# Why continuum hypothesis implies the unique hyperreal system, ${}^{\ast}{\Bbb R}$?

On page 33, Robert Goldblatt, Lectures on Hyperreals(1998):

Now it has been shown under certain set-theoretic assumption called continuum hypothesis the choice of $\mathcal F$ is irrelevant: All quotients of $\Bbb {R}^{\Bbb N}$ with respect to nonprincipal ultrafilters on $\Bbb N$ are isomorphic as ordered fields.

I don't know why this is true. It seems to me if we want to prove continuum hypothesis implies the unique hyperreal system, we have to show that, for any two free ultrafilters $\mathcal {F}_{1}$ and $\mathcal {F}_{2}$ on $\Bbb N$, there is an bijection $f: \Bbb N \to \Bbb N$ such that there it's also a bijection from $\mathcal {F}_{1}$ to $\mathcal {F}_{2}$. How to show this holds with $\bf CH$ and fails without it.

• You seem to think that the uniqueness of the hyperreals is equivalent to the continuum hypothesis. No one said that. (That been said, it might still be true) – Asaf Karagila May 25 '13 at 12:42
• @AsafKaragila:So $\bf CH$ implies uniqueness of the hyperreals, but it's an open problem that whether the negation of $\bf CH$ implies non-uniqueness of the hyperreals, right? – Metta World Peace May 25 '13 at 12:49
• I don't know... – Asaf Karagila May 25 '13 at 12:49
• ...such that what is a bijection from $\mathcal{F}_1$ to $\mathcal{F}_2$? – Trevor Wilson May 25 '13 at 13:59
• @TrevorWilson: Sorry for not phrasing it properly. I mean, given the bijection $f$, we require that for all $A \in \mathcal {F}_{1}$, we have $f(A) \in \mathcal {F}_{2}$, for all $B \in \mathcal {F}_{2}$, we have $f^{-1}(B) \in \mathcal {F}_{1}$. Since $f$ is a bijection from $\Bbb N$ onto $\Bbb N$, for each $C \neq D$ such that $C, D \in \mathcal {F}_{1}$, $f(C) \neq f(D)$, thus the induced mapping from $\mathcal {F}_{1}$ to $\mathcal {F}_{2}$ is one-on-one. – Metta World Peace May 25 '13 at 14:20

As bestian tang suggested, you can use approach similar to the one used here.

First, notice that any countable ultrapower of ${\bf R}$ has cardinality continuum, so any two such ultrapowers are equipotent.

Then, by standard argument, a countable ultrapower of any structure in a countable language is $\aleph_1$-saturated (you can just recursively construct an element realizing a given type), so under CH, any countable ultrapower of ${\bf R}$ is saturated.

Finally, it is a well-known result that any two structures which are saturated and have the same cardinality are isomorphic (by a standard back-and-forth argument), so any two countable ultrapowers of ${\bf R}$ are isomorphic.

I don't think this approach will work for the whole hyperreal construction (with all real sets, relations, functions), as the language will, in this case, have cardinality strictly larger than ${\mathfrak c}$

This is not an answer; it’s just showing that your approach can’t work.

The Rudin-Keisler order on $\beta\omega\setminus\omega$ is defined as follows. For $p,q\in\beta\omega\setminus\omega$ write $p\le_{RK}q$ iff there is an $h:\omega\to\omega$ such that $A\in p$ iff $h^{-1}[A]\in q$ for each $A\subseteq\omega$, i.e., $q=\{h^{-1}[A]:A\in p\}$. It turns out that $p\le_{RK}q\le_{RK}p$, written $p\equiv_{RK}q$, iff $h$ can be taken to be a bijection. Your conjecture, then, is that $\mathsf{CH}$ implies that $p\equiv_{RK}q$ for all $p,q\in\beta\omega\setminus\omega$. However, this is false. There are only $2^\omega$ bijections from $\omega$ to $\omega$, but $|\beta\omega\setminus\omega|=2^{2^\omega}>2^\omega$: with or without $\mathsf{CH}$ there aren’t enough bijections in ${}^\omega\omega$ for all free ultrafilters on $\omega$ to be Rudin-Keisler equivalent. (In fact under $\mathsf{CH}$ it’s extremely easy to construct non-equivalent ultrafilters.) Thus, the uniqueness of the ordered field $\Bbb R^\omega/p$ under $\mathsf{CH}$ must come from some other source.

If Goldblatt doesn’t cite a reference for the result, you might look for one in H.J. Keisler’s survey The hyperreal line in P. Ehrlich (ed.) Real numbers, generalizations of reals, and theories of continua, Kluwer Academic Publishers, $1994$, pp. $207$-$237$; I’ve not seen it, but I know that it at least mentions some closely related results.

• You opened a parenthesis... but you didn't close it! – Asaf Karagila May 25 '13 at 21:02
• @Asaf: ))))))))) Worse yet, I omitted the full stop and the space before the next sentence! – Brian M. Scott May 25 '13 at 21:04
• Well, I'm not a native English speaker, but Common Lisp did leave quite an impression on my mind during the little time that I worked with it. :-) – Asaf Karagila May 25 '13 at 21:05