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Let $\mathcal{U}$ be a non-trivial ultrafilter on the set of integers $\mathbb{N}$, and let $C(K)$ denote the Banach space of continuous functions on a compact $K$. Under the continuum hypothesis CH, the Banach ultrapower $C(K)_\mathcal{U}$ is isomorphic to $\ell_\infty/c_0$ for many metrizable $K$. I wonder if it is possible to get this result for some $\mathcal{U}$ without assuming CH.

More precisely, given an infinite, metrizable, totally disconnected compact space $K$ having no isolated points, and a nontrivial ultrafilter $\mathcal{U}$ on $\mathbb{N}$, the ultracoproduct $K^\mathcal{U}$ is a Parovicenko space. Under CH, the only Parovicenko space is $\beta N\setminus N$, so $$ C(K)_\mathcal{U} \equiv C(K^\mathcal{U}) \equiv C(\beta N\setminus N) \equiv\ell_\infty/c_0. $$ Without CH, there are many Parovicenko spaces. I want to know if it is posible to show without CH that $K^\mathcal{U}=\beta N\setminus N$, or that $C(K^\mathcal{U})$ is isomorphic to $\ell_\infty/c_0$, for some $\mathcal{U}$.

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What exactly do you mean by $K^\mathcal{U}$? You said it is a set-ultrapower, but then what topology are you putting on it to talk about $C(K^\mathcal{U})$? If $K^\mathcal{U}$ is the metric space ultrapower, then it is isometric to $K$ so it is certainly not a Parovicenko space. – Ramiro de la Vega Oct 10 '13 at 9:10
@Ramiro de la Vega: You are right: $K^\mathcal{U}$ is not the set-ultraproduct, but the ultracoproduct (following $\mathcal{U}$) in the sense of Bankston [J. Symbolic Logic 52 (1987) 404-424]. The set ultrproduct is dense in $K^\mathcal{U}$. – M.González Oct 10 '13 at 9:45
I suppose you are also assuming that $K$ has no isolated points (and hence $K$ is just the Cantor space). Otherwise $K^\mathcal{U}$ would have isolated points. – Ramiro de la Vega Oct 10 '13 at 15:03
@Ramiro de la Vega: You are right again. Anyway, an answer when $K$ is the Cantor space would be very interesting for me. – M.González Oct 11 '13 at 7:25
If CH fails, then you can find $2^{\mathfrak{c}}$ ultrapowers of $C(K)$ which are not isometric to $\ell_\infty / c_0$. As for the isomorphic case, it seems that there are also $2^{\mathfrak{c}}$ isomorphic types by a version of Theorem 3 from adjusted to the setting of – Tomek Kania Sep 14 '14 at 20:02

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