# 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.

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.