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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.