In the book Axiomatic Set Theory (Takeuti, G; Zaring, W.M. - 1973) the theorem 6.4 states that if $\varphi$ is a closed formula of a given language then it is satisfied in every boolean valued structure iff is logically valid. One side is trivial, given that $\varphi$ is logically valid iff it is satisfied in every $\{0,1\}$ structure and $\{0,1\}$ is a complete boolean algebra. What I don't understand is the converse side of the proof, he uses the following result:
(Rasiowa-Sikorski) If B is a Boolean algebra, if $a_0\in \textbf{B}\setminus\{0\}$ and $b_n = \sup A_n \in \textbf{B}$, $A_n\subset \textbf{B}$ and $n\in \omega$, then there exists a Boolean homomorphism $f: \textbf{B}\rightarrow \{0,1\}$ such that $f(a_0)=1$ and $f(b_n)=\sup f(A_n)$.
In the following way: if $\varphi$ is not satisfied by some $\textbf{B}$-structure $A$ i.e., $\| \varphi \|_A=b\neq 1 $ then the computation of $\| \varphi \|_A$ requires only a finite number of application of the definition of $\|\cdot \|_A$, say: $b_1=\inf A_1,...,b_n=\inf A_n$ and $b'_1=\sup B_1,\dots,b'_n=\sup B_n$ and now he jumps to the conclusion and I can't get why he can do it (maybe it's trivial, but I can't see it): By the Rasiowa-Sikorski Theorem, there is a homomorphism $f:\textbf{B}\rightarrow \{0,1\}$ such that $f(b)=0$. The final result follows easily from that, my problem is with that passage.
Hint
The proof assume that $b \ne 1$ and we can consider the case that $b=b'_n=\sup B_n$.
By the R-S Th there is an homomorphism $h: \textbf{B}\rightarrow \textbf{2}$ such that: $h(b'_n)=\sup h(B_n)$.
But - see proof of R-S Th, page 29 - an homomorphism between boolean algebras maps $0$ onto $0$ and $1$ onto $1$. This means that, if $b'_n=b \ne 1$, also $h(b)\ne 1$.
But $h(b) \in \textbf{2}$ and thus, if it is not $1$, it must be $0$.