Externality of a set

Question

In nonstandard analysis (for example in $^*ZFC$ developed there), how can one easily see that $^\sigma X$ is external if $X$ is a standard infinite set ?

Answer 1

$\omega$ is external because otherwise $^*\omega\setminus\omega$ would be internal and would have a least element.

Let $f:A\to\omega$ be a map which is onto. Then $^*f:^*A\to^*\omega$ is also onto. Let us suppose that $^\sigma A$ is internal, then $^*f[^\sigma A]$ is also internal.

$^*f[^\sigma A]=\lbrace ^*f(^*a)\mid a\in A\rbrace=\lbrace ^*(f(a))\mid a\in A\}=\{f(a)\mid a\in A\rbrace=\omega$

But $\omega$ is external and this is a contradiction.

Answer 2

In Nelson's IST this is a fairly immediate consequence of the axiom of idealisation. I am copying the comment from wiki: If $S$ is infinite, then we take for the relation $R(g,f)$: $g$ and $f$ are not equal and $g$ is in $S$. Since "For every standard finite set F there is an element g in S such that g ≠ f for all f in F" (the infinite set $S$ is not a subset of the finite set $F$), we may use Idealisation to derive "There is a G in S such that G ≠ f for all standard f." In other words, every infinite set contains a non-standard element (many, in fact).

In Robinson's framework, the general proof that an internal subset of $\mathbb R$ is necessarily finite is somewhat involved and occupies nontrivial space in Goldblatt's book. However, the fact that the natural extension of any infinite set necessarily contains nonstandard elements is simpler.

If $X\subseteq\mathbb R$ is an infinite subset then there is a strictly monotone sequence $u=\langle u_n\colon n\in\mathbb N\rangle$ of elements of $X$. The equivalence class $[u]\in{}^\ast\mathbb R$ is then infinitely close to but not equal to the real number $\lim_{n\to\infty} u_n\in\mathbb R$. Thus $[u]\in{}^\ast\!X\setminus\mathbb R$.

If $X$ is a member of a fixed level in the superstructure the proof is similar.