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For forcing notions $\mathbb{P}$ and $\mathbb{Q}$ let us write $\mathbb{P}\triangleleft\mathbb{Q}$ if forcing with $\mathbb{Q}$ always adds a $\mathbb{P}$-generic filter over $V$. In other words, $\mathbb{P}\triangleleft\mathbb{Q}$ holds if there is a $\mathbb{Q}$-name $\tau$ such that $\tau[H]$ is a $\mathbb{P}$-generic filter over $V$ whenever $H$ is a $\mathbb{Q}$-generic filter over $V$. If $\mathbb{P}\triangleleft\mathbb{Q}$ and $\mathbb{Q}\triangleleft\mathbb{P}$ need $\mathbb{P}$ and $\mathbb{Q}$ be forcing isomorphic, ie must they produce the same generic extensions? I expect not, but haven't been able to think of a counterexample.

I believe this question can be phrased in terms of Boolean algebras; ie if two complete Boolean algebras are each isomorphic to a complete Boolean subalgebra of the other, need they be forcing isomorphic?

(Incidentally, if the generics the forcings add are done in a 'reversible' way, ie if $\tau,\sigma$ are the $\mathbb{Q},\mathbb{P}$ names for the added generics and always $\tau[\sigma[G]]=G$ and $\sigma[\tau[H]]=H$ then the forcing notions will be equivalent. This is mentioned in the beginning of Shelah's text Proper Forcing).

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I find the following question to be related: mathoverflow.net/questions/70954/… –  Joel David Hamkins Oct 27 '11 at 22:29

4 Answers 4

up vote 9 down vote accepted

Here is a counterexample, which works in ZFC without any additional large cardinal or other extra hypothesis. This argument, which verifies the guess I made in my original answer, is the result of a conversation I had with Arthur Apter.

The example involves the forcing $\mathbb{S}$ to add a stationary non-reflecting subset of $\omega_2$, that is, a stationary set $S\subset\omega_2$, such that $S\cap\gamma$ is not stationary for any $\gamma\lt\omega_2$ of cofinality $\omega_1$. Conditions in $\mathbb{S}$ consist of bounded sets $s\subset\omega_2$ satisfying the condition for all $\gamma\leq\sup(s)$. The forcing is $\lt\omega_2$-strategically closed, since in the game where players play a game of length $\omega_2$, with player II playing at limit stages, player II may invent an imaginary club set $c$ which is extended as play proceeds, and she ensures that this club remains disjoint from the conditions $s$ that are played. Thus, the forcing adds no new subsets of $\omega_1$ and in particular, preserves $\omega_2$. Also, it follows that the generic set $S\subset\omega_2$ added by $\mathbb{S}$ is indeed stationary. Note that $\mathbb{S}$ has no $\leq\omega_1$-closed dense subset, since with such a highly closed dense subset we would be able to construct an initial segment of $S$ that contains a club of order type $\omega_1$, which would violate the non-reflecting property.

Next, let $\mathbb{T}$ be the forcing to destroy the stationarity of the set $S$ added by $\mathbb{S}$, by adding a club set $C\subset\omega_2$ with $S\cap C=\emptyset$, using closed initial segments. A bootstrap argument shows that the combined forcing $\mathbb{S}\ast\mathbb{T}$ has a dense subset that consists essentially of $(s,c)$, where $s\subset\gamma=\sup(s)$ and $c$ is a closed set containing $\gamma=\sup(c)$ with $s\cap c=\emptyset$. This dense set is $\leq\omega_1$-closed, and thus the combined forcing $\mathbb{S}\ast\mathbb{T}$ is forcing equivalent to $\text{Add}(\omega_2,1)$. So the situation is that $\mathbb{S}$ makes the regrettable faux pas of creating a stationary non-reflecting set, but $\mathbb{T}$ apologizes, and the combination $\mathbb{S}\ast\mathbb{T}$ is completely mild.

So now we can build the counterexample to Cantor-Bernstein for forcing. Let $\mathbb{P}=\text{Add}(\omega_2,1)$, and let $\mathbb{Q}=\mathbb{P}\ast\mathbb{S}$. Clearly $\mathbb{P}$ is a factor of $\mathbb{Q}$, and $\mathbb{Q}$ is a factor of $\mathbb{P}$, precisely because $\mathbb{Q}\ast\mathbb{T}=\text{Add}(\omega_2,1)\ast\mathbb{S}\ast\mathbb{T}=\text{Add}(\omega_2,1)\ast\text{Add}(\omega_2,1)\cong\text{Add}(\omega_2,1)=\mathbb{P}$. So each embeds completely into the other, but they are not forcing equivalent, because $\mathbb{P}$ has a $\leq\omega_1$-closed dense subset, but $\mathbb{Q}$ does not, and this is a property preserved by forcing equivalence.

