6
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ No, yes and yes. $\endgroup$
    – Ramiro de la Vega
    Oct 26, 2012 at 11:38
  • $\begingroup$ Thanks, any hint for the family $F$? Can it have cardinality $2^{\mathfrak{c}}$? $\endgroup$
    – Slavoj Žižek
    Oct 26, 2012 at 11:43
  • 4
    $\begingroup$ 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
    Oct 26, 2012 at 12:13
  • 2
    $\begingroup$ Welcome to MO! Are you by any chance this Slavoj Žižek? en.wikipedia.org/wiki/Slavoj_%C5%BDi%C5%BEek $\endgroup$
    – HJRW
    Oct 26, 2012 at 12:51

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.