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 $(\omega,{\cal U}\cup \{\emptyset\})\cong (\omega,{\cal U}\cup \{\emptyset\})^2$?



Products of spaces with an underlying free ultrafilter as topology
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 $A,B\in\mathcal{U}$.


