edited | Julian Barathieu | CC BY-SA 3.0 |
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| Feb 28, 2018 at 20:39 | history | edited | Julian Barathieu | CC BY-SA 3.0 |
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| Jan 31, 2018 at 16:28 | comment | added | Paul Larson | There's a different version of the history paper on my webpage which has references in it : users.miamioh.edu/larsonpb/Cabal_Determinacy.pdf | |
| Jan 30, 2018 at 5:15 | comment | added | Mohammad Golshani | @JulianBarathieu The reference is The consistency strength of successive cardinals with the tree property | |
| Jan 25, 2018 at 13:35 | comment | added | Julian Barathieu | If every regular cardinal $\geq\kappa$ has the tree property, then there will be infinitely many pairs of successive cardinals above the continuum with the tree property, whatever $\kappa$ you chose. So it always have high consistency strength by the result above. | |
| Jan 24, 2018 at 22:34 | comment | added | Tim Campion | As I think about it more, I think what I really wanted to know about is the assertion "There exists a cardinal $\kappa$ such that for all regular $\lambda \geq \kappa$, $\lambda$ has the tree property". You're saying the consistency strength of this statement (or rather, a variant just for successors) for $\kappa = \aleph_2$ is very high (and possibly inconsistent). I suspect it doesn't make much difference to allow $\kappa$ to vary... | |
| Jan 24, 2018 at 12:43 | history | answered | Julian Barathieu | CC BY-SA 3.0 |