# Canonical name for the Boolean algebra in a Boolean-valued model

Following Jech's Set Theory, fix a complete Boolean algebra $B$ and form the Boolean-valued model $V^B$ of ZFC. We then have the name $\check{B}\in V^B$. Apparently, it is the case that $$\|\check{B}\text{ is a Boolean algebra}\|=1$$ How come? Certainly we can define $\check{p}\wedge\check{q}=(p\wedge q)\check{\phantom{p}}$ for $p,q\in B$ etc., but is this enough? Or do we have to specify $B$-names for functions, which will represent the Boolean operations on $\check{B}$?

This question is motivated by the later discussion of the name of a generic object $\dot{G}$, defined by $\text{dom}(G)=\{\check{p};p\in B\}$ and $\dot{G}(\check{p})=p$. Various sources state that $$\|\dot{G} \text{ is a generic ultrafilter on }\check{B}\|=1$$ but for this to hold, of course, $\check{B}$ has to be a Boolean algebra (in $V^B$), and this is what confuses me. The (outline of a) proof I've seen of this, seems only to consider canonical names of elements of $B$ when proving, for example, upward closedness, and I'm not sure why this should work.

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Indeed it is the case that $\|\check B\text{ is a Boolean algebra}\|=1$. This is because one can prove that for any $\Delta_0$ formula $\varphi$, we have $V\models\varphi[a]$ if and only if $\|\varphi(\check a)\|=1$ in $V^B$. This can be proved by induction on the complexity of the formula. It now follows since $B$ really is a Boolean algebra, and this is expressible by a $\Delta_0$ formula, that $\|\check B\text{ is a Boolean algebra}\|=1$. Note that it is not generally true that $||\check B\text{ is a complete Boolean algebra}||=1$, even when $B$ is a complete Boolean algebra in $V$, since the completeness of forcing with a nontrivial $B$ is lost in the forcing extension.

Nevertheless, it is true that $\|\check B\text{ is a complete Boolean algebra in }\check V\ \|=1$. More generally, without the restriction to $\Delta_0$ formulas, one can prove by induction on formulas that $V\models\varphi[a]$ if and only if $\|\varphi^{\check V}(\check a)\|=1$, where the formula is relatized to the class $\check V$, defined so that $\|\tau\in\check V\|=\bigvee_{x\in V}\|\tau=\check x\|$.

This leads to a quibble with your last displayed equation, where you should say instead that $$\|\dot G\text{ is }\check V\text{-generic}\|=1,$$ since of course $\dot G$, being an element of $V^B$, will not meet all the dense sets in $V^B$.

This is all discussed in detail in the lecture notes for my recent tutorial on the Boolean ultrapower for the Young Set Theory Workshop, a tutorial that was based on an extensive article on the topic, which I am preparing with Dan Seabold.

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Is this also going to be a part of that book that you are hoping to finish someday? :-) – Asaf Karagila Dec 28 '11 at 16:43
Well, it would probably appear in the appendix, since it is fundamental to my way of thinking about forcing. – JDH Dec 28 '11 at 16:50
Well, I'll keep on waiting for you to have the time to finish writing. – Asaf Karagila Dec 28 '11 at 21:41
I think I understand. My problem was in understanding how to interpret the Boolean structure on $\check{B}$. Let me see if I have this right: by using $\Delta_0$-absoluteness, we get a names $\check{B}$ and $\check{\leq}$. These give us a Boolean algebra in $V^B$, on which $\dot{G}$ is an ultrafilter. – Miha Habič Dec 28 '11 at 22:28
I've checked out your (very fine) notes. Since you mentioned my mistake about $\check{V}$-genericity, would you mind adding a couple of lines explaining why, as on page 6, we consider only dense sets which are check names when proving $\check{V}$-genericity? As you state elsewhere, we could have $\|D\in\check{V}\|=1$ without $D$ being a check name. I don't think we can use the relativization result, since there is an errant $\dot{G}$ in the formula. – Miha Habič Dec 28 '11 at 22:37