A common forcing extension obtained via different forcing notions

I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.

I have always assumed that this fact was part of the classical forcing folklore results, but it doesn’t seem to be mentioned explicitly in the usual forcing literature (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.)

Theorem 1. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.

The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.

Theorem 2. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.

By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.

Proof: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.

I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.
In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.

Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that

  • $d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$.
  • $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$.
  • $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.

Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.

Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.

To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.

Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.QED

Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions.

14 thoughts on “A common forcing extension obtained via different forcing notions

  1. Joel, this is a great argument to know and I am glad that my unfounded skepticism prompted you to write it up :). Although, I might claim later that the argument we needed this fact for could be carried out differently as well.
    p.s. I am becoming more and more convinced of the advantages of forcing with Boolean algebras.

    • Oh, yes, I’d be interested to see your alternative argument in the context of our application. Meanwhile, I view theorems 1 and 2 here as basic facts about forcing, and so it is of course also fine to use them.

  2. Joel,

    This is an old result of mine. I don’t think I ever published it. It may appear in:

    Serge Grigorieff in Intermediate Submodels and Generic Extensions in Set Theory [Annals of Mathematics 101 (1975), 447-490].

    — Bob

    • Bob, Great! Thanks for the reference. The need for this result pops up once in a while—I used it in my “As you like it” paper, and now Vika and I just used it again.

    • Grigorieff’s paper always scared me for some reason.

      Maybe it’s the outdated notation, maybe it’s the fact that I was never a big fan of relative constructibility (go symmetries!), or maybe it’s just the fact that each section had its own counter for theorems and remarks (resulting in a handful of Theorem 1 within the same paper).

      Still, I’d rank this as one of the most important papers in choiceless set theory, maybe the most important. And I guess that it’s not without a reason that this paper appeared on Annals of Mathematics.

      • Asaf,

        Could you check whether the result is stated in that paper? For completeness, I’d like to track down the earliest reference.

        • Joel,

          I didn’t sit to read the whole thing. But it seems to me that you will be wise to take a look at sections 3.5 and 3.6 (Each include a single Theorem 1 and proof thereof).

          This would be pages 460-461 of the paper.

    • Thanks, Mohammad!

      Bagaria states the result there as lemma 0.11, but he doesn’t actually prove it, instead he cites Jech’s book Lemma 25.5. So I looked it up in Jech, and there it is! Somehow, we missed the fact that it was in Jech’s book all along! And Jech’s proof is essentially the same as what I give here.

  3. Very interesting, thanks for sharing this insight!

    I have some questions about the proof of theorem 2.

    (1) If you strengthen the assumption to “For each $ (V, \mathbb{B}) $-generic $ G $, there exists some $ (V, \mathbb{C}) $-generic $ H $ such that $ H \in V[G] $.”, then you have $ 1_\mathbb{B} \vDash \exists x (x \text{ is } (\check{V}, \check{\mathbb{C}})\text{-generic}) $. By the maximal principle, fix some name $ \dot{H} $ such that $ 1_\mathbb{B} \vDash (\dot{H} \text{ is } (\check{V}, \check{\mathbb{C}})\text{-generic}) $. Also, fix some $ (V, \mathbb{B}) $-generic $ G $, and let $ H := \dot{H}_G $. Then $ H $ is indeed $ (V, \mathbb{C}) $-generic, right? So, using your notation from above, you have $ b = 1_B $?

    (2) In the next paragraph, I don’t understand what you mean by “If $ c \in H $ forces that every $ d $ in the generic filter has that property, then …”. Could you explain how to find that condition $ c $?

    (3) As I didn’t understand how to find $ c $, I also don’t see why $ \pi(c) = b $ holds. In other words, why does $ b = [\![\check{c} \in \dot{H}]\!]^\mathbb{B} $ hold?

    Thanks in advance!

    • Good questions.

      Yes, I agree with (1). Except that ultimately the isomorphism is between $\mathbb{C}\upharpoonright c$ and $\mathbb{B}\upharpoonright \pi(c)$, rather than $\mathbb{B}\upharpoonright b$.

      For (2), let $D$ be the set of $d$ with $b\wedge[\![\check d\in\dot H]\!]\neq 0$. So $H\subset D$, and $D$ is in the ground model. Let $c\in H$ be a condition forcing that every condition in the general filter is in $D$. It follows that every $d\leq c$ has $d\in D$, since otherwise, $c$ couldn’t force that.

      For (3), we don’t necessarily have $\pi(c)=b$ in the proof. Rather, we have $\pi(c)\leq b$, which is clear, since we took a meet with $b$ in the definition of $\pi$. So the isomorphism is between $\mathbb{C}\upharpoonright c$ and $\mathbb{B}\upharpoonright \pi(c)$.

      • Thanks for your quick reply!

        So that $ b $ in Theorem 2 is $ \pi(c) $ and not the $ b $ used in the proof. I was a bit confused about this. 😉

        I wanted to apply your Theorem 2 (or some version of it) to the following problem:

        Suppose that, for each $ (V, \mathbb{B}) $-generic $ G $, there exists some $ (V, \mathbb{C}) $-generic $ H $ such that $ H \in V[G] $. Also assume that $ \mathbb{C} $ is isomorphic to each of its cones. Then $ \mathbb{C} $ is isomorphic to some complete subalgebra of $ \mathbb{B} $.

        In other words, can I get rid of the restriction to cones by the two strengthened assumptions?

        I think the above statement is widely applicable (if its provable after all) as the situation naturally occurs, for example, with amoeba tree forcing notions (and possible others).

        • Yes, that’s right about $\pi(c)$ and $b$; is it confusing? We may assume that $\pi(c)=b$ without loss of generality, though, simply by strengthening $b$ to $\pi(c)$.

          Your proposed theorem is correct. What you get is that a cone of $\mathbb{C}$ and hence $\mathbb{C}$ itself is isomorphic to a complete subalgebra of various cones $\mathbb{B}\upharpoonright b$ in $\mathbb{B}$. But there will be a dense set of these $b$. So take a maximal antichain $A$ of those $b$’s, and then map any element of $\mathbb{C}$ to the join of the corresponding copies of $\mathbb{C}$ below each $b\in A$. This will be a complete embedding of $\mathbb{C}$ into $\mathbb{B}$.

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