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In

http://arxiv.org/pdf/math/9704205.pdf

they define the ultracoproduct of a sequence of compact Hausdorff spaces, $\sum_\mathcal{U}X_i$ along an ultrafilter $\mathcal{U}$ as the Wallman-Frink compactification of the ultraproduct

$\prod_\mathcal{U}A_i$

where each $A_i$ is the normal lattice base of the closed sets of $X_i$. More generally, I it appears that you can do this for more general topological spaces (at least Tychonoff?):

http://www.susanjkleinart.com/compactification/Wsp3.pdf .

I'm wondering: is it possible to drop (or at least some how relax) the condition that the base is separating (condition 4, Definition P3.1) and have some sort of ultracoproduct construction?

If we consider compact Hausdorff spaces again, then for a sequence $(X_i)$ of spaces, I believe it is also true that

$C(\sum_{\mathcal{U}}X_i)\cong\prod_{\mathcal{U}}C(X_i)$

where $C(X)$ is the Banach space of continuous, real-valued functions on $X$ and $\prod_{\mathcal{U}}C(X_i)$ is the Banach ultraproduct. Would it make sense (for more general topological spaces $X_i$) to just define the ultracoproduct as

$C_b(\sum_{\mathcal{U}}X_i)\cong\prod_{\mathcal{U}}C_b(X_i)$

where $C_b(X_i)$ is the Banach space of bounded, continuous, real-valued functions on $X_i$? I suppose that this would make sense if there was a good way to recover the space $\sum_{\mathcal{U}}X_i$ from $C_b(\sum_{\mathcal{U}}X_i)$.

Also, Can an ultracoproduct be (perhaps naively) be seen as a kind of Kolmogorov quotient of an ultraproduct?

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  • $\begingroup$ Ah, the answer might be that the ultracoproduct of general topological spaces is homeomorphic to the ultracoproduct of their Stone-Čech compactification. $\endgroup$
    – greg
    Commented Jul 16, 2013 at 0:42

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