Is it possible to have numbers that are to Hyperreal numbers what Hyperreals are to Reals numbers?

There are Hyperreal numbers that are smaller than any real number , also those that are larger than any real, they have properties analogous to those of Real numbers thanks to the Transfer principle and so on.

Then, It seems I could think of another kind of numbers that can be smaller or larger than any Hyperreal, and another kind that can be smaller or larger than any number of the former kind and one more ad infinitum. Is it correct or is it not? If not, how it'spossible to prove that ?

I recently took science of the existence of Hyperreal numbers after started studying calculus using Jerome Keisler's book, which is available here for free

• perhaps beyond $dx$ there is $dxdx$? And then $dxdxdx$ and... – James S. Cook Feb 1 '14 at 3:59
• Does this really fall under number theory (as tagged)? I'm not sure, but I would have went with analysis or something similar. – joeA Feb 5 '14 at 18:58
• @joeA I think I fixed the tags. – Mark S. Feb 15 '14 at 4:01

Let $D$ be a non-principal ultrafilter on the natural numbers. We start from the non-standard model $\mathbb{R}^N/D$.
Let $L$ the language $L$ which has a constant symbol for every element of $\mathbb{R}^N/D$, and function symbols, relation symbols of the appropriate arity for every function, relation on $\mathbb{R}^N/D$. Let $T$ be the theory whose axioms are all sentences of $L$ that are true in $\mathbb{R}^N/D$. Extend $L$ by adding a new constant symbol $c$, and extend $T$ to a theory $T'$ by adding the axioms $c\gt \alpha$ for all constant symbols $\alpha$ in $L$.
Then by compactness $T'$ has a model which is an elementary extension of $\mathbb{R}^N/D$ and has elements that are "infinite" with respect to all the elements of $\mathbb{R}^N/D$, and hence positive elements $\epsilon$ which are less than any positive element of $\mathbb{R}^N/D$.
• I was probably too formal in my answer, but did not want to just say "we can repeat the kind of construction that gives us a non-standard model." Part of the reason is that we can also have non-standard models $U$, $V$ such that $U\subseteq V$ but $U$ is cofinal in $V$ (for any $v\in V$, there is a $u\in U$ such that $u\gy v$. Such a pair of models would not give what you asked for. – André Nicolas Feb 1 '14 at 18:45