The argument easily generalizes to higher cardinals than $\omega_2$ (but not for $\omega_1$).


Here is the original answer:

Here is a counterexample, but it uses a large cardinal. I expect that we will be able to eliminate the large cardinal, perhaps by constructing a similar example down low.

Suppose that $\kappa$ is weakly compact. Let $\mathbb{P}=\text{Add}(\kappa,1)$ be the forcing to add a Cohen subset of $\kappa$ by initial segment. Let $\mathbb{Q}=\text{Add}(\kappa,1)*\mathbb{T}$ be the forcing that first adds a Cohen subset to $\kappa$, and then forces to create a $\kappa$-Suslin tree.

Clearly, $\mathbb{P}$ is explicitly a forcing factor of $\mathbb{Q}$. For the converse direction, observe that the forcing $\mathbb{T}$ to create the $\kappa$-Suslin tree can be followed by the forcing that destroys this Suslin tree $T$, by forcing $\mathbb{D}$ to cover it with $\kappa$-many branches. The combined forcing $\mathbb{T}\ast\mathbb{D}$ is actually isomorphic to the forcing consisting of trees of height less than $\kappa$ that are already covered by the branches. This forcing is ${\lt}\kappa$-closed and hence isomorphic to $\text{Add}(\kappa,1)$. It follows that $\mathbb{Q}\ast\mathbb{D}$ is the same as $\mathbb{P}\ast\mathbb{T}\ast\mathbb{D}$, which is the same as $\text{Add}(\kappa,1)\ast\text{Add}(\kappa,1)$, which is forcing equivalent to $\text{Add}(\kappa,1)$, which is $\mathbb{P}$. Thus, we have argued that a further forcing extension of $\mathbb{Q}$ is isomorphic to $\mathbb{P}$, and so $\mathbb{Q}$ is a factor of $\mathbb{P}$.

But the forcing notions $\mathbb{P}$ and $\mathbb{Q}$ are not always equivalent. For example, it is possible to make the weak compactness of $\kappa$ indestructible by $\text{Add}(\kappa,1)$, that is, by $\mathbb{P}$, but the weak compactness of $\kappa$ is always destroyed by $\mathbb{Q}$, since this adds a $\kappa$-Suslin tree, which is incompatible with $\kappa$ being weakly compact.

This forcing was the basis of Kunen's argument that weak compactness is not downwards absolute. In the forcing extension where the $\kappa$-Suslin tree is created, $\kappa$ is not weakly compact, but the weak compactness is recovered once one destroys the tree, since the combined forcing is just $\text{Add}(\kappa,1)$.

By making the preparatory forcing part of $\mathbb{P}$ and $\mathbb{Q}$, one can show that whenever $\kappa$ is weakly compact, then there are forcing notions that are factors of each other, but not forcing equivalent.

I suspect that a similar example can be made down low at the level of $\omega_2$, but I have to think it through. (If someone else can do this, please post an answer.)

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It seems that one might make a similar example with the forcing to add a stationary non-reflecting set, followed by the forcing to shoot a club through the complement (thereby destroying stationarity), a combination which is mild. –  Joel David Hamkins Oct 28 '11 at 0:17
    
Thanks, this is a nice answer. Though I'm going to hold out on accepting for a little while to see about a ZFC example.. –  Justin Palumbo Oct 28 '11 at 16:34
    
Yes, do please wait, since I agree that we should hope for a ZFC example. –  Joel David Hamkins Oct 28 '11 at 17:02
    
Excellent as always. Thanks. –  Justin Palumbo Oct 29 '11 at 17:54
    
Thanks, Justin; it is a really great question! I wonder if there is a simpler example? –  Joel David Hamkins Oct 30 '11 at 16:31

I was asked this question years ago by David Fremlin.
I came up with two nonlinear iterations of Sacks forcing that are not equivalent since they give different structures of degrees of reals over the ground model. Still, each completely embeds into the other one.

Then Ilijas Farah observed that there is an example that is even ccc: Koppelberg and Shelah showed that there is a complete subalgebra of a Cohen algebra (completion of a finite support iteration of Cohen forcing) which is not Cohen. This non-Cohen algebra however has a complete subalgebra that is isomorphic to the big Cohen algebra.

