show/hide this revision's text 3 Added ZFC counterexample at omega_2

Here is a counterexample, which works in ZFC without anyadditional large cardinal or other extra hypothesis. Thisargument, which verifies the guess I made in my original answer, is the result of a conversation I had with ArthurApter.

The example involves the forcing $\mathbb{S}$ to add astationary non-reflecting subset of $\omega_2$, that is, astationary 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 boundedsets $s\subset\omega_2$ satisfying the condition for all$\gamma\leq\sup(s)$. The forcing is$\lt\omega_2$-strategically closed, since in the game whereplayers play a game of length $\omega_2$, with player IIplaying at limit stages, player II may invent an imaginaryclub set $c$ which is extended as play proceeds, and sheensures that this club remains disjoint from the conditions$s$ that are played. Thus, the forcing adds no new subsetsof $\omega_1$ 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 beable to construct an initial segment of $S$ that contains aclub of order type $\omega_1$, which would violate thenon-reflecting property.

Next, let $\mathbb{T}$ be the forcing to destroy thestationarity of the set $S$ added by $\mathbb{S}$, byadding a club set $C\subset\omega_2$ with $S\capC=\emptyset$, using closed initial segments. A bootstrapargument shows that the combined forcing$\mathbb{S}\ast\mathbb{T}$ has a dense subset that consistsessentially of $(s,c)$, where $s\subset\gamma=\sup(s)$ and$c$ is a closed set containing $\gamma=\sup(c)$ with $s\capc=\emptyset$. This dense set is $\leq\omega_1$-closed, andthus the combined forcing $\mathbb{S}\ast\mathbb{T}$ isforcing equivalent to $\text{Add}(\omega_2,1)$. So thesituation is that $\mathbb{S}$ makes the regrettable fauxpas 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-Bernsteinfor forcing. Let $\mathbb{P}=\text{Add}(\omega_2,1)$, andlet $\mathbb{Q}=\mathbb{P}\ast\mathbb{S}$. Clearly$\mathbb{P}$ is a factor of $\mathbb{Q}$, and $\mathbb{Q}$is a factor of $\mathbb{P}$, precisely becauseSo each embeds completely into the other, but they are notforcing equivalent, because $\mathbb{P}$ has a$\leq\omega_1$-closed dense subset, but $\mathbb{Q}$ doesnot, and this is a property preserved by forcingequivalence.

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

show/hide this revision's text 2 added 6 characters in body

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 $\mathbb{Q}\mathbb{D}$ \mathbb{Q}\ast\mathbb{D}$ is the same as $\mathbb{P}\mathbb{T}\ast\mathbb{D}$, \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.)
show/hide this revision's text 1

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 $\mathbb{Q}\mathbb{D}$ is the same as $\mathbb{P}\mathbb{T}\ast\mathbb{D}$, 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.)