I was trying to understand better boolean valued models for forcing and I have the following question : Why is the this formula an inequality and not an equality ?
$$(u(x)\leq\parallel x \in u \parallel^B )$$
For instance if i have the name $$p_0= <\emptyset, 0>$$ and $$u=<{p_0,0}>$$
then i would get $$u(p_0) =0 $$ and $$ \parallel p_0 \in u \parallel^B=0$$ right ? could u give me an example where $$u(x)< \parallel u \in x \parallel^B $$?? Furthermore which property of first order logic does this inequality reflect ?
It is in fact possible that $$ u(x) <_B \parallel x \in u \parallel_B. $$
Consider for example a $B$-name $u$ with $\mathrm{dom}(u) = \{x,y\}$ such that $$\parallel x = y \parallel_B \wedge u(x) = 0 \text{ and } \parallel x = y \parallel_B \vee u(x) = 1$$ and $u(y) = 1$. Then $$ \begin{align*} \parallel x \in u \parallel_B & = ( \parallel x = x \parallel_B \wedge u(x)) \vee ( \parallel x = y \parallel_B \wedge u(y)) \\ &= u(x) \vee \parallel x = y \parallel_B \\ &= 1 \end{align*} $$ but $u(x) <_B 1$.
Roughly speaking: If $g$ is a generic filter such that $u(x) \in g$ then, in $V[g]$, we have that $x^g \in u^g$. However, it is possible to have a generic filter $h$ such that $u(x) \not \in h$, yet we still have $x^h \in u^h$.