The following is a folklore result.

Suppose $P$ is a countable support iteration of nontrivial forcings, $\langle P_\alpha, \dot{Q}_\alpha : \alpha < \omega_1 \rangle$. Then there is a complete embedding of $\mathrm{Add}(\omega_1)$ into $P$.

The forcing to add a Cohen subset of $\omega_1$ fails the $\omega_1$-approximation property, since it produces a “fresh” sequence— a sequence such that all initial segments are in the ground model.

In the 1979 paper, “Iterated perfect-set forcing,” Baumgartner and Laver seem to make a contrary claim. Lemma 6.2 states that the countable support iteration of Sacks forcing produces no fresh sequences of length some ordinal of uncountable cofinality. This is key to their argument that iterating Sacks forcing up to a weakly compact forces the tree property at $\omega_2$.

I do not see a flaw in their argument. Is the folklore claim correct? How is this resolved?

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    $\begingroup$ I've not seen that "folklore result" before - does it actually appear used in the literature? $\endgroup$ Aug 4, 2020 at 23:44
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    $\begingroup$ @NoahSchweber It’s stated as Remark 0.3 in Goldstern’s “Tools for your forcing construction.” $\endgroup$ Aug 5, 2020 at 7:28
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    $\begingroup$ Goldstern was an idiot. (But that was long ago, perhaps he has learned something since then?) $\endgroup$
    – Goldstern
    Aug 5, 2020 at 11:32
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    $\begingroup$ This is similar to "aren't the csi and the csp of ($\omega$-) Cohen forcing the same?". This holds true in the finite support case. But in the countable case, one is proper and the other collapses the continuum. $\endgroup$ Aug 7, 2020 at 19:41

2 Answers 2


I don't see why the "folklore result" holds. The analogous result for finite support iteration is as follows.

$\textbf{Fact}$: If $\langle P_i, Q_j: i \leq \alpha + \omega, j < \alpha + \omega \rangle$ is a finite support iteration of non-trivial forcings, then $P_{\alpha + \omega}$ adds a Cohen real over $V^{P_{\alpha}}$.

Its proof goes as follows. WLOG, $\alpha = 0$. Fix $q_k, r_k \in V^{P_{\alpha + k}}$ such that $\Vdash_{P_{\alpha + k}} q_k, r_k \in Q_k \wedge q_k \perp_{Q_k} r_k$. Define $c \in V^{P_{\omega}} \cap 2^{\omega}$ by $c(k) = 1$ iff $(\exists p \in G(P_{\omega}))(p(k) = q_k)$. To see that $c$ is Cohen over $V$, suppose $D \in V$ dense in $2^{< \omega}$ and $p \in P_{\omega}$. Fix $n$ such that $p \in P_n$. Extend $p$ to $p' \in P_n$ in $n$ steps such that for each $k < n$, $p \upharpoonright k \Vdash_{P_k} (p(k) \perp_{Q_k} q_k) \text{ or } (p_k \leq_{Q_k} q_k)$. Let $s \in 2^n$ be determined by this $p'$. Choose an extension $s'$ of $s$ in $D$. The rest should be clear.

In the case of countable support iteration, the passage from $p$ to $p'$ could be problematic since one might have to extend $p$ infinitely many times. This also shows that the folklore result would work if we iterate countably closed forcings. In any case, Baumgartner-Laver result implies that the folklore result is false. The closest related result about countable support iteraton is the following. Any countable support iteration of non-trivial forcings of length $\omega_1$ collapse the continuum to $\omega_1$. This is the Sublemma inside Lemma 6.3 in the Baumgartner-Laver paper.

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    $\begingroup$ In fact this fails in the following example. Let $c\in 2^{\omega_1}$ be the name defined analogously to the case of finite support iteration. Let $Q_0$ be Cohen forcing, adding a Cohen real $x\in 2^\omega$. Now let $p\in P_\omega$ be defined as follows: $p(0)=$ trivial, and $p(n+1)$ is a name for ``if $x(n)=0$, then $q_n$, else $r_n$.'' -- Clearly [don't believe Goldstern if he says "clearly"!] $p$ forces that $c$ is not fresh, as its initial $\omega$-segment $x$ is not in $V$. - See also Johannes Schürz' more general answer. $\endgroup$
    – Goldstern
    Aug 5, 2020 at 11:29

Exactly. The rumour (not folklore ;) ) is even wrong if you iterate Cohen forcing (on $\omega$ !!) with countable support $\omega_1$ many times. Let $P$ denote the iteration of Cohen forcings. It follows that any complete embedding $F$ from $\text{Add}(\omega_1)$ into $P$ is an independent permutation of coordinates and $0$ and $1$'s. Indentify $2^\omega$ with the corresponding max antichain in $ \text{Add}(\omega_1)$ and wlog assume that for every $x \in 2^\omega$ it holds that $F(x)= \inf_{n \in \omega} F(x \restriction n)$. But then $F[2^\omega]$ has a uniform, countable support $A$. Wlog let $A=\omega$. But there is $p\in P$ such that $p$ codes the first Cohen-real horizontally into $p\restriction \omega\, (\cdot)\, (0)$. As the Cohen-real differs from any ground model real, the embedding cannot be complete.

  • $\begingroup$ Wanted to post this as a 'sketchy' comment, but turned out to be too long... $\endgroup$ Aug 4, 2020 at 23:11

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