Timeline for A Question Regarding Weak Diamond
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 10, 2015 at 15:37 | comment | added | Paul McKenney | Thomas: the full $MA$ kills all Souslin trees. The point of $MA(S)$ is that it is the forcing axiom for exactly those ccc forcings which don't kill $S$. | |
| Feb 10, 2015 at 14:42 | comment | added | Thomas Benjamin | So I guess the question is (for me, at least), then, under what conditions does $MA$ kill or preserve Souslin trees? | |
| Feb 10, 2015 at 14:16 | comment | added | Paul McKenney | Thomas: Of course! (Assuming the consistency of ZFC, of course.) $MA_{\omega_1}(S)[S]$ implies the negation of CH, so all you need is $MA_{\omega_1}(S)[S]$, which you can get by forcing with a suborder of the standard $MA$ forcing. | |
| Feb 10, 2015 at 14:10 | comment | added | Thomas Benjamin | McKenny: So $ZFC$+$MA_{\omega_1}$($S$)[$S$]+$\lnot$$CH$ is then consistent if $ZFC$ is? | |
| Feb 7, 2015 at 15:02 | comment | added | Paul McKenney | @ThomasBenjamin: What was the problem here? "All Whitehead groups are free" is, according to Assaf Rinot's article, equivalent to "no ladder system on a stationary set has the uniformization property." But this is implied by $MA_{\omega_1}(S)[S]$ (according to the Larson-Tall article). | |
| Sep 20, 2014 at 22:46 | comment | added | Thomas Benjamin | (Perhaps the property of normal first countable spaces being "collectionwise Hausdorf" in a model of $MA_{\omega_1}$(S)[S] might have something to do with it, at least as far as $\Phi_{\omega_1}$ failing yet all Whitehead groups being free goes....) | |
| Sep 20, 2014 at 22:21 | comment | added | Thomas Benjamin | Could one, for example, have a model of ZFC+$MA_{\omega_1}(S)[S]$+$\lnot$CH in which all Whitehead groups are free? If not, why not? | |
| Sep 20, 2014 at 22:11 | comment | added | Thomas Benjamin | Mc'Kenney: And yet, in their paper "Locally Compact Perfectly Normal Spaces May All Be Paracompact", Larson and Tall show that (in their Theorem 3) $MA_{\omega_1}$(S)[S] implies that all Whitehead groups are free, even though $MA_{\omega_1}$(S)[S] also implies (as you point out) that $2^{\omega}$=$2^{\omega_1}$. Thus you have a situation in which $\Phi_{\omega_1}$ fails and yet all Whitehead groups are free. Why is that? | |
| Sep 20, 2014 at 11:58 | comment | added | Paul McKenney | No, it's a ZFC result. | |
| Sep 19, 2014 at 22:03 | comment | added | Thomas Benjamin | Paul McKenny: Does Larson and Todorcevic's result depend in any way, shape or form on $AD^{L(\mathscr R)}$?| | |
| Sep 19, 2014 at 19:34 | history | answered | Paul McKenney | CC BY-SA 3.0 |