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Aug 17, 2023 at 12:55 comment added KP Hart In fact, one can pretend that the boundary is the Cantor set; the idea of the example is that the sets $P$ and $Q$ are so entwined that the closures of corresponding open sets shoudl intersect. ($P$ and $Q$ are like the sets of irrational and rational numbers respectively.)
Aug 17, 2023 at 12:32 comment added KP Hart That will not work: in $\beta\omega\setminus\omega$ disjoint open $F_\sigma$-sets have disjoint closures. The sets you describe are of this nature. The open set $V$ in my example is not an $F_\sigma$-set.
Aug 17, 2023 at 11:04 comment added HenrikRüping I think this example is easier to visualize by identifying the boundary of the tree with the Cantor set $C\subset [0,1]$. Then I am looking for a point $a$ that is in the closure of $(a-\varepsilon,a)$ and $(a,a+\varepsilon)$, e.g. we have an increasing sequence and a decreasing sequence converging to $a$. We can now use the self-similarity and use that $C$ consists of four copies of $C$. Let $f$ be the increasing map mapping $C$ to the second copy of $C$. Then the sequences are given by $f^n(0)$ and $f^n(1)$.
Aug 17, 2023 at 9:10 history answered KP Hart CC BY-SA 4.0