Is it true that $\Bbb Z^*\setminus\Bbb Z$ has no finite elements?

Question

If we consider the hyperreals, we know that there exist non-zero infinitesimals so $\mathbb R^*\setminus\mathbb R$ has finite elements. However, it seems like that is not true for $\mathbb Z^*\setminus\mathbb Z$ -i.e. it doesn't have any finite elements. Is this correct? Is there a formal proof for that?

Answer 1

$\forall n\in\Bbb Z(n\ge 0\to n\ge 1)$ transfers as $\forall n\in\Bbb Z^*(n\ge 0\to n\ge 1)$, so there’s nothing in $\Bbb Z^*$ between $0$ and $1$.

This is very clear if you think in terms of ultrapowers. Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$, and let $\Bbb Z^*$ be $\prod_{\mathscr{U}}\Bbb Z$. If

$$[\langle 0,0,0,\ldots\rangle]<[\langle n_k:k\in\Bbb N\rangle]<[\langle 1,1,1,\ldots\rangle]\;,$$

then

$$\varnothing=\{k\in\Bbb N:0<n_k<1\}=\{k\in\Bbb N:0<n_k\}\cap\{k\in\Bbb N:n_k<1\}\in\mathscr{U}\;,$$

which is absurd.

Added: More generally, if $n\in\Bbb Z^+$, the interval $[0,n]$ is the same in $\Bbb Z^*$ as it is in $\Bbb Z$. Thus, any positive member of $\Bbb Z^*$ that is not in $\Bbb Z$ must be infinite.