Timeline for A Weakening of the Tree Property
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 15, 2014 at 7:26 | vote | accept | Danielle Ulrich | ||
| Jul 14, 2014 at 7:34 | answer | added | Yair Hayut | timeline score: 4 | |
| Jul 14, 2014 at 7:28 | answer | added | Péter Komjáth | timeline score: 2 | |
| Jul 14, 2014 at 7:23 | comment | added | Péter Komjáth | Douglas: here's Yair's argument. Assume that $P$ is $\sigma$-closed and adds a $b$ as described. First, if $p$ forces that $b|\alpha=\beta$, then there are extsnions $p'$ and $p''$ of $p$ which force $b|\beta=g',g''$ such that $g'$ and $g''$ differ. (otherwise $b$ is in $V$.) Notice that you can make $\beta$ arbitrarily large. Now build the decreasing sequences $p_0\geq p_1\geq\cdots$ and $q_0\geq q_\geq\cdots$ such that $p_i$ forces that $b$ restricted to [\alpha_i,\alpha_{i+1}$ is $g_i$, q_i$ forces that | |
| Jul 14, 2014 at 1:23 | comment | added | Danielle Ulrich | @YairHayut, why doesn't $\sigma$-closed forcing add threads? My understanding is that $\sigma$-closed forcing won't add new $\omega$-sequences, but may add new $\kappa$-sequences. | |
| Jul 13, 2014 at 23:42 | comment | added | Monroe Eskew | @YairHayut, can you elaborate a bit? Do you not have to consider coherent sequences $\langle b_\alpha : \alpha < \kappa \rangle$ that are new? | |
| Jul 13, 2014 at 11:56 | comment | added | Paul McKenney | @DouglasUlrich: No problem! As for when that principle holds, it turns out that the principle was studied before Velickovic by Charlie Gray, and proven by him to hold in $L$. Velickovic has a remark to this effect at the end of his paper. (Side-note: the principle is pronounced "square with built-in diamond", which I think is awesome.) | |
| Jul 13, 2014 at 8:19 | comment | added | Yair Hayut | Since $\sigma$-closed forcing can't add a thread $b$ to the sequence $\langle b_\alpha |\alpha < \kappa\rangle$ (whenever $\text{cf }\kappa > \omega$), we get that $Col(\mu,<\kappa)$ forces the weak tree property at $\mu^{+}$ whenever $\mu$ is regular and $\kappa$ is weakly compact, and if $\kappa$ is strongly compact $Col(\omega_1,<\kappa)$ forces the weak tree property at every regular cardinal. It's interesting to see if the weak tree property at $\kappa$ implies $\neg \square(\kappa)$. | |
| Jul 13, 2014 at 5:04 | comment | added | Danielle Ulrich | (A minor point.) In that paper it is just conjectured that the principle holds in L. Is there a more recent paper in which the conjecture is proven? | |
| Jul 13, 2014 at 5:02 | history | edited | Danielle Ulrich | CC BY-SA 3.0 |
edited title
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| Jul 13, 2014 at 3:57 | history | edited | Danielle Ulrich | CC BY-SA 3.0 |
added 69 characters in body; edited title
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| Jul 13, 2014 at 3:51 | comment | added | Danielle Ulrich | Wow, that was a fast response. Thanks! | |
| Jul 13, 2014 at 0:57 | comment | added | Paul McKenney | It is certainly consistent that the weak tree property fails at $\aleph_2$; see, for instance, Theorem 7 from B. Velickovic, "Jensen's square principles and the Novak number of partially ordered sets". The assumption used there is a strengthening of $\square_{\omega_1}$ and $\diamondsuit_{\omega_2}$ which holds in $L$. His theorem is stronger than what you're looking for, however; there are probably easier proofs out there in the literature. | |
| Jul 13, 2014 at 0:06 | history | asked | Danielle Ulrich | CC BY-SA 3.0 |