# When are two forcing posets "the same"?

Let $$B$$ and $$C$$ be complete Boolean algebras. To avoid triviality I may also want them to be atomless. For $$b\in B$$ nonzero, denote $$B\upharpoonright b=\{p\in B:p\leq b\}$$, which can be viewed as a complete Boolean algebra with maximal element $$b$$. By $$B^+$$ I mean the nonzero elements of $$B$$. The following is well-known, see e.g. this post by Hamkins.

TFAE:

(i) For every $$C$$-generic filter $$H$$, $$V[H]$$ contains a $$B$$-generic $$G$$ such that $$V[G]=V[H]$$, and vice versa.

(ii) $$\{c\in C^+:\exists b\in B^+(C\upharpoonright c\simeq B\upharpoonright b)\}$$ is dense in $$C^+$$, and vice versa.

Questions:

1. Can (ii) be strengthened to the following: there exist maximal antichains $$A_1\subseteq B$$ and $$A_2\subseteq C$$ such that for every $$b\in A_1$$, $$B\upharpoonright b\simeq C\upharpoonright c$$ for some $$c\in A_2$$, and vice versa.

2. Consider (iii) $$B$$ completely embeds into $$C$$ and vice versa. How is this related to (i) and (ii)? What is an example showing $$B$$ and $$C$$ may not be isomorphic?

The motivation for the first question is that $$B$$ and $$C$$ are not necessarily isomorphic, since for example $$C$$ could be a direct product of a large number of $$B$$'s, so they cannot be isomorphic for cardinality reason; direct product of Boolean algebras corresponds to disjoint sum (or lottery sum) of posets. I wonder if this is "the only obstacle" in some sense (if $$B$$ is a complete Boolean algebra and $$A\subseteq B$$ is a maximal antichain, then $$B$$ is isomorphic to $$\prod_{a\in A}B\upharpoonright a$$).

1. I shall refer to the condition that you describe in Question 1 as condition (a). If you have condition (a) then you get back (ii), since if $$B\upharpoonright b\simeq C\upharpoonright c$$ then for all $$b'\leq b$$ there is $$c'\leq c$$ such that $$B\upharpoonright b'\simeq C\upharpoonright c'$$. Just look at the image of $$b'$$ under the isomorphism.

Condition (a) is strictly stronger than condition (ii), though. Let $$B$$ be a rigid atomless complete boolean algebra, and let $$E$$ be the lottery sum of $$B\upharpoonright b$$ for all $$b\in B$$. Then $$B$$ and $$E$$ satisfy condition (ii), since for all $$e\in E\backslash\{\mathbf{1}\}$$, $$E\upharpoonright e\simeq B\upharpoonright b$$ for some $$b\in B$$, and vice versa for all $$b\in B$$. However, if $$A\subseteq E$$ is a maximal antichain, then let $$e\in A$$. Let $$b\in B$$ be such that $$e$$ is in the $$B\upharpoonright b$$ summand of the lottery sum, let $$b'\leq b$$, and let $$e'\in A$$ be such that $$e'$$ is in the $$B\upharpoonright b'$$ summand of the lottery sum. Then by rigidity, the unique $$b''\in B$$ such that $$E\upharpoonright e'\simeq B\upharpoonright b''$$ satisfies $$b''\leq b'$$. Hence, no antichain $$A'\subseteq B$$ will be sufficient to capture all $$e\in E$$.

Thanks to Andreas Lietz for pointing out this construction.

1. Condition (iii) is independent of conditions (i) and (ii).

((i) but not (iii)). Consider the example of a large lottery sum: $$B$$ is a countable atomless complete boolean algebra, and $$C$$ is the (boolean completion of the) lottery sum of $$\aleph_1$$-many copies of $$B$$. Then cardinality shows that we cannot have $$B$$ and $$C$$ be isomorphic, but we do have conditions (i) and (ii). On the other hand, we cannot have a complete embedding from $$C$$ into $$B$$. Let $$b\in B$$ and let $$\langle\alpha,b\rangle\in C$$ be the $$\alpha$$th copy of the element $$b$$ in $$C$$. Then if $$f\colon C\to B$$ is any function, there must be $$\alpha<\beta$$ such that $$f(\alpha,b)=f(\beta,b)$$, so $$\langle\alpha,b\rangle\bot\langle\beta,b\rangle$$, but $$f(\alpha,b)\|f(\beta,b)$$, and hence $$f$$ is not a complete embedding.