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Stefan, could you give the details for the Farah example? –  Joel David Hamkins May 2 '12 at 10:28
    
The Koppelberg-Shelah paper is <a href="arxiv.org/pdf/math/9610227.pdf">here</a>;. For every $\kappa\geq\aleph_2$ they construct a two-step forcing iteration where the first step is a ccc forcing consisting of certain finite partial function whose domains are subsets of $\kappa$. The two step iteration is equivalent to Cohen forcing (for adding $\kappa$ Cohen reals), but the first step alone is not equivalent to Cohen forcing. But it is easily checked that the Cohen forcing for adding $\kappa$ reals completely embedds into the first step. –  Stefan Geschke May 2 '12 at 11:51

Here is another, perhaps simpler, counterexample. Let $\mathbb{D}$ be Hechler forcing, the usual ccc forcing to add a dominating real which consists of conditions $\langle s,f\rangle$ where $s\in\omega^{<\omega}$ and $f\in\omega^\omega$. Let $\mathbb{D}_T$ be the tree variant of Hechler forcing; this forcing consists of nonempty trees $T⊆ω^{<ω}$ with the property that there is a unique stem $s∈T$ so that for every $t∈T$ extending $s$, $t⌢n∈T$ for all but finitely many n∈ω. The forcing is ordered by inclusion.

I claim that $\mathbb{D}$ and $\mathbb{D}_T$ are not forcing equivalent though each adds a generic for the other. The reason they are not forcing equivalent is that $\mathbb{D}$ adds an unbounded real which is eventually dominated by every dominating real, while $\mathbb{D}_T$ does not. These facts, are (hopefully) due to me. See Theorems 1,2 and Proposition 5 in a recent preprint: http://arxiv.org/abs/1201.2932

To come completely clean, this example was my original motivation for asking this question. I discovered this counterexample to the Cantor-Bernstein for forcing, and wondered whether previous counterexamples were known (or perhaps even that the Cantor-Bernstein for forcing was known to be true, thus meaning there was an error in my result). So Joel's answer was very reassuring, and I have cited it in the paper. I wanted to wait for everything to be written up before posting this answer.

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I realize I'm a bit late to the party, but I wanted to add that counterexamples can be easily constructed out of almost any type of forcing. I'll give a particular example, but the method is flexible. All we need are three forcings, $\mathbb{R}_1$, $\mathbb{R}_2$ and $\mathbb{R}_3$, pairwise incomparable by the $\lhd$ relation described in the question. For concreteness, let's take $\mathbb{R}_1 = Add(\omega,1)$, $\mathbb{R}_2 = Add(\omega_1,1)$ and $\mathbb{R}_3 = Add(\omega_2,1)$.

Let $\mathbb{P}$ be the lottery sum of $\mathbb{R}_1$ and $\mathbb{R}_1 \times \mathbb{R}_2$ (so $\mathbb{P}$ first chooses generically whether to force with just $\mathbb{R}_1$ or with the product $\mathbb{R}_1 \times \mathbb{R}_2$, and then forces accordingly). Similarly, let $\mathbb{Q}$ be the lottery sum of $\mathbb{R}_1$ and $\mathbb{R}_1 \times \mathbb{R}_3$.

$\mathbb{P}$ and $\mathbb{Q}$ are clearly not forcing equivalent, since the first has the option of adding a generic for $\mathbb{R}_2$, which the second cannot (and vice versa, adding a generic to $\mathbb{R}_3$). However, forcing with one will always add a generic for the other, since either one will add a generic for $\mathbb{R}_1$, which is one option in the lottery in each case.

This does not answer the second question, about Boolean algebras, and I think it demonstrates that the order relation on Boolean algebras given by "embedding into a subalgebra" is really distinct from the $\lhd$ order -- here we have two forcings satisfying $\mathbb{P} \lhd \mathbb{Q}$ and $\mathbb{Q} \lhd \mathbb{P}$, but neither Boolean algebra embeds into a subalgebra of the other.

This suggests a natural additional condition, namely that $H$ give the entire extension $V[G]$. That is, in every extension $V[G]$ by $\mathbb{P}$ there is $H$ generic for $\mathbb{Q}$ such that $V[G]=V[H]$. I believe this guarantees forcing equivalence in the case where $\mathbb{P} \lhd \mathbb{Q}$ and $\mathbb{Q} \lhd \mathbb{P}$, since every extension by $\mathbb{P}$ equals and extension by $\mathbb{Q}$ and vice versa.

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Great, Jonas! –  Joel David Hamkins Nov 24 '12 at 3:13
    
You could also take $\mathbb{R}_1$ to be trivial forcing, and your observation amounts to the fact: if $\mathbb{P}$ and $\mathbb{Q}$ each have an atom, then even though they may be very different below other conditions, nevertheless $\mathbb{P}\lhd\mathbb{Q}\lhd\mathbb{P}$. –  Joel David Hamkins Nov 24 '12 at 16:38
    
Joel, great example -- this drives home the point the $\lhd$ relation really says hardly anything about the posets at all. –  jonasreitz Nov 24 '12 at 23:05

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