$\DeclareMathOperator\Card{Card}$The book *Forcing Eine Einführung in die Mathematik der Unabhängigkeitsbeweise* by Hoffmann provides an intuition behind boolean valued models of set theory which I will explain below. But when I try to make use of the intuition, as I understand it, I run into problems.


**On pages 272-275 the following intuition is provided:**  

Consider the boolean algebra $B:=\mathcal{P}(\lbrace 1,2,3 \rbrace)$.  

The single elements $1$, $2$ and $3$ can be thought of as three possible "worlds".  

When considering a boolean set, i.e. an element of $V^{(B)}$ like $x \mathrel{:=} \lbrace(\emptyset,\lbrace2,3\rbrace),(\lbrace(\emptyset,\lbrace1,3\rbrace)\rbrace,\lbrace1,3\rbrace)\rbrace$ it can be written as $x = \begin{cases}
 1: \lbrace\lbrace\emptyset\rbrace\rbrace\\
 2: \lbrace\emptyset\rbrace\\
 3: \lbrace\emptyset,\lbrace\emptyset\rbrace\rbrace
 \end{cases}$
showing which "normal" set hides behind x according to each world.  

Also the boolean truth value $\lVert\varphi\rVert \in B$ of a sentence $\varphi$ in set-language (including constants for each boolean set) can now be seen as the set of worlds in which $\varphi$ holds. The following example is provided:  

Along with $x$, consider the boolean set $y \mathrel{:=} \lbrace(\emptyset,\lbrace1,3\rbrace),(\lbrace(\emptyset,\lbrace1,2,3\rbrace)\rbrace,\lbrace1,2,3\rbrace)\rbrace = \begin{cases}
 1: \lbrace\emptyset,\lbrace\emptyset\rbrace\rbrace\\
 2: \lbrace\lbrace\emptyset\rbrace\rbrace\\
 3: \lbrace\emptyset,\lbrace\emptyset\rbrace\rbrace.
 \end{cases}$  

Then from the intuition behind $\Vert \cdot \Vert$, it immediately follows:  

$\lVert x \in y\rVert = \lbrace2\rbrace$,  

$\lVert x \subseteq y\rVert = \lbrace1,3\rbrace$,  

$\lVert x = y\rVert = \lbrace3\rbrace$.


**From this, I understood the given intuition as follows:**  

For a given boolean algebra $B$ we consider the set $W\_B$ of worlds.  

This should be possible by the representation theorem of Stone.  

(And each of these worlds can be seen as a "copy of the universe $V$".)  

A boolean set $x \in V^{(B)}$ then induces a map $x': W\_B \rightarrow V$ mapping each world to the set that "hides behind $x$".  

(As different boolean sets can induce the same map, I doubt that the use of the equality symbol between a boolean set and the respective case distinction, as done in the book, is fine.) For an arbitrary sentence $\varphi$ we would get (according to the intuition) $\lVert\varphi\rVert = \lbrace w \in W\_B \mid \text{"$V \models \varphi\_w$"} \rbrace$, where $\varphi\_w$ is obtained by replacing the constants in $\varphi$ with the appropriate set according to world $w$ (e.g. $x \in V^{(B)}$ as constant-symbol in $\varphi$ is replaced by $x'(w)$).


Analogously, the same can be done when considering the boolean set universe $\mathcal{M}^{(B)}$
of a countable, transitive standard model $\mathcal{M} \models ZFC$, where $B \in \mathcal{M}$ is a boolean algebra that is complete in $\mathcal{M}$. Now worlds can be seen as "copies of $\mathcal{M}$" and $x \in \mathcal{M}^{(B)}$ induces a map $x': W\_B \rightarrow \mathcal{M}$. As above, we get $\lVert\varphi\rVert^{\mathcal{M}^{(B)}} = \lbrace w \in W\_B \mid \mathcal{M} \models \varphi\_w \rbrace$.


**The problem:**  

I am well aware that my understanding of the intuition given in the book is lacking of something. This can be seen when considering results like (page 367):
If B satisfies CCC in $\mathcal{M}$, then $\forall x \in \mathcal{M}: \mathcal{M} \models \Card(x) \Leftrightarrow \lVert \Card(\check{x}) \rVert^{\mathcal{M}^{(B)}} = 1\_B$. We would get
the conclusion directly from our intuition (without needing $B$ to satisfy CCC in $\mathcal{M}$) as the standard representative $\check{x}$ induces a constant map and $1\_B$ is the set of all worlds.


As the book does not go into much further detail concerning this intuition, I would be very glad if somebody could help me out. What has to be changed in order for everything to be sound again? Is the intuition only valid for very particular sentences, perhaps only atomic sentences? Is my understanding of 'world' too simple? Are particular boolean algebras necessary? Is the intuition presented in the book limited to very special cases? (This is the first time I have encountered this intuition.)


I'm a set-theory beginner and appreciate any help — even if it is obvious.



Intuition behind Boolean-valued models of set theory
Due to the insightful comments made by @AlexKruckman and @AndreasBlass, we can give an answer to the question.


Let $\mathcal{M}$ be a countable, transitive standard model of ZFC.  

Further, let $B \in \mathcal{M}$ be a boolean algebra that is complete in $\mathcal{M}$.  

Then we can understand the intuition given in the book as follows:  

A "world" is an ultrafilter $U \subset B$. (Note that $U$ induces the model $\mathcal{M}^{(B)}/U$.)  

Therefore, in the example given in the book, the "worlds" $1,2$ and $3$ actually stand for their corresponding principal ultrafilters $U\_1, U\_2$ and $U\_3$. And the case distinction just shows that e.g. $x$ names the set $[x]\_{U\_1} \in \mathcal{M}^{(B)}/U\_1$ with respect to the first "world" etc.


In order to make sense of the statement "the value $\Vert\varphi\Vert \in B$ can be seen as the set of worlds in which $\varphi$ holds.", we recall that by the representation theorem of Stone we have that $B$ is isomorphic to $\sigma(B)$.
The map $\sigma: B \rightarrow \mathcal{P}(S)$ is defined via $\sigma(b) := \lbrace U \in S \mid b \in U \rbrace$, where $S$ is the set of all ultrafilters on $B$. With intersection and union the set $\sigma(B)$ forms a boolean algebra.  

Now we can translate the intuition (here $\Vert\cdot\Vert$ is defined with respect to $\mathcal{M}^{(B)}$): $\lbrace \text{worlds in which } \varphi \text{ holds} \rbrace = \lbrace U \in S \mid \mathcal{M}^{(B)}/U \models \varphi\rbrace = \lbrace U \in S \mid \Vert\varphi\Vert \in U \rbrace = \sigma(\Vert\varphi\Vert) \cong \Vert\varphi\Vert.$


As a side note: This intuition lives 'outside' of $\mathcal{M}$, because whenever we consider ultrafilters of $B$ or when we construct $\sigma(B)$ we do that 'outside' of $\mathcal{M}$. The reason is that the 'interesting' ultrafilters are those which are not members of $\mathcal{M}$.


Feel free to correct me if anything is wrong.