((iii) but not (i)). For $$n<\omega$$, let $$P_n=\prod_{m\leq n}\text{Add}(\omega_m,1)$$. Let $$P=\bigoplus_{n<\omega}P_{2n}$$, and $$Q=P_0\oplus\bigoplus_{n<\omega}P_{2n+1}$$, and let $$B$$ and $$C$$ be their respective boolean completions. For $$k, $$P_n$$ completely embeds into $$P_k$$, so we have complete embeddings $$B\to C$$ and $$C\to B$$. However, if $$G\subseteq B$$ is $$V$$-generic, then $$V[G]\neq V[H]$$ for any $$V$$-generic $$H\subseteq C$$, since $$B$$ will add new subsets of the $$\omega_m$$s after some odd $$n$$, whereas $$C$$ will add new subsets of the $$\omega_m$$s after some even $$n$$.

You may be interested in a stronger notion of morphisms: dense embeddings. $$\pi\colon B\to C$$ is a dense embedding if

1. For all $$b,b'\in B$$, $$b'\leq b$$ if and only if $$\pi(b')\leq\pi(b)$$; and
2. For all $$c\in C$$ there is $$b\in B$$ such that $$\pi(b)\leq c$$.

Then there is a dense embedding $$B\to C$$ for complete boolean algebras $$B,C$$ if and only if $$B$$ is isomorphic to $$C$$.

• Why is (iii) stronger than (i) and (ii)? In (i) there is the extra requirement that $V[G]=V[H]$, which isn't clearly implied by (iii). Jul 26 at 10:26
• I like your (iii) but not (i) example—I had been looking at more complicated examples. A similar perhaps simpler example would be to let $P$ be the lottery sum of the collapses of $\omega_k$ for even $k$ and $Q$ the same for odd $k$. Jul 26 at 13:16
• @JoelDavidHamkins Yes that's a much cleaner fix, d'oh! Jul 26 at 13:55
• I've swapped back the construction and implemented Joel's fix instead, since my original "fix" still had the problem of having a summand of $B$ not be embed-into-able, so you can have a $B$-generic with no $C$-generic inside. Jul 27 at 8:07
• @CalliopeRyan-Smith If Im not mistaken, your strategy to show (a) to be strictly stronger than (ii) should work if you take $B$ a rigid atomless cBa and set $E=B$. Aug 1 at 14:33

Let me augment Calliope's excellent answer by providing a slightly stronger example. What I want to provide is an example exhibiting property (iii) without property (i), but in a strong way, in that the two forcing notions have no common extensions at all.

(Note that Calliope's example does not have this stronger feature since her two lotteries have the first lottery option in common, and so the two forcing notions are isomorphic below the respective conditions opting for that lottery winner.)

In short, what I want are forcing notions $$\newcommand{\P}{\mathbb{P}}\P$$ and $$\newcommand\Q{\mathbb{Q}}\Q$$ such that every forcing extension arising from either of them is a submodel of a forcing extension by the other, but the two forcing notions have no common forcing extension.

To achieve this stronger feature, we simply modify the forcing slightly. Let $$\P=\oplus_{n>1}\ \text{Coll}(\omega,\beth_n)\ast\dot{\text{Add}}(\check\beth_n^+,1)$$ be the lottery sum of the collapse forcing notions of $$\beth_n$$ to $$\omega$$ for $$n<\omega$$, followed by the forcing of CH. The generic filter will therefore pick one $$n$$ and collaspse $$\beth_n$$ to $$\omega$$, and then force CH.

Next, let $$\Q=\oplus_{n>1}\ \text{Coll}(\omega,\beth_n)\ast \dot{\text{Add}}(\omega,\check\beth_n^{++})$$ be the similar lottery sum, except that in each lottery option we also force $$\neg$$CH afterward.

It follows that $$\P$$ and $$\Q$$ can have no forcing extensions in common, since $$\P$$ forces CH but $$\Q$$ forces $$\neg$$CH. This is a strong violation of (i), since the two posets force different theories.

Nevertheless, property (iii) holds, since every factor in either of the $$\P$$ and $$\Q$$ lotteries embeds as a complete subalgebra of a later factor of the other, one that collapses a large enough cardinal.

We can strengthen (ii) to obtain an isomorphism of Boolean algebras, but we first need to make copies of the Boolean algebras $$B,C$$ to eliminate the obstructions explained in the other answers. We may also reformulate (ii) in terms of partitions.

