# The Boolean-valued model $V^B$ is extensional.

I am currently studying forcing from Jechs Set Theory and i have encountered this seemingly innocent lemma which says:$\\$

Lemma 14.17. $V^B$ is extensional.
Proof. Let $X,Y \in V^B$. By definition of $a \Rightarrow b$ we observe that if $a \le a'$, then $(a' \Rightarrow b) \le (a \Rightarrow b)$.
Thus for any $u \in V^B$ we have $(\|u \in X\| \Rightarrow \|u \in Y\|) \le (X(u) \Rightarrow \|u \in Y\|)$ and therefore:
$\hspace{2cm} \prod\limits_{u \in V^B}(\|u \in X\| \Rightarrow \|u \in Y\|) \le \prod\limits_{u \in V^B}(X(u) \Rightarrow \|u \in Y\|)$
While the left-hand side is equal to $\|\forall u(u \in X \rightarrow u \in Y)\|$, the right-hand side is easily seen to equal $\|X \subset Y\|$. Consequently,
$\hspace{2cm} \|\forall u(u \in X \leftrightarrow u \in Y)\| \le \|X = Y\|\hspace{7.7cm}\Box$

What confuses me here is the fact that $dom(X)$ can't be all of $V^B$ since there exists some $\alpha$ such that $X \in V^B_{\alpha + 1}$ and $dom(X) \subset V^B_\alpha$. So how do we evaluate $X(u)$ for $u \not\in dom(X)$? Do we substitute with $0$? How is the proof in the book valid?

• The same thing happens in the next lemma and we use $a_u(t)$ but in that lemma t can be outside of $dom(a_u)$ too. Commented Jul 19, 2018 at 14:45

Your guess is correct: $X(u)$ should be interpreted as $0$ if $u$ is not in the domain of $X$. This makes sense intuitively: if you think of $X(u)$ as representing a "truth value" to which $u$ is an element of $X$, then any $u$ not in the domain should be assigned a truth value of $0$.

With this correction, the proof then works, since $X(u)\Rightarrow \|u\in X\|=1$ when $u\not\in dom(X)$ and these terms do not affect the meet and so $$\prod\limits_{u \in V^B}(X(u) \Rightarrow \|u \in Y\|)=\prod\limits_{u \in dom(X)}(X(u) \Rightarrow \|u \in Y\|)=\|X\subset Y\|.$$

• But something isn't still ok, we work in ZFC then looking at X(u) from a truth value perspective then it becomes a class function and our reasoning fails in ZFC since we don't have classes. Commented Jul 19, 2018 at 17:42
• There is no difficulty whatsoever here. Just read $X(u)$ everywhere as an abbreviation for "if $u\in dom(X)$, the unique $v$ such that $(u,v)\in X$, and otherwise $0$". Commented Jul 19, 2018 at 17:45
• Now that i am thinking more about it i can't formalize this in first-order logic. Can you please write this out for me? (The expanded form of: $\prod\limits_{u \in V^B}(X(u) \Rightarrow \|u \in Y\|) = T$.) Commented Jul 19, 2018 at 18:04
• Well, writing it out fully in the language of set theory would be excruciatingly long. But, $\prod\limits_{u \in V^B}(X(u) \Rightarrow \|u \in Y\|) = T$ means that $T\in B$, $T\leq v\Rightarrow\|u\in Y\|$ for all $(u,v)\in X$, $T\leq 0\Rightarrow \|u\in Y\|$ for all $u\not\in dom(X)$, and $T\geq b$ for all other $b$ with these same properties. Commented Jul 19, 2018 at 18:09
• (It is not entirely trivial that such a $T$ must exist, but this follows from completeness of $B$, since the collection of elements of the form $X(u)\implies\|u\in Y\|$ is a subset of $B$, even if $u$ is allowed to range over a proper class.) Commented Jul 19, 2018 at 18:15

So how do we evaluate $X(u)$ for $u \not\in dom(X)$? Do we substitute with $0$?

In this proof (1) that's exactly what we do. It's a slight abuse of notation (which is very common in all of mathematics).

(1) I wouldn't be surprised if the same abuse of notation occurs in a situation where evaluating it as $0$ does not yield the desired result. Hence I don't claim generality.

• How would we proceed if we didn't want to use this notation? How would we write our proof formally? Commented Jul 19, 2018 at 17:43
• @ShervinSorouri You can add an extra step and prove that there is some set $x$ such that $\prod_{u \in V^B}(X(u) \Rightarrow \|u \in Y\|)= \prod_{u \in x}(X(u) \Rightarrow \|u \in Y\|)$ (this you can formalize without referring to classes by showing that the infimum can't get any lower (any counterexample to that would be a single element)). And then show that you can replace $V^B$ with $\mathrm{dom}(X)$ on the right hand side and verify that the inequality still holds true. Commented Jul 19, 2018 at 17:48
• I understand what you're trying to say but the moment you write $\prod\limits_{V^B}(X(u) ....$ you are giving $X$ parameters that are out of its domain. Commented Jul 19, 2018 at 17:52
• I think what Eric suggested should do the trick. Thanks! Commented Jul 19, 2018 at 17:54
• @ShervinSorouri Sorry, I actually copied the wrong part of the equation. What I meant to write was: $\prod\limits_{u \in V^B}(\|u \in X\| \Rightarrow \|u \in Y\|) = \prod\limits_{u \in x}(\|u \in X\| \Rightarrow \|u \in Y\|)$ for some set $x$ (of size $\le B$). Commented Jul 19, 2018 at 17:58

$X$ is a partial function, since it may not defined for every argument from $V^B$ and this is not explained well in Jech's book. For a partial function, all notions involving it are contingent on the function being defined. For example, for partial functions $f$ and $g$ we have that $f = g$ if for all $x$ we have that whenever $f(x)$ is defined and $f(x) = y$, then $g(x)$ is defined and $g(x) = y$ and vice versa.

• Can you provide a reference for me to study please? Commented Jul 19, 2018 at 17:44
• You shouldn't say that $X$ is a partial function defined on all of $V$ or $V^{B}$ (since this leads to other issues -- like quantifying over all $B$-names). Rather, you should say that $X$ will be regarded as a partial function whenever we talk about its values on elements not in the domain. Commented Jul 19, 2018 at 17:52
• Indeed; I will update my answer to reflect this. Commented Jul 19, 2018 at 17:55
• See Chapter 2 of J. Gallier's Discrethe Mathematics (Springer Universitet, 2011) for more on partial functions. Commented Jul 20, 2018 at 21:04