7
$\begingroup$

This question is related to my question "Forcing with c.c.c forcing notions, Cohen reals and Random reals".

A natural way to answer Prikry's conjecture is to build a c.c.c. forcing notion which adds a minimal generic real.

I arrived to the idea of constructing such a forcing notion by considering the following:

1) Mathis forcing at $\omega$ is essentially the same as Prikry's forcing at some measurable cardinal,

2) Recently, a minimal Prikry type forcing at a measurable cardinal is constructed by Koepke-Räsch-Schlicht (see "A minimal Prikry-type forcing for singularizing a measurable cardinal, J. Symbolic Logic Volume 78, Issue 1 (2013), 85-100." and also Is Prikry forcing minimal? ).

So a natural idea is to define a version of the above forcing for $\omega$. It turns out to me that such a forcing is constructed many years ago by Judah-Shelah in "Forcing minimal degree of constructibility" and there it is shown that under $CH$ we can choose nice ultrafilters so that the resulting forcing adds a minimal generic real which is minimal.

Let me describe the forcing I have in mind: Let $(A_n: n<\omega)$ be a partition of $\omega$ into infinite sets and for each $n$ let $U_n$ be a non-principal ultrafilter over $A_n$. Let $\mathbb{P}$ consists of all pairs $(t, T)$ where $T \subseteq [\omega]^{<\omega}$ is a tree with trunk $t$ and for all $t \unlhd u\in T$ (where $\unlhd$ denotes end-extension relation) $Suc_T(u) \in U_{max(u)}.$ We define $(t, T) \leq (s, S)$ iff $T \subseteq S$ and $(t, T) \leq^* (s, S)$ iff $T \subseteq S$ and $t=s.$ The following can be proved easily:

1) $(\mathbb{P}, \leq)$ satisfies the c.c.c.,

2) $(\mathbb{P}, \leq, \leq^*)$ satisfies the Prikry property,

3) Let $G$ be $\mathbb{P}-$generic over $V$ and let $f_G=\bigcup \{t: \exists T, (t, T)\in G \}.$ Then $f_G$ is a real,

4) Suppose $d \subseteq f_G$ is infinite, then $V[d]=V[f_G]$ (the proof is the same as in the proof of Proposition 4.4 in Koepke-Räsch-Schlicht paper).

Question. Is it consistent with $ZFC$ that there is no sequence of ultrafilters as above such that the corresponding forcing defined above adds a minimal generic real?

Note that in such a model $CH$ or $MA+\sim CH$ should fail.

Remark 1. If we can show that the forcing does not add a minimal generic real when all of our ultrafilters are on the same class of near-coherence ultrafilters, then the answer to question 1 will become positive, as it is consistent that there is just one class of near-coherence non-principal ultrafilters.

Remark 2. As pointed out by Dorais, if all of the ultrafilters are RK equivalent, then the resulting forcing does not add a minimal generic real.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Groszek studied this and related forcings in Combinatorics on ideals and forcing with trees (JSL 52 (1987), 582–593; MR0902978). (Note that Groszek allows each node $\sigma$ to have a different ideal, so yours is a special case of what Groszek denotes $L(\Sigma^+)$. Also note that $L(\Sigma^+) = L(\Sigma^1)$ when all ideals are maximal.)

Theorem 7 gives a sufficient condition — the simultaneous tree property — for the generic $L(\Sigma^+)$-real to have minimal degree. I'm not sure the exact strength of the existence of a $\Sigma$ with the simultaneous tree property is but it feels like it should be possible to construct such a thing under CH. (The paper contains a construction under V = L.)

$\endgroup$
7
  • $\begingroup$ @MohammadGolshani: Since you formulated question 1 in the negative form, I should also have pointed out that "yes" that is possible. For example, if all the ultrafilters are RK equivalent. $\endgroup$ Nov 20, 2013 at 6:15
  • $\begingroup$ But what I am mainly interested is to see if it is consistent with ZFC that there is no sequence of ultrafilters as above such that the corresponding forcing adds a minimal generic real. I will edit the question to make this precise. $\endgroup$ Nov 20, 2013 at 6:31
  • $\begingroup$ @MohammadGolshani: I see. It might be possible to generalize Groszek's isomorphism result to near-coherence. Then near-coherence of ultrafilters would give you what you want. $\endgroup$ Nov 20, 2013 at 6:37
  • $\begingroup$ That sounds reasonable, thanks. Would you please add more details in your answer. $\endgroup$ Nov 20, 2013 at 6:38
  • $\begingroup$ @MohammadGolshani: Which details? $\endgroup$ Nov 20, 2013 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.