# Show that there is a hyperreal number system of the same size as the real numbers?

The hyperreal number system is defined as one that contains the real numbers, satisfies the first order properties of real numbers, and contain infinitesimals. It can't be as simple as stating the reals are a subset of the hyperreals. Do I need to prove an isomorphism between the two? Or am I completely missing something here?

Any help would be appreciated.

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Since the hyperreals are an extension of the real numbers of course they cannot have a smaller size. You need to show that there is such extension which does not change the size, namely there is a hyperreal field $^\ast\mathbb R$ such that $|^\ast\mathbb R|=|\mathbb R|$.

Hint: Recall that one of the canonical ways to construct such field is using a non-principal ultrafilter $\mathcal U$ over $\mathbb N$, and taking $^\ast\mathbb{R=R^N}/\mathcal U$.

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The first-order properties of the reals are what we call "real closed field" ... so start with the reals, add an infinite element to that ordered field [say use the rational functions $\mathbb R(t)$ over $\mathbb R$ where $t$ is large and positive]. Then take its real-closure. So it is enough to show that the real-closure of an ordered field of power $\mathfrak c$ still has power $\mathfrak c$. [Concretely, perhaps, a space of Puisieux series.] Maybe this approach has a less "Axiom-Of-Choice" feel than either ultrafilter or Lowenheim-Skolem.

I am not convinced by André's comment. But anyway, my answer is for the first-order theory of the ordered field of the reals with only these usual relation symbols: $+, \cdot, 0, 1, \lt$. Only the OP can tell us if this is what he really intended.
There is ambiguity about what "first-order properties" of the reals may mean. For non-standard analysis, it is convenient to throw into the language function symbols, predicate symbols for all functions $\mathbb{R}^n\to \mathbb{R}$, all relations. Use as axioms all sentences of this extended language that are true in the reals. The ultrapower preserves everything. – André Nicolas Dec 13 '12 at 23:21
Wait ... "predicate symbols for all functions $\mathbb R^n \to \mathbb R$" ... that's $2^{\mathfrak c}$ different predicate symbols? Won't you need a higher ultrapower to "preserve" all that? – GEdgar Dec 14 '12 at 1:41
As long as all the function and relation symbols in the signature have finite arity, the ultrapower construction is good enough, no matter how many symbols you have. The trouble with this gigantic signature is that the Löwenheim-Skolem theorem can no longer push down the cardinality of the model to what we want ($\mathfrak{c}$)... – Zhen Lin Dec 14 '12 at 15:08