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In a comment to Miha's question in Forcing PFA with ccc forcing, I suggested that if such situation is even possible, it might be achieved by screwing with PFA by some ccc forcing (e.g. adding a Cohen real), and then "fixing all the problems" via a ccc forcing.

And that raises the question:

Apart of adding a Cohen real, what sort of ccc forcing adds a Suslin tree on $\omega_1$?

Of course, take any ccc forcing $\Bbb P$, and consider $\Bbb P\times\operatorname{Add}(\omega,1)$ is an answer. So to avoid this triviality, we only consider forcings which do not add a Cohen real themselves. So we can rephrase the question:

Suppose that $\Bbb P$ is a ccc forcing which adds a Suslin tree on $\omega_1$. Does $\Bbb P$ add a Cohen real?

If the answer is negative (namely it is possible to add a new Suslin without adding a Cohen), does the answer change if we assume the ground model has no Suslin trees?

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  • $\begingroup$ Along with Miha's question, it seems that today is "ccc day"! :-P $\endgroup$ – Asaf Karagila Jul 11 '16 at 16:49
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    $\begingroup$ By a result of Laver, if MA holds, then adding a Random real does not add a Souslin tree. But maybe one can show the consistency of "there are no Souslin trees and adding a random real adds a Souslin tree"? If so, we can get a consistent negative answer to your question! $\endgroup$ – Mohammad Golshani Jul 13 '16 at 7:35
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    $\begingroup$ @MohammadGolshani I don't know whether Tennenbaum's poset for a Suslin tree always adds a Cohen real, but it can consistently do so. Specifically, by a theorem of Pawlikowski, any poset of size $<\mathbf{add}(\mathcal{M})$ that adds a real adds a Cohen real. So over a model of MA where CH fails, for example, Tennenbaum's poset will add a Cohen real. $\endgroup$ – Miha Habič Jul 13 '16 at 15:13
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    $\begingroup$ How about the following idea: Let $T$ be a Suslin tree such that forcing with its nodes as conditions will not only kill the Susliness of $T$, but will introduce a fresh Suslin tree $S$ to the universe. (I think such $T$ should exist if we twist the usual '$\diamondsuit$ yields Suslin tree' construction). This would give us a negative solution to the question as the forcing does not add any reals at all. $\endgroup$ – Stefan Hoffelner Jul 14 '16 at 12:32
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    $\begingroup$ @MihaHabič Tennenbaum's poset for adding a Suslin tree is same as adding $\aleph_1$ Cohen reals. $\endgroup$ – Ashutosh Jul 15 '16 at 1:15
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It is consistent that the answer is no.

If we start with $L$ as our ground model then whenever $T$ is a Suslin tree, the forcing $\mathbb{P}_T$ which shoots a branch through $T$ will always introduce new Suslin trees. Thus in $L$, $\mathbb{P}_T$ is an example of a forcing which has the ccc, adds Suslin trees and does not add a Cohen real.

To see that $\mathbb{P}_T$ adds a new Suslin tree, note first that whenever $G$ is $L$-generic for $\mathbb{P}_T$, $\diamondsuit$ still holds in the extension $L[G]$: if $(a_{\alpha} \, : \, \alpha < \omega_1)$ is $\diamondsuit$-sequence in $L$ then $(a_{\alpha}^G \, : \, \alpha < \omega_1)$ will be a $\diamondsuit$-sequence in $L[G]$ as can be seen using the ccc of $\mathbb{P}_T$ and the fact that $\mathbb{P}_T$ has size $\aleph_1$.

Next observe that $\mathbb{P}_T$ will add a new subset of $\omega_1$ and hence a new stationary subset of $\omega_1$ by a theorem of Solovay, saying that each stationary subset of $\omega_1$ can be partitioned into $\omega_1$-many stationary sets.

Finally fix a new stationary subset $B \subset (\omega_1 \cap Lim)$ and, working in $L[G]$, use the restricted diamond sequence $\diamondsuit_B := (a_{\alpha}^G \, : \, \alpha \in B)$ to define a new Suslin tree $S$: Build the tree via induction on its height. If $\alpha$ is a limit ordinal then let $S_{\alpha}:= \bigcup_{\beta< \alpha} S_{\beta}$. If $\alpha$ is a double successor ordinal let $S_{\alpha+1}$ be the extension of $S_{\alpha}$ which adjoins to every top level $x \in S_{\alpha}$ infinitely many immediate successors. If $\alpha$ is limit we distinguish two cases in the definition of $S_{\alpha+1}$.

If $\alpha \notin B$ let $S_{\alpha+1}$ be the $<_L$-least normal extension of $S_{\alpha}$ (note here that we can use the $L$-wellorder as $\mathbb{P}_T$ is $\omega$-distributive, hence no new countable sets are added by $\mathbb{P}_T$).

If $\alpha \in B$ and $a_{\alpha}^G$ is a maximal antichain in $S_{\alpha}$ let $S_{\alpha+1}$ be the $<_L$-least normal extension of $S_{\alpha}$ such that $a_{\alpha}^G$ remains maximal in $S_{\alpha+1}$. And finally if $\alpha \in B$ but $a_{\alpha}^G$ is not a maximal antichain in $S_{\alpha}$, let $S_{\alpha+1}$ be a normal extension of $S_{\alpha}$ which is not the $<_L$-least.

The standard argument shows that $S:= \bigcup_{\alpha < \omega_1} S_{\alpha}$ is a Suslin tree. However the way we defined $S$ ensures that $S$ can not be in $L$.Indeed if we assume that $S \in L$ then one could define $B \subset \omega_1$ via just looking at the stages of $S$ where $S_{\alpha+1}$ is not the $<_L$-least normal extension of $S_{\alpha}$. Thus if $S \in L$, $B$ would be in $L$ as well which is a contradiction.

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  • $\begingroup$ Great! This was anticipated, however, which is why I already included the follow-up question, what if we assume that there are no Suslin trees in the universe? $\endgroup$ – Asaf Karagila Oct 13 '16 at 10:57
  • $\begingroup$ Have not thought about it at all. But Lyubomyr Zdomskyy just mentioned to me that in models of $\mathsf{PFA}(S)$, where $S$ is a Suslin tree, after forcing with $S$ the $PID$ holds, hence there are no Suslin trees anymore, so the opposite is consistent as well. $\endgroup$ – Stefan Hoffelner Oct 13 '16 at 11:10
  • $\begingroup$ That's a good step forward. I wonder, now, if the answer is straightout provable modulo "There exists a Suslin tree". My regards to Lyubomyr! $\endgroup$ – Asaf Karagila Oct 13 '16 at 11:16

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