Vacuous truths in Superstructure approach to Nonstandard Analysis Good evening everybody,
at the moment I'm studying non-standard analysis, specifically the superstructure approach to it. This approximately works as described in chapter 3 of http://people.dm.unipi.it/dinasso/papers/20.pdf. So far I've mostly understood this construction, and also worked through the detailed construction in chapter 4.4 of Chang & Keisler's Model Theory. But there is one unmentioned detail that has been bugging me, concerning the interpretation of formulas in $V(X)$, the superstructure over a base set $X$.
One of the central points of nonstandard analysis is the transfer principle, which states that for a bounded quantifier formula $\varphi (x_1,...,x_n)$, if $a_1,...,a_n \in V(X)$, then $$(V(X), \in) \models \varphi[a_1,...,a_n] \iff (V(^* \! X), \in) \models \varphi[^*a_1,...,{} ^*a_n].$$
Since we only interpret these formulas in $V(X)$, this means that when considering a bounded quantifier, say the quantifier in $\forall w \in u (w \in v)$ (or, phrased differently, $u \subseteq v$), the quantifier ranges only over elements in $u \cap V(X)$. In most cases, this will not be a problem, since $V(X)$ is transitive, so if $u \in V(X) \setminus X$, then $u \cap V(X) = u$. 
But if we consider any element $x$ of the base set $X$ (so $x \cap V(X) = \emptyset$) and an arbitrary $y \in V(X)$, then $(V(X), \in)$ 'believes' $\forall w \in x (w \in y)$ vacuously, although this usually isn't true in our normal universe. I think this is suboptimal, since the goal should be proving valid formulas (in our usual set theoretic setting) by proving them inside $V(X)$ and $V(^* \! X)$, using the transfer principle and other tools.
I first tried the naive fix of just allowing assignments $a_1,...,a_n \in V(X) \setminus X$, but this does not handle nested quantifications, for example in the statement  $\forall x \in X(\forall z \in x(z \in y))$, where $y \in V(X)$ is arbitrary and $X$ is our base set. Also, apart from when they are the range of a quantifier, we need to be able to assign individuals (relative to $V(X)$) to free variables.
It seems like we need to avoid any quantification over elements of rank $0$ to be able to infer truth of $\varphi$ (in our set theoretic universe) from truth of $\varphi$ in $(V(X), \in)$. So, my questions are as follows:
$1.$ First, and foremost: Am I missing something? Is this an actual problem or just a misunderstanding?
$2.$ If not, why wasn't this mentioned in the texts I considered? Also, is there a simpler characterization of the formulas that will be 'misinterpreted' by $(V(X), \in)$?
Thank you for your time.
 A: This has a lot to do with language. $V(X)$ is an $\mathcal{L}$-structure of language $\mathcal{L}=\{=,\in,V(X)\}$ where $"="$ and $"\in"$ are predicates and $V(X)$ is a set of constants. 
For $V(X)$ to be an $\mathcal{L}$-structure you need an interpretation of your language. So for every constant $x\in V(X)$ we need to associate an element $y\in V(X)$, clearly we can pick $y=x$. For our predicates $"\in"$ and $"="$ we have to a find subsets $M_{\in},M_{=}\subset V(X)\times V(X)$. We then say for $"a\in b"$ or $"c=d"$ are true if and only if $(a,b)\in M_{\in}$ and $(c,d)\in M_{=}$ respectively. Again you can pick the obvious subsets $M_{=}=\{(x,x)\in V(X)\times V(X)$ and $M_{\in}=\{(x,y)\in V(X)\times V(X):x\in y\}$. These symbols along with logical symbols as $\rightarrow,\wedge,\neg,\vee,\forall$, etc are the only symbols we may use to write statements. 
The statement $(\forall x\in X)(\forall z\in x)(z\in y)$ where $y$ is some constant would be rewritten as
$$(\forall x)(x\in X\rightarrow (\forall z)(z\in x\rightarrow z\in y))$$
where $y$ and $X$ are constants. 
Now the question remains, is this formula true. This does depend on your interpretation of the $"\in"$ predicate. But if you interpret it in such a way that $(x,y)\not\in M_{\in}$ for all $y\in X$, then this is a true statement. $z\in x$ is not a true for $x\in x$, so $(\forall z)(z\in x\rightarrow z\in y)$ is true by definition of $\rightarrow$ This then makes $(\forall x)(x\in X\rightarrow (\forall z)(z\in x\rightarrow z\in y))$ true again due to the definition of $\rightarrow$.
Basically this statement is true in the same way that the statement: "every element of the empty set is a vampire" is true. You will never be able to avoid these kind of silly true formula's, they are just a consequence of writing a rigorous logical language, but they are true in both $V(X)$ and set theory.
