# Quickie on Boolean valued models

Bell writes on page 21 (you may use the search in the preview to search for "21" to view the page):

"..., we show that, for any complete Boolean algebra $B$, all the theorems of $ZFC$ are true in $V^{(B)}$. ..."

On Wikipedia it says: "For any poset $P$ there is a complete Boolean algebra $B$ and a map $e$ from $P$ to $B^+$ (the non-zero elements of $B$) such that the image is dense, $e(p)\le e(q)$ whenever $p \le q$, and $e(p)e(q)=0$ whenever $p$ and $q$ are incompatible. This Boolean algebra is unique up to isomorphism. It can be constructed as the algebra of regular open sets in the topological space of $P$ (with underlying set $P$, and a base given by the sets $U_p$ of elements $q$ with $q\le p$)."

Question 1: Is it the case that if we extend $V$ then we can use any Boolean algebra but if we use a model $M \subset V$ then we have to use the unique Boolean algebra as described on Wikipedia?

Question 2: Assume I have a model $M$ of some theory, not necessarily $ZF$, and I want to extend it. Then I first want to construct a Boolean valued model. To this end, I first want to construct a suitable Boolean algebra satisfying the requirements mentioned on Wikipedia. (Is this correct so far?) Then what do regular open sets look like in the topology generated by $U_p$? (A set is called regular open if $U = \mathring{\overline{U}}$)

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A boolean algebra is in particular a poset, so you can embed it in a complete boolean algebra as described. (Alternatively, you could take its completion, using Stone duality.) You can construct boolean-valued models of any first-order theory, but there are extra complications when working with set theory in particular. – Zhen Lin Dec 27 '12 at 10:47
@ZhenLin Thank you for the comment! – Rudy the Reindeer Dec 27 '12 at 10:59

Q1. You can use any complete Boolean algebra in your ground model. The construction on Wikipedia gives a method of embedding an arbitrary poset $P$ in a complete Boolean algebra $B$ via a map $e : P \hookrightarrow B$.

Q2. Don't worry too much about what regular open sets look like. The open sets of a poset are generated by down-sets $\downarrow(p) = \{ q \in P\, :\, q \le p \}$; that is, the open sets are unions of the down-sets. If $P$ were a total order then all the open sets would be either down-sets, $P$ or $\varnothing$, and they'd all be regular. If $P$ is a more general poset then it's a bit more complicated... try drawing a picture.

If you want a more general picture of what a regular open set looks like, just think: they're those open sets which don't change when you close and open them. Equivalently, you lose no information by taking the closure. It's good to consider $\mathbb{R}$: for instance, $(0,1) \cup (1,2)$ isn't regular open since its closure is $[0,2]$ (so the interior of its closure is $(0,2)$). In this case you lose the information that $1$ doesn't lie in the set. On the other hand, $(0,2)$ is regular open.

I recently wrote this introduction to Boolean-valued models and forcing that you might be interested in reading.

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Just to check that I understand what you're saying: when making a Boolean valued model (BVM) we take a poset $P$ and embed it into a complete Boolean algebra $B$ and then use this particular Boolean algebra to make the BVM because then we know what properties the BVM has (because verifying properties of BVMs is a pain in the neck but easy to do for posets and Boolean algebras containing posets?)? – Rudy the Reindeer Dec 27 '12 at 11:30