10
$\begingroup$

We work in ZFC.

Let $C(X)$ be the ring of continuous functions $f:X\to\mathbb{R}$, and $M$ a maximal ideal. We call $C(X)/M$ a hyperreal field if it's not isomorphic to $\mathbb{R}$.

A field $E$ is called real-closed if it has unique ordering $x\geq 0$ iff $\exists_{y\in E}\ x = y^2$ and all polynomials of odd degree have a root.

An ordered field $E$ is called an $\eta_1$-field if for all countable $A, B\subseteq E$ such that $A < B$ there is $x\in E$ such that $A < v < B$.

All hyperreal fields are real-closed $\eta_1$-fields.

We have the following result:

Theorem. Assume CH. Then all real-closed $\eta_1$-fields of size continuum are isomorphic.

Corollary. Assume CH. Then all hyperreal fields of size continuum are isomorphic.

Let P be "All hyperreal fields $C(\mathbb{R})/M$ are isomorphic".

I'm wondering if one can get some insight into what kind of statement is P in relation to known set-theoretic principles.

Is P independent of ZFC+$\lnot$CH? Above theorem shows that ZFC+CH implies P.

Note: The choice of $X = \mathbb{R}$ was arbitrary, to not consider every possible topological space $X$ such that $C(X)/M$ is a hyperreal field of size continuum. If this is helpful, please consider a different space $X$ instead.

Improve this question
$\endgroup$
16
  • 1
    $\begingroup$ Oh, I think I see what you mean, all hyperreal fields of the form $C(\mathbb{R})/M$ are of size continuum. $\endgroup$
    – Jakobian
    Commented Aug 18, 2023 at 17:51
  • 1
    $\begingroup$ Re, I was confused because your original version referred only to $C(\mathbb R)$. I now see that the restriction to size continuum in the Theorem is necessary because you want to speak of hyperreal fields obtained by taking quotients of rings of continuous functions on other spaces. $\endgroup$
    – LSpice
    Commented Aug 18, 2023 at 17:51
  • 1
    $\begingroup$ @LSpice yes, I'm sorry, I was wondering how general should my question be, and decided on restricting to $X = \mathbb{R}$ after all. It was a typo I didn't clean from the initial version $\endgroup$
    – Jakobian
    Commented Aug 18, 2023 at 17:52
  • 3
    $\begingroup$ Since ZFC+CH is equiconsistent with ZFC alone, what is the sense of strength that you intend in the question? Your hypothesis has no extra consistency strength, if it is implied by CH, since CH has none. $\endgroup$
    – Joel David Hamkins
    Commented Aug 18, 2023 at 18:00
  • 2
    $\begingroup$ @JoelDavidHamkins my answer below applies to many spaces, certainly normal ones: if you have an infinite closed and discrete subset then you have a closed copy of $\mathbb{N}$ in your space and every ultrafilter $u$ on $\mathbb{N}$ yields a maximal ideal $M_u$ in $C(X)$. And $C(X)/M_u$ is the ultrapower of $\mathbb{N}$ by $u$. $\endgroup$
    – KP Hart
    Commented Aug 19, 2023 at 14:21

1 Answer 1

Reset to default
14
$\begingroup$

In On ultra powers of Boolean algebras (Topology Proceedings 9 (1984) 269-291) Alan Dow proved (Corollary 2.3) that $\neg\mathsf{CH}$ implies there are two fields of the form $C(\mathbb{N})/M$ that are not isomorphic (as ordered sets, and hence as ordered fields).

To be a bit more precise: Alan constructed two ultrafilters $u$ and $v$ on $\mathbb{N}$ with non-isomorphic ultrapowers. These yield maximal ideals $M_u$ and $M_v$ in $C(\mathbb{N})$ whose quotients are (isomorphic to) the respective ultrapowers. And because $\mathbb{N}$ is a closed discrete subspace of the metric space $\mathbb{R}$ these determine maximal ideals $N_u$ and $N_v$ in $C(\mathbb{R})$ whose quotients are isomorphic to those by $M_u$ and $M_v$ respectively.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Wow. Thank you so much! So the statement "all hyperreal fields of size continuum are isomorphic" is actually equivalent to CH. That's surprising. $\endgroup$
    – Jakobian
    Commented Aug 19, 2023 at 12:09
  • $\begingroup$ I meant $\mathbb{N}$ because the question allowed for different spaces. But in this particular case an ultrafilter on $\mathbb{N}$ determines maximal ideals in $C(\mathbb{N})$ and in $C(\mathbb{R})$, whose quotients are isomorphic anyway. $\endgroup$
    – KP Hart
    Commented Aug 19, 2023 at 14:56
  • $\begingroup$ I seem vaguely to remember that Shelah has a paper where he estimates the number of non-isomorphic ultrapowers modulo the negation of CH, but I can't seem to locate this right now. Does this ring the bell? $\endgroup$
    – Mikhail Katz
    Commented Aug 20, 2023 at 7:46
  • $\begingroup$ Yes, the paper is cited in math.stackexchange.com/a/836784/72694 $\endgroup$
    – Mikhail Katz
    Commented Aug 20, 2023 at 8:10
  • $\begingroup$ Apparently, according to one of the answers in the cited question, in ZFC+$\lnot$CH there is always $2^\mathfrak{c}$ non-isomorphic hyperreal fields $C(\mathbb{N})/M$. So when there is no uniqueness, the uniqueness fails completely! $\endgroup$
    – Jakobian
    Commented Aug 20, 2023 at 15:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .