I'm not going to try and quote to what I'm responding -- I don't think it would be helpful.
I confess that I made up the word "hyperarchmedian" -- I'm going along with the convention of prefixing the *-transfer of various concepts with the word "hyper". (e.g. hyperreal, hypernatural, hyperfinite)
One statement of Dedekind completeness is given by:
<br />
\forall S, T \in P(\mathbb{R}):<br />
(S \neq \emptyset \wedge T \neq \emptyset \wedge S \cup T = \mathbb{R} \wedge S \cap T = \emptyset \wedge<br />
\forall s \in S: \forall t \in T: s < t)<br />
\implies<br />
\exists r \in \mathbb{R}:<br />
\forall s \in S: \forall t \in T: s \leq r \leq t<br />
This is a true statement in standard analysis. Therefore, by the transfer principle, the *-transfer of this statement must also be true. In other words:
<br />
\forall S, T \in P({}^\star \mathbb{R}):<br />
(S \neq \emptyset \wedge T \neq \emptyset \wedge S \cup T = {}^\star \mathbb{R} \wedge S \cap T = \emptyset \wedge<br />
\forall s \in S: \forall t \in T: s < t)<br />
\implies<br />
\exists r \in {}^\star \mathbb{R}:<br />
\forall s \in S: \forall t \in T: s \leq r \leq t<br />
So, internally speaking, the hyperreals
are Dedekind complete.
The catch is that
P here is the
internal power-set function.
P(R) contains only the
internal subsets of
R. Of course, for the standard model, every subset of
R is internal.
But in the nonstandard model,
P(*R) is "missing" some subsets. For example \mathbb{N} \notin P({}^\star \mathbb{R}), because
N isn't an internal subset of
*R.
(Incidentally, this whole issue about the internal power-set operation was one of my biggest stumbling blocks in understanding this stuff!)
(I actually have a gripe about the usual treatment of second-order logic that relates to this context -- but that's for another thread!)
Incidentally, this fact is useful to prove certain sets are not internal. For example, I can prove that the set of finite hyperreals is not an internal set, because I can use it to construct a Dedekind cut around one of the gaps!
When I'm working in
Q, I can still use all the normal definitions of things like continuity, infinite sums, et cetera. Of course, in
Q, things like:
<br />
\sum_{n = 0}^{+\infty} \frac{1}{n!}<br />
do not converge, despite the fact the sequence of partial sums is Cauchy.
But, it can still be shown (using the exact same proof as one would use in the reals) that the sum
<br />
\sum_{n = 1}^{+\infty} 9 \cdot 10^{-n}<br />
converges to 1.
A sequence, incidentally is nothing more than a function written with different notation. The domain of this function is called the "index" set.
In general, if we do not say otherwise, we assume that all infinite sequences are indexed by
N. (and finite sequences indexed by something of the form {a, a+1, a+2, ..., b-1, b}, usually with a=0 or a=1)
But in the nonstandard case, it becomes more appropriate to index sequences by *
N, and
not by
N. In particular, this means that when we speak about a hyperdecimal, its digits should be indexed by *
N and
not by
N.
You are correct in observing that the
N-indexed sequence {1 / (n+1)} does not converge in the hyperreals. However, the *
N-indexed sequence {1 / (n+1)} does converge to zero, which can be shown either through the transfer principle, or directly with an epsilon-N argument.
I would also like to point out that st()
does not necessarily exist in a nonstandard model of
R! It is another example of something external -- a function we can define set-theoretically, but have absolutely no reason to think it is a part of the model we're studying.
(Actually, models of
R typically have no functions whatsoever -- what I mean to say is "model of real analysis", but that's too wordy.
)
)isn't enough to quench all doubt, perhaps a question:
I can't believe I said R*. I meant *R. Me is embarrassed. 
