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According to this paper of Chaz Schlindwein, any countable support iteration of semi-proper forcings is semi-proper. This seems like a breakthrough simplification, and I wonder why it is not more well-known. Before diving into the arguments I want to ask the forcing community whether this work is widely viewed as correct? Thanks for your help.

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    $\begingroup$ I may give the sad news that Prof. Chaz Schlindwein died on August 19, 2015. See here $\endgroup$ Sep 25, 2015 at 16:13
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    $\begingroup$ Let me add: all my respect to Professor Schlindwein, whom I unfortunately did not have the chance to meet. $\endgroup$ Sep 25, 2015 at 17:26
  • $\begingroup$ @Mohammad I'm sorry to hear about this, I was not aware $\endgroup$ Sep 25, 2015 at 20:33
  • $\begingroup$ @MohammadGolshani I too am very sorry to hear this. $\endgroup$ Sep 26, 2015 at 9:59

2 Answers 2

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I think that this claim is false, at least in the way that I understand "countable support". The following was explained to me by Menachem Magidor in the 1990s; it may be folklore, and I suspect it is the reason for introducing RCS in the first place.)

Consider the following countable support iteration $(P_\alpha, Q_\alpha: \alpha < \kappa)$ of length $\kappa$, where $\kappa$ is either a measurable cardinal or $\omega_2$ in a universe where Namba forcing is semiproper:

  • $Q_0$ is Prikry forcing on $\kappa$, or Namba forcing.
  • For each $\alpha>0$, $Q_\alpha$ is an antichain on $\omega_1^V$ (with an additional largest element $1_\alpha$ thrown in). (Really: the canonical $P_\alpha$-name for this set.)
  • Each $Q_\alpha$ is semiproper. (Even proper, except for $Q_0$.)
  • For $0\le \alpha\le \kappa$, $P_\alpha$ is defined in the usual way, as the set of all countable partial functions $p$ with domain $\subseteq \alpha$ such that for all $\beta\in dom(p)$ we have $p\restriction \beta$ is in $P_\beta$ and forces $p(\beta)\in Q_\beta$.

Then the following statements are true:

  • Every condition in $P_\kappa$ has bounded domain. (This would not be true if we had used revised countable support iteration.)
  • $Q_0$ introduces a new sequence $(\kappa_n: n \in \omega)$ which is cofinal in $\kappa$. (I.e., there is a $Q_0$-name for such a sequence. Hence there is also a $P_\kappa$-name for such a sequence.)

We define functions $g$ and $h$ as follows:

  • Let $g$ be a $P_\kappa$-name for the function from $\kappa\setminus \{0\}$ to $\omega_1^V$ which assigns to each ordinal $\alpha$ the generic element of the antichain $Q_\alpha$.
  • Let $h$ be the $P_\kappa$-name for the function $n\mapsto g(\kappa_n)$.

Then the empty condition forces that $h$ is a surjective function from $\omega$ onto $\omega_1$. (Hence $P_\kappa$ is not semiproper.)

Proof: Assume that $\gamma< \kappa$, and that $p\in P_\kappa$ is a condition. It is enough to find a stronger condition forcing that $\gamma$ is in the range of $h$.

First find $\alpha$ such that $p\in P_\alpha$.
Note that $p$ "knows nothing" about the values of $g$ above $\alpha$, and that $p$ knows very little about the sequence $(\kappa_n:n\in\omega)$.

Let $n$ be the length of the stem of $p(0)$, the $Q_0$-component of $p$. By extending the stem, we can find $p'(0)$ stronger than $p(0)$, such that $p'(0)$ forces a value (say $\beta$) to $\alpha_n$, and such that $\beta> \alpha$.

Define $p'$ as follows: $dom(p')=dom(p)\cup \{\beta\}$. $p'(\beta)=\gamma$. (And $p'(i)=p(i)$ for all $i\not=0,\beta$.)

Now $p'$ forces that $h(n)=g(\alpha_n)=g(\beta)=\gamma$. QED


In defense of Schlindwein's paper I should add that he might have had a different (nonequivalent) definition of "CS iteration" in mind.

