If ${\cal U}$ is any ultrafilter on $\omega$, the pair $(\omega,{\cal U}\cup \{\emptyset\})$ is a connected topological space. Is there a non-principal ultrafilter ${\cal U}$ on $\omega$ such that we have $(\omega,{\cal U}\cup \{\emptyset\})\cong (\omega,{\cal U}\cup \{\emptyset\})^2$?
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asked Jul 20, 2017 at 13:31
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$\begingroup$ Perhaps it might be worth mentioning that the somewhat similar product of ultrafilters is mentioned in this post: Product of ultrafilters, is it an ultrafilter? $\endgroup$– Martin SleziakCommented Jul 20, 2017 at 14:08
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1 Answer
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No. Note that in the space $X = (\omega, \mathcal{U}\cup\{\emptyset\})$, every set is either open or closed. However in the space $X^2$, neither $\overline{\Delta} = \{(m,n) : n\ge m\}$ nor $\underline{\Delta} = \{(m,n) : n < m\}$ can be open, since neither of these sets contains a rectangle of the form $A\times B$ with $A,B\in\mathcal{U}$.
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$\begingroup$ Very nice argument -- it escaped me totally! $\endgroup$ Commented Jul 20, 2017 at 15:55