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In Assaf Rinot's survey article "Jenson's diamond principle and its relatives", he proves the following fact:

Fact 2.5:For every stationary set S, $\Phi_{S}$...entails that no ladder system <$L_{\alpha}$| $\alpha$$\in$S> has the uniformization property.

Can one prove the converse, that is, for every stationary set S, if no ladder system <$L_\alpha$| $\alpha$$\in$S> has the uniformization property then $\Phi_{S}$? If not, why not?

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    $\begingroup$ Can you tell us exactly what $\Phi_S$ asserts? $\endgroup$
    – Joel David Hamkins
    Commented Sep 17, 2014 at 17:09
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    $\begingroup$ I think Assaf's paper is here: arxiv.org/pdf/0911.2151v2.pdf. $\Phi_{S}$ asserts, for a stationary subset $S$ of $\lambda^{+}$, that for every function $F$ from $^{< \lambda^{+}} 2$ into $2$ there exists a function $g$ from $\lambda^{+}$ to $2$ such that for every $f$ from $\lambda^{+}$ to $2$, the set $\lbrace \alpha \in S : F(f|\alpha) = g(\alpha) \rbrace$ is stationary. $\endgroup$
    – Avshalom
    Commented Sep 17, 2014 at 22:24
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    $\begingroup$ I believe that in a model of $PFA(S)[S]$, no ladder system $\left<L_\alpha: \alpha \in \omega_1 \right>$ has the uniformization property while $2^{\aleph_0}=2^{\aleph_1}$ (and hence $\lnot \Phi_{\omega_1}$). But I might be wrong. $\endgroup$
    – Ramiro de la Vega
    Commented Sep 18, 2014 at 13:04
  • $\begingroup$ @RamirodelaVega: Have you a reference? $\endgroup$
    – Thomas Benjamin
    Commented Sep 18, 2014 at 16:41
  • $\begingroup$ What happened to the answer? I made a hard copy of the reference the answer gave (thanks, by the way), went away from the computer to read through the reference and when I came back to the computer,the answer was gone. I had some questions to ask regarding the relation between the reference and the answer, so I would like to see the answer again. $\endgroup$
    – Thomas Benjamin
    Commented Sep 19, 2014 at 21:56

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This answer is just an elaboration of Ramiro de la Vega's comment above. In "Chain conditions in maximal models" (which you can find here) Larson and Todorcevic show that after forcing with a Suslin tree, no ladder system (on $\omega_1$) in the extension has the uniformization property.

However, there are many models for which there exists a Suslin tree $S$ such that after forcing with $S$, $2^\omega = 2^{\omega_1}$ holds, and hence $\Phi_{\omega_1}$ fails. For instance, a model of $MA(S)$ (the forcing axiom for all ccc posets which preserve $S$) satisfies $2^\omega = 2^{\omega_1}$, and forcing with $S$ does not change this.

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  • $\begingroup$ Paul McKenny: Does Larson and Todorcevic's result depend in any way, shape or form on $AD^{L(\mathscr R)}$?| $\endgroup$
    – Thomas Benjamin
    Commented Sep 19, 2014 at 22:03
  • $\begingroup$ No, it's a ZFC result. $\endgroup$
    – Paul McKenney
    Commented Sep 20, 2014 at 11:58
  • $\begingroup$ 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? $\endgroup$
    – Thomas Benjamin
    Commented Sep 20, 2014 at 22:11
  • $\begingroup$ 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? $\endgroup$
    – Thomas Benjamin
    Commented Sep 20, 2014 at 22:21
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    $\begingroup$ 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 $S$. $\endgroup$
    – Paul McKenney
    Commented Feb 10, 2015 at 15:37

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