# Internal Set Theory: extensionality (set equality)

In Internal Set Theory we can extend extensionality (set equivalence) via transfer. I'd like to work this out explicitly and then ask my question.

Extensionality: $$\forall z(z\in x\iff z\in y)\implies x=y$$

Transfer: Let $$A(z,t_1,...,t_k)$$ be an internal formula with free variables $$z,t_1,...,t_k$$ then,

$$\forall^s t_1 ... \forall^s t_k (\forall^s z A\implies\forall z A)$$

So in this case my $$t$$'s are $$x$$ and $$y$$ in the above formula. Let's do an example. Let $$\nu$$ be a nonstandard natural number and the notation $$^SA$$ is a standard set $$\{a\in A | a\mbox{ is standard}\}$$. Then $$^S[0,\nu]$$ and $$\mathbb{N}$$ contain the same standard elements, the standard natural numbers. So for any standard $$n\in\phantom{ }^S[0,\nu]$$ we are guaranteed $$n\in\mathbb{N}$$ and vice versa.

Here is my question. Am I then interpreting the implication in the Tranfer axiom correctly when I deduce, that for any element $$x\in\phantom{ }^S[0,\nu]$$ it is the case that $$x\in\mathbb{N}$$ and vice versa?

• I want to provide an example from Edward Nelson's unfinished book on Internal set theory on this topic. Let $X=^S\{z\in\mathbb{R} | z\simeq 0\}$. Then $X=\{0\}$ since 0 is the only standard infinitesimal. in that case no nonstandard elements made it. Then consider $Y=^S\{z\in\mathbb{R} | z \mbox{ is limited}\}$ Then $Y=\mathbb{R}$ because all standard reals are limited. "Thus we an have $z\in Y$ without $z$ being limited." So in that second case Nelson seems to be saying that the sets Y and R are equal because, as standard sets, they have the same standard elements and THEN all the rest – kevin roberge Apr 15 at 2:15
• users: I have a question. I find relatively little on this site about Internal Set Theory (little anywhere really) and some of those limited resources are at a fairly advanced level (as opposed to the wealth of introductions to Robinson's nonstandard analysis). Is there a better way for me to ask 'elementary' questions about IST in a way that creates a resource for future beginners to refer to? – kevin roberge Apr 16 at 15:44

So the set $$^S[0,\nu]$$ I reference above, really is the same as $$\mathbb{N}$$. Likewise for subset. For two standard sets A and B, $$A\subset B$$ as soon as for all standard $$x\in A$$ it is the case that $$x\in B$$.