I'm reading a short section about internal set theory(see here), in which
$x$ is limited in case for some standard $r$ we have $|x| ≤ r$.
while the predicate “standard” is not defined. I'm interpreting an element $x \in A$ is standard, if it dosn't belong to $^*A \setminus A$.
One exercise in it asks:
Can one prove that every standard positive real number is limited?
I think not. Because either $\{x \in {\bf{R}} : x >0 \land x \text{ is standard} \}$ or $\{x \in {\bf{R}} : x \text{ is limited} \}$ are "illegal set formations" in the lauguage of set theory. so there's no way to tell whether an element in one set necessarily belongs to the other.
On the other hand, I'm doubtful about such reasoning, since if $x$ is a standard positive real number, so is $x+1$. Because $x < x+1$, we have every standard positive real number must be limited.
What's wrong?
There is a difference between saying that the set of limited numbers exists, and that the set of standard numbers exists, and that every standard number is limited.
To see the full meaning of this, let me give you a close example. In $\Bbb R$ we cannot define $\Bbb Z$, but we can define each of the integers, and therefore we can define every natural number if we wish to define it. Still we cannot define $\Bbb N$ itself.
Similarly here, we cannot define the set of standard reals, nor the set of limited reals. But we can prove that every standard real is limited. Simply by arguing that if $r$ is standard then $|r|$ is also standard, and therefore $|r|\leq|r|$ as wanted.