3
$\begingroup$

It is well known that for any compact metric space $(X, d)$, and any ultrafilter $\mu$ there is a map $i_\mu:\prod_\mu (X, d) \to (X_d)$ in the category of metric spaces and Lipschitz maps where $i_\mu$ is an isomorphism in this category (and in particular an inverse to the constant map $\Delta: (X, d) \to \prod_\mu (X, d)$).

It is also known that for any compact topological space $(X, O(X))$ and ultrafilter $\mu$ there is a map $i_\mu: \prod_\mu X \to X$ such that for all $U \in O(X)$:

$$ (\forall (x_n) \in \prod_\mu X)\ i_\mu(x_n) \in U \leftrightarrow \{j : x_j \in U\} \in \mu. $$

Further the ultraproduct $\prod_\mu (X, O(X))$ in the category of (compact Hausdorff) topological spaces has $\prod_\mu X$ as its underlying set.

My questions are: For which compact Hausdorff topological spaces is $i_\mu$ always continuous? For which compact Hausdorff topological spaces is $i_\mu$ always a homeomorphism?

$\endgroup$
  • 1
    $\begingroup$ Maybe I'm understanding your definitions wrong, but doesn't any nonprincipal ultrapower in the category of spaces have the indiscrete topology? The quotient map $\prod X\to \prod_\mu X$ is surjective on any nonempty open set (because an open set depends on only finitely many coordinates, and $\mu$ is nonprincipal). $\endgroup$ – Eric Wofsey Mar 28 '15 at 3:37
  • $\begingroup$ Ah, of course, that is exactly what I wasn't seeing. Thanks Eric. If you want to write your comment as an answer I can accept it. $\endgroup$ – Nate Ackerman Mar 28 '15 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.