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  • $\begingroup$ Correct me if I'm wrong, but for non-finite support iteration the definition with $p\restriction\beta\Vdash_\beta p(\beta) \in Q_\beta$ is not equivalent to $1_\beta\Vdash_\beta p(\beta) \in Q_\beta$. And the latter is "the usual" definition for iteration anyway. $\endgroup$
    – Asaf Karagila
    Sep 26, 2015 at 1:36
  • $\begingroup$ Jech requires $1_\beta\Vdash p(\beta)\in Q_\beta $ (Def 16.29, page 280). Kunen requires only $p\mathord\restriction\beta \Vdash p(\beta)\in Q_\beta$. (Def 5.8, p. 273) - But Jech's conditions are dense in Kunen's: Let $p$ be a "Kunen" condition, and define a ("Kunen", for the moment) condition $p'$ with the same support as follows: $p'(\beta):= p(\beta) $ if that happens to be in $Q_\beta$, and $1_\beta$ otherwise. (Use existential completeness, of course.) By induction on $\beta$, show $p'\mathord\restriction \beta \le p\mathord\restriction \beta$. - Now $p'$ satisfies Jech's condition. $\endgroup$
    – Goldstern
    Sep 26, 2015 at 12:52
  • $\begingroup$ Yes, and Kunen points out that for finite support it's the same; but for larger supports it isn't the same (Kunen 1980 edition, the chapter about iterations, exercises K2-K3), at least in terms of Boolean completions. And if the Boolean completion is not the same, then the forcings are not the same. Or am I missing something? (I'm probably missing something.) $\endgroup$
    – Asaf Karagila
    Sep 26, 2015 at 12:56
  • $\begingroup$ When I wrote about "Kunen" conditions, I did not mean to restrict only to conditions $p$ such that each $p(\beta)$ is in $dom(Q_\beta)$ (as Kunen does). This would lead to "Kunen's pathology" (exercise E4; one could almost call it the "Kunen inconsistency":-) that a CS iteration of proper (even ccc!) forcings might collapse $\omega_1$. - I only use "full names $Q_\beta$", or equivalently, ALL names $q(\beta)$ are allowed, at least below a certain rank. - I think that then the discrepancy disappears. $\endgroup$
    – Goldstern
    Sep 28, 2015 at 15:08
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Following up on Martin's answer:

Definition 10 of the paper defines the concept "hemi-properness", shows that semi-proper forcings are hemi-proper, and then claims (Theorem 16) that hemi-properness is preserved under countable support iterations. This last is demonstrably false, as we will show that "hemi-properness" is equivalent to "preserves $\omega_1$", and it is well known that this property is not preserved by countable support iteration.

Definition [Definition 10 from the paper]

A poset $P$ is hemi-proper if and only whenever $\lambda$ is a sufficiently large regular cardinal, $M$ and $N$ are countably elementary submodels of $H(\lambda)$, $P\in M\in N$, and $q\in P\cap M$, then $q\nVdash``\omega_1\cap M[G_P]>\omega_1\cap N$''.

Lemma 11 in his paper shows that hemi-proper forcings do not collapse $\omega_1$.

Claim: If forcing with $P$ preserves $\omega_1$, then $P$ is hemi-proper.

Let $\lambda$, $M$, $N$, and $q$ be given. Any condition forces that $M[G_P]$ is countable (it is just the interpretations of the countably many $P$-names living in $M$). In particular, any condition forces that $M[G_P]\cap\omega_1$ is an ordinal $\alpha<\omega_1$ since $\omega_1$ is preserved. The model $N$ can see $M$, and so $N$ will contain a name $\dot\alpha$ for the countable ordinal $M[G_P]\cap\omega_1$ from $M[G_P]$. In $N$, we can extend $q$ to a condition $r$ deciding a particular value $\alpha$ for $\dot\alpha$. Notice that since $r$ and $\dot\alpha$ are in $N$, the countable ordinal $\alpha$ is in $N$ as well. Necessarily we have $\alpha<N\cap\omega_1$. Since $r$ extends $q$, it follows that $q$ did not force $\omega_1\cap M[G_P]$ to be greater than $N\cap \omega_1$. This establishes that $P$ is hemi-proper.

Schlindwein lectured on this at a few meetings in the 1990s. The paragraph at the top of page 9 was added in an attempt to meet my objections that his notion was the same as "preserves $\omega_1$'', but there are no such $M$ and $N$ as he describes there.

Chaz was a nice guy. I'm sad to hear the news that he has passed away.

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  • $\begingroup$ Thank you for addressing some of the mathematics. Can you recommend something that would encourage students to read the paper, or that you took away from the paper? ("No" would be a reasonable answer.) Gerhard "Looking For The Bright Side" Paseman, 2015.09.25 $\endgroup$ Sep 25, 2015 at 21:55
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    $\begingroup$ Not with this paper, Gerhard. I think he may have even withdrawn it eventually, but things live forever on Arxiv. $\endgroup$ Sep 25, 2015 at 23:49

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