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.


1 Answer 1


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.)

  • $\begingroup$ @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. $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented Nov 20, 2013 at 6:37
  • $\begingroup$ That sounds reasonable, thanks. Would you please add more details in your answer. $\endgroup$ Commented Nov 20, 2013 at 6:38
  • $\begingroup$ @MohammadGolshani: Which details? $\endgroup$ Commented Nov 20, 2013 at 6:40

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