I'm mainly concerned with countable support iterations of proper forcings that add reals of some large cardinal length. It is known that countable support iteration of Sacks forcing/Cohen forcing of weakly compact ($\kappa$) length forces $\kappa=\omega_2$ has the tree property. Is there a general theorem, like: for any $\langle P_i, \dot{Q}_j: i\leq \kappa, j<\kappa\rangle$ countable support iteration of proper forcings that add reals for some large cardinal $\kappa$ (weakly compact), then the tree property at $\omega_2$ holds in the forcing extension? Note $2^\omega=\kappa=\omega_2$ in the extension.

Edit: for larger cardinals as Sean Cox pointed out, the answer is positive.

  • $\begingroup$ I'd argue that for things like weakly-compact cardinals, the natural embeddings associated with them are enough to witness the tree property in the extension. The idea being that some fragment of $\Pi_1^1$ reflection remains in the extension. Which since you collapsed everything between it and $\omega_1$ would imply there would be a witness below $\omega_2$. $\endgroup$ – Not Mike Feb 15 '18 at 15:32
  • $\begingroup$ But you want the tail to not add any cofinal branches $\endgroup$ – Jing Zhang Feb 15 '18 at 15:35
  • $\begingroup$ Sacks and countably closed forcings don't add Branches to $\omega_1$-trees. $\endgroup$ – Not Mike Feb 15 '18 at 15:39
  • $\begingroup$ Sure. The case for Sacks is known, countably closed forcing a don't add reals. My question is about the general proper forcing s that add reals $\endgroup$ – Jing Zhang Feb 15 '18 at 15:40
  • $\begingroup$ It might have something to do with separating models. When you add a real, every countably closed ground model forcing is no longer countably closed in the extension (however it's still $\omega_1$-baire.) The idea here would be that the new real forced a set of new paths for countably closed partial orders, which they are happy to go down (stationarily often anyway.) $\endgroup$ – Not Mike Feb 15 '18 at 16:00

If $\kappa$ is huge, then any countable support iteration $\mathbb{P}$ of proper forcing up to $\kappa$ (where each component is of size $<\kappa$) that forces $2^\omega = \kappa = \omega_2$, must also force the tree property at $\omega_2$.

To see this, suppose $j: V \to N$ is a huge embedding with critical point $\kappa$. The assumptions on $\mathbb{P}$ ensure that if $G$ is $(V,\mathbb{P})$-generic, the quotient $j(\mathbb{P})/G$ is proper from the point of view of $N[G]$, and by closure of $N$ in $V$, this quotient is proper from the point of view of $V[G]$ as well (because $V[G]$ sees that it really is a CS iteration of proper forcings). Then $V[G]$ sees that the poset $j(\mathbb{P})/G$ is a proper forcing that introduces a generic elementary embedding with critical point $\omega_2$. By the proof of Theorem 5 in my Chang's Conjecture and semiproperness of nonreasonable posets, this implies that $V[G]$ satisfies a strong form of Chang's Conjecture that I call $\text{SCC}^{\text{cof}}_{\text{gap}}$ (this is just a minor variation of an argument of Hiroshi Sakai). By a result of Torres-Perez and Wu (``Strong Chang’s Conjecture and the tree property at $\omega_2$"), together with failure of CH this implies the tree property at $\omega_2$.

(Edit: I see now that, due to a newer theorem of Torres-Perez and Wu in "Strong Chang's Conjecture, Semi-Stationary Reflection, the Strong Tree Property and two-cardinal square principles", in $V[G]$ you actually get the Strong Tree Property at $(\omega_2,\lambda)$ for all $\lambda < j(\kappa)$, since the kind of parameterized Strong Chang's Conjecture they use in that paper holds in $V[G]$ below $j(\kappa)$. In particular, $V_{j(\kappa)}[G]$ models the full Strong Tree Property. Also, as Jing pointed out in the comments below, you don't really need hugeness for this; strong compactness of $\kappa$ is enough. NOTE: the abstract in the Torres-Perez and Wu paper fails to mention failure of CH, which is of course required for their proof in Section 3).

Even if $\mathbb{P}$ is an RCS iteration of semiproper posets (each of size $<\kappa$), you get the same result. The only difference is that the generic elementary embedding is obtained by a semiproper (rather than proper) forcing, but that's still enough (by Sakai's argument) to get what I call $\text{SCC}^{\text{cof}}$ and apply the Torres-Perez and Wu theorem.

Also note that in many cases, measurability of $\kappa$ is enough to run the argument. You just need that the individual posets used in the $j(\mathbb{P})/G$ iteration are not only proper in the generic extension of $N$, but also in the corresponding generic extension of $V$. There should be plenty of scenarios where $\kappa$-closure of $N$ in $V$ (rather than $j(\kappa)$-closure) is enough to obtain that (and similarly for the RCS iteration of semiproper).

  • $\begingroup$ With respect to properness, wouldn't it simply be enough to witness winning strategies for the properness-game? Given the size assumptions on the iteration $\mathbb{P}$ that seems to only require countable closure. What am I missing? $\endgroup$ – Not Mike Feb 15 '18 at 21:24
  • $\begingroup$ Yeah, for the strong tree property, the length you need is strongly compact (on the other hand, by Viale-Weiss it is necessary) since such iteration will always force the Semi-stationary reflection (which is equivalent to the strong Chang's Conjecture used in Torres-Perez and Wu's paper). So I think the question is only interesting in the case $\kappa$ is properly weakly compact. (there exists a $\kappa$-Aronszajn tree in the ground, the iteration can't add a path as it is $\kappa$-Knaster). $\endgroup$ – Jing Zhang Feb 15 '18 at 22:19
  • $\begingroup$ @JingZhang Could you clarify what you meant in your last sentence? I don't understand your comment about "properly weakly compact". $\endgroup$ – Sean Cox Feb 15 '18 at 22:27
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    $\begingroup$ @NotMike: I'm confused. Are you saying $V\models M^\omega\subset M$ then any proper forcing in $M$ is proper in $V$? I must have misunderstood you as that's not the case. Say $T$ is a distributive $\omega_1$ tree with no branch in $V$, the specializing forcing $S(T)$ is c.c.c, so proper in $V$. Now force with $T$, by distributivity $V^T\models V^\omega\subset V$ but then $S(T)$ is not proper in $V^T$ as forcing with it collapses $\omega_1$. $\endgroup$ – Jing Zhang Feb 16 '18 at 1:39
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    $\begingroup$ @NotMike: that's not how you negate. there is $p\in j(P)$, for all strategy $\pi$, there exists names $\dot{\alpha}_i$, such that $p\Vdash \exists n \forall m \dot{\alpha}_n\neq \beta_m$, $\beta_i's$ are played according to the strategy. The witnesses $\dot{\alpha}_n$ change for each strategy. $\endgroup$ – Jing Zhang Feb 16 '18 at 3:15

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