In Davis' Applied nonstandard analysis a proof of the following, often seen, proposition is presented:
For a sequence $S_n$
$S_n \rightarrow L$ iff $S_n \approxeq L$ for all infinite n.
He then states that 'let us choose some $\epsilon \in R^+$, corresponding to this there exist some $n_0$' and then he gives the following formula for defining the limit in the usual way: $(\forall n\in N)(n>N\implies |s_n - L|<\epsilon)$.
He then say that using the transfer theorem one gets that for any $n\in *N$ for which we have $n>n_0$ we have that $|s_n - L|<\epsilon$.
Then he says that since $n_0$ is finite this inequality holds forall infinite *N. And tells us to note that epsilon was any real positive number so we can conclude $S_n \approxeq L$ for any integer $n$.
Going the other way he tells us to let $S_n \approxeq L$ and again choose $\epsilon \in R^+$ then proceeds to reconstruct the classical definition written in $*R$.
My question is why in applying the transfer theorem hasn't $\epsilon$ become an element of $*R^+$ and similairly why is $n_0$ finite? Surely after the application of the transfer theorem one has $n_0$ as an element of *N so there is no reason to assume it's finite.
I just can't see why the reasoning isn't revolving around manipulating the elements of *R, and why standard elements like $\epsilon$ have escaped being transferred.
Furthermore I note when he goes the other way he constructs the statement $(\exists n_0\in *N)(\forall n\in N)(n>N\implies |s_n - L|<\epsilon)$ so the $(\exists n_0\in *N)$ term has now appeared this time as an element of the hypernaturals but the reference to $\epsilon \in R^+$ remains.
I have found I've similar issues in other proofs so I suspect I have misunderstood some crucial point. Any help would be greatly appreciated.