The following version of the Schroder-Bernstein theorem was generalized to $$\sigma$$-complete Boolean algebras by Sikorski and Tarski.

Theorem: Let $$B,C$$ be $$\sigma$$-complete Boolean algebras. If $$B\upharpoonright b\simeq C$$ and $$C\upharpoonright c\simeq B$$, then $$B\simeq C$$.

Proposition: Let $$B,C$$ be complete Boolean algebras. Then the following are equivalent:

1. There are sets $$I,J$$ where the direct powers $$B^I=\{f\mid f:I\rightarrow B\}$$ ($$B^J$$ is the direct power of $$B$$ in the category and also the variety of Boolean algebras) and $$C^J$$ are isomorphic.

2. $$\{b\in B^+:\exists c\in C,B\upharpoonright b\simeq C\upharpoonright c\}$$ is dense in $$B$$ and $$\{c\in C^+:\exists b\in B,B\upharpoonright b\simeq C\upharpoonright c\}$$ is dense in $$C$$.

3. There is a partition $$p$$ of $$B$$ and a partition $$q$$ of $$C$$ so that if $$b\in p$$, then there is some $$c\in C$$ where $$B\upharpoonright b\simeq C\upharpoonright c$$ and if $$c\in p$$, there is some $$b\in B$$ where $$B\upharpoonright b\simeq C\upharpoonright c$$.

Proof:

$$1\rightarrow 3.$$ Let $$B_0,C_0$$ be complete Boolean algebras and suppose that $$p$$ is a partition of $$B_0$$ and $$q$$ is a partition of $$C_0$$ where $$B_0\upharpoonright b \simeq B$$ for $$b\in p$$ and $$C_0\upharpoonright c\simeq C$$ for $$c\in q$$. Let $$\phi:B_0\rightarrow C_0$$ be a Boolean algebra isomorphism. Then for each $$b\in p$$, the set $$\{b\wedge\phi^{-1}(c):c\in q\}^+$$ is a partition of $$b$$. However, if $$c\in q$$, then $$B_0\upharpoonright(b\wedge\phi^{-1}(c))\simeq C_0\upharpoonright(\phi(b)\wedge c)\simeq C\upharpoonright c_0$$ for some $$c_0\in C$$. Since $$B_0\upharpoonright b\simeq B$$, we conclude that there is a partition $$r$$ of $$B$$ where for each $$b\in r$$, there is some $$c_0\in C$$ where $$B\upharpoonright b\simeq C\upharpoonright c_0$$.

$$3\rightarrow 2$$. This is trivial.

$$2\rightarrow 3.$$ Set $$L=\{b\in B^+:\exists c\in C,B\upharpoonright b\simeq C\upharpoonright c\}$$ and assume that $$L$$ is dense in $$B$$. Then let $$p$$ be a maximal subset of $$L$$ subject to the condition that $$b_0\wedge b_1=0$$ whenever $$b_0,b_1\in p$$. Such a set $$p$$ exists by Zorn's lemma and is necessarily a partition of $$B$$.

$$3\rightarrow 1$$. There is a partition $$p$$ of $$B$$ and a function $$\phi_p:p\rightarrow C^+$$ where $$B\upharpoonright a\simeq C\upharpoonright \phi_p(a)$$ for $$a\in p$$.

In this case, $$B\simeq\prod_{a\in p}B\upharpoonright a\simeq\prod_{a\in p}C\upharpoonright\phi_p(a)\simeq C^p\upharpoonright(\phi_p(a))_{a\in p}$$. Therefore, if $$|I|\geq|p|$$, then $$B^I\simeq C^I\upharpoonright c$$ for some $$c\in C^I$$. By an analogous argument, we can conclude that $$C^I\simeq B^I\upharpoonright b$$ for some $$b\in B^I$$ as long as $$I$$ is large enough. Therefore, by Sikorski and Tarski's version of the Schroder-Bernstein theorem, we conclude that $$C^I\simeq B^I$$. $$\square$$

• Can you clarify the meaning of "direct power"? Does this indicate finite support? Aug 2 at 21:22
• I meant the direct power in the category of Boolean algebras which is just the direct power in the variety of Boolean algebras. This means we have arbitrary support (and the direct power is complete so we don't need to take a completion after taking the product). Aug 2 at 22:42
• Thanks. That is the same as the lottery sum of I copies of B. Aug 2 at 23:06