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?

  • $\begingroup$ 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 $\endgroup$ – kevin roberge Apr 15 '19 at 2:15
  • $\begingroup$ 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? $\endgroup$ – kevin roberge Apr 16 '19 at 15:44

I'm not sure why it took me so long to fully grasp this idea. But yes, transferred extensionality implies that once two standard sets share the same standard elements they must share the same elements, standard or not.

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 $x\in B$.

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