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Let $p,q\in \beta \mathbb{N}\setminus \mathbb{N}$. Must always the spaces $\beta \mathbb{N}\setminus \{p\}$ and $\beta \mathbb{N}\setminus \{q\}$ be homeomorphic? If no, can we for each point $p\in \beta \mathbb{N}\setminus \mathbb{N}$ find $q\in \beta \mathbb{N}\setminus \mathbb{N}$ ($q\neq p$) such that $$\beta \mathbb{N}\setminus \{p\}\approx \beta \mathbb{N}\setminus \{q\}?$$

Or maybe there exists an uncountable family $F\subset \beta \mathbb{N}\setminus \mathbb{N}$ such that the spaces corresponding to points in $F$ are mutually non-homeomorphic?

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    $\begingroup$ No, yes and yes. $\endgroup$
    – Ramiro de la Vega
    Commented Oct 26, 2012 at 11:38
  • $\begingroup$ Thanks, any hint for the family $F$? Can it have cardinality $2^{\mathfrak{c}}$? $\endgroup$
    – Slavoj Žižek
    Commented Oct 26, 2012 at 11:43
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    $\begingroup$ Yes. A homeomorphism of $\beta\mathbb N\setminus\{p\}$ and $\beta\mathbb N\setminus\{q\}$ is uniquely determined by its restriction to the isolated points, i.e., by a permutation of $\mathbb N$. Thus, for any $p$ there are at most $\mathfrak c$ such $q$, whereas there are $2^{\mathfrak c}$ ultrafilters in total. $\endgroup$
    – Emil Jeřábek
    Commented Oct 26, 2012 at 12:13
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    $\begingroup$ Welcome to MO! Are you by any chance this Slavoj Žižek? en.wikipedia.org/wiki/Slavoj_%C5%BDi%C5%BEek $\endgroup$
    – HJRW
    Commented Oct 26, 2012 at 12:51

1 Answer 1

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As Emil pointed out in his comment, there is a correspondence between permutations of $\mathbb{N}$ and self-homeomorphisms of $\beta\mathbb{N}$, and that pretty much answers the OP´s original questions.

If we restrict ourselves to $\mathbb{N}^\ast=\beta\mathbb{N} \setminus \mathbb{N}$ the questions get more interesting (although the answers remain the same): When you mod out $\mathbb{N}^\ast$ by its group of self-homeomorphisms you get $2^\mathfrak{c}$ many orbits. This follows from Frolik´s theory of types of ultrafilters.

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