Assume $\kappa > \aleph_1$ is regular and let $P=(P_\alpha, \dot{Q}_\beta: \alpha \leq \kappa^+, \beta< \kappa^+)$ be an iteration such that:

1) For even $\beta, \dot{Q}_\beta$ is forced to be $\aleph_1$-closed, $\kappa$-c.c. and a subset of $V_\kappa$.

2) For odd $\beta, \dot{Q}_\beta$ is forced to be $\kappa$-closed,

3) For $p \in P,$ if $Supp(p)$ denotes the support of $p$, then $|Supp(p) \cap E| < \aleph_1$ and $|Supp(p) \cap O|< \kappa$ (where $E$ and $O$ are the class of all even and odd ordinals respectively).

From $P$ we can naturally define the forcing notions $P^E$ and $P^O$, where $$P^E=\{p \in P: \forall \beta \in O, p\restriction \beta \Vdash p(\beta)=1 \}$$ and $$P^O=\{p \in P: \forall \beta \in E, p\restriction \beta \Vdash p(\beta)=1 \}$$. Note that there is a natural map $$\pi: P^E \times P^O \to P$$ which is defined as follows: $\pi(p, q)=r$, where for even $\beta, r(\beta)$ is forced to be $p(\beta)$ and for odd $\beta$ it is forced to be $q(\beta)$.

Clearly $P^O$ is $\kappa$-closed.

Question. Is $\pi$ a projection of forcing notions, assuming $P^E$ is $\kappa$-c.c.?

Remark. I think the extra assumption ``$P^E$ is $\kappa$-c.c.'' is necessary for the question, as otherwise, maybe the arguments from "The $\aleph_2$-Souslin Hypothesis" may be used to show that for some suitable choice of the forcings, $P$ may collapse $\kappa,$ while this can not happen for the product $P^E \times P^O$ by Easton's lemma.

  • $\begingroup$ Now I'm intrigued as to what the construction is supposed to be doing. Looks like some tree property related thing... $\endgroup$
    – Asaf Karagila
    Commented Mar 7, 2017 at 8:15

1 Answer 1


You can drop the hypothesis of $P^E$ being $\kappa$-c.c.

So take $(p_1,p_2)\in P^E\times P^O$, and set $p=\pi(p_1,p_2)$. Let $q\in P$ be such that $q\leq p$.

We shall build $q_1\in P^E$ and $q_2\in P^O$ such that $(q_1,q_2)\leq(p_1,p_2)$ and $\pi(q_1,q_2)\leq q$. We define $q_1\upharpoonright\beta$ and $q_2\upharpoonright\beta$ by induction on $1\leq\beta\leq\kappa^+$.

For $\beta=1$ set $q_1\upharpoonright\beta=q\upharpoonright\beta$ and $q_2\upharpoonright\beta=\{(0,1)\}$.

Suppose for some $\beta\leq\kappa^+$ both $q_1\upharpoonright\alpha$ and $q_2\upharpoonright\alpha$ have been defined for all $\alpha<\beta$, with $q\upharpoonright\alpha\leq q_1\upharpoonright\alpha, q_2\upharpoonright\alpha$ and $q_i\upharpoonright\alpha\leq p_i\upharpoonright\alpha$ for $i=1,2$.

If $\beta$ is odd, say $\beta=\gamma+1$, let $q_1(\gamma)$ be a $ P\upharpoonright\gamma$-name such that $q\upharpoonright\gamma\Vdash q_1(\gamma)=q(\gamma)$ and for all $q'\leq q_1\upharpoonright\gamma$ in $ P\upharpoonright\gamma$ with $q'\perp q\upharpoonright\gamma$ we have $q'\Vdash q_1(\gamma)=p(\gamma)$. Finally, let $q_2(\gamma)$ be a $ P\upharpoonright\gamma$-name such that $q_2\upharpoonright\gamma\Vdash q_2(\gamma)=1$. It is easy to verify that $q\upharpoonright\beta\leq q_1\upharpoonright\beta,q_2\upharpoonright\beta$ and $q_i\upharpoonright\beta\leq p_i\upharpoonright\beta$.

Similarly, if $\beta$ is even and successor, we define $q_2(\gamma)$ in such a way that $q\upharpoonright\gamma\Vdash q_2(\gamma)=q(\gamma)$ and $q_2\upharpoonright\gamma-q\upharpoonright\gamma\Vdash q_2(\gamma)=p(\gamma)$, and $q_1(\gamma)$ with $q_1\upharpoonright\gamma\Vdash q_1(\gamma)=1$, again we have $q\upharpoonright\beta\leq q_1\upharpoonright\beta,q_2\upharpoonright\beta$ and $q_i\upharpoonright\beta\leq p_i\upharpoonright\beta$.

For $\beta$ limit, $q_i\upharpoonright\beta$ is just the union of $q_i\upharpoonright\alpha$, $\alpha<\beta$.

We have $q_i\in P$ as $supp(q_i)\subseteq supp(q) \cup supp(p)$.

It is clear that $(q_1,q_2)\leq (p_1,p_2)$.

It is easy to show by induction on $1\leq\beta\leq\kappa^+$ that $\pi(q_1,q_2)\upharpoonright\beta\leq q\upharpoonright\beta$.

Therefore $\pi$ is a projection.

  • $\begingroup$ @MohammadGolshani in that case we have $q\upharpoonright\gamma\leq q_2\upharpoonright\gamma$ by inductive hypothesis, but as $q_2\upharpoonright\gamma\Vdash q_2(\gamma)=1$, we then have $q\upharpoonright\gamma\Vdash q(\gamma)\leq q_2(\gamma)$. $\endgroup$ Commented Apr 3, 2017 at 16:55
  • $\begingroup$ You are right, I made a mistake with the order relation. $\endgroup$ Commented Apr 5, 2017 at 3:25
  • $\begingroup$ Your claim "It is easy to show by induction on $1\leq\beta\leq\kappa^+$ that $\pi(q_1,q_2)\upharpoonright\beta\leq q\upharpoonright\beta$" is not clear. $\endgroup$ Commented Apr 8, 2017 at 4:27
  • $\begingroup$ As at successors we are forcing with the projected condition and not the condition itself. $\endgroup$ Commented Apr 8, 2017 at 5:20
  • $\begingroup$ @Mohammad You prove $\pi(q_1,q_2)\upharpoonright\beta\leq q\upharpoonright\beta$ by induction on $\beta\geq 1$. Case $\beta=1$ is obvious. If $\beta=\gamma+1$ is odd, then $1\Vdash_{P\upharpoonright\gamma} \pi(q_1,q_2)(\gamma)=q_1(\gamma)$, and thus as by induction $\pi(q_1,q_2)\upharpoonright\gamma\leq q\upharpoonright\gamma$, we get $\pi(q_1,q_2)\upharpoonright\gamma\Vdash \pi(q_1,q_2)(\gamma)=q(\gamma)$, and therefore $\pi(q_1,q_2)\upharpoonright\beta\leq q\upharpoonright\beta$. $\endgroup$ Commented Apr 8, 2017 at 17:20

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