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$\DeclareMathOperator\bso{\beta^*\!\omega}\DeclareMathOperator\Homeo{Homeo}$Let $\bso$ be the complement of the countable discrete space $\omega$ in its Stone-Čech compactification $\beta\omega$ (some authors denote it $\omega^*$).

Question. Assume ZFC+CH. Does there exist a self-homeomorphism of $\bso$ with an isolated fixed point (i.e., a fixed point that is not a limit of other fixed points)?

Bonus question: if yes: it it true that for every $x\in\bso$ there exist a self-homeomorphism of $\bso$ with $x$ an isolated fixed point?

Other bonus question: what about asking the self-homeomorphism to have order 2?


Contextual remarks:

  1. Say that a self-homeomorphism of $\bso$ is smooth if it is induced by a bijection between two subsets of $\omega$ with finite complement. If $f\in\Homeo(\bso)$ is smooth then the set of fixed points of $f$ is clopen, hence has no isolated point.

  2. Shelah in the early 1980s showed the existence of models of ZFC in which every self-homeomorphism of $\bso$ is smooth (and in particular in which $\#(\Homeo(\bso))=\mathfrak{c}$). Hence in such models the above question has a negative answer.

  3. Rudin proved in the 1950s that under ZFC+CH, $\#(\Homeo(\bso))=2^\mathfrak{c}$ and its action on $\bso$ is not transitive.

  4. If there's a self-homeomorphism with an isolated fixed point $x$, it is easy to modify it to get another one with unique fixed point $x$.

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2 Answers 2

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The answer to your main question is yes. In fact, there is (under $\mathsf{CH}$) a self-homeomorphism of $\omega^*$ with exactly one fixed point. Such a mapping is constructed in the proof of Theorem 5.7 in my paper:

"$P$-sets and minimal right ideals in $\mathbb N^*$," Fundamenta Mathematicae 229 (2015), pp. 277-293. (pdf)

Sketch of the construction: Begin with the shift map $\sigma$ on $\omega^*$, which is the (unique) self-homeomorphism of $\omega^*$ that is induced by the mapping $n \mapsto n+1$ on $\omega$. (That is, $\sigma$ maps every ultrafilter $\mathcal U$ to the ultrafilter generated by $\{ A+1 :\, A \in \mathcal U\}$.) This map has no fixed points. Using some ideas from earlier in the paper, there is a nowhere-dense, closed $P$-set $X \subseteq \omega^*$ such that $X$ is closed under $\sigma$. (Recall that $X$ is a $P$-set if the intersection of any countably many neighborhoods of $X$ is again a neighborhood of $X$.) Now modify the topology of $\omega^*$ by collapsing $X$ to a point. Assuming $\mathsf{CH}$, this quotient space is again homeomorphic to $\omega^*$. ($\mathsf{CH}$ implies that collapsing a nowhere dense $P$-set to a point always results in a copy of $\omega^*$; see Corollary 1.2.4 in van Mill's survey article about $\beta \omega$ for a proof.) The map $\sigma$ induces a self-homeomorphism of the quotient space, and the collapsed copy of $X$ is its unique fixed point. $\quad \square$

Notice that the construction is quite flexible. If you wanted exactly $17$ fixed points, this could be arranged. If you wanted no fixed points but a point of order $42$, then this could be arranged as well. (In the above argument, instead of collapsing $X$ to a point, collapse each of the $42$ sets $X \cap (k+42\mathbb N)^*$ to a point.) I think the most general thing one could achieve along these lines is: for any function $f: \mathbb N \rightarrow \mathbb N \cup \{\aleph_0\}$, there is (assuming $\mathsf{CH}$) a self-homeomorphism of $\omega^*$ having $f(n)$ points of order $n$ for every $n$.

For the bonus questions, I think this should be possible using a different technique (but I'd have to check the details to be sure). Under $\mathsf{CH}$ I think that any point $u \in \omega^*$ is a bowtie point, meaning that $\omega^* \setminus u$ can be partitioned into two nonempty clopen sets. And I think these two sets are homeomorphic, so permuting them (while leaving $u$ fixed) gives you the kind of map you want. Note that this also gives you a different way of getting a positive answer to your main question as well.

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  • $\begingroup$ Great! Of course if you can achieve one fixed point you can achieve $n\ge 1$ fixed points (using that the space is homeomorphic to $n$ copies of itself). In this direction one could ask whether every closed subset is the fixed point set of a self-homeomorphism. In this direction and also involving periodicities one could then ask if for every family $(K_n)_{n\ge 1}$ of closed subsets, satisfying $K_m\subset K_n$ whenever $n$ divides $n$, there exists a self-homeo $f$ for which $\mathrm{Fix}(f^n)=K_n$ for every $n$ (unless I'm missing an obvious obstruction). $\endgroup$
    – YCor
    Commented May 3, 2021 at 13:20
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    $\begingroup$ See below: a fixed point of an autohomeomorphism of $\omega^*$ is necessarily a P-point. $\endgroup$
    – KP Hart
    Commented May 6, 2021 at 11:17
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As an answer to the bonus question: no, see K. P. Hart and J. Vermeer. Fixed-point sets of autohomeomorphisms of compact F-spaces, Proceedings of the American Mathematical Society, 123 (1995), 311–314: every fixed-point set of an autohomeomorphism of $\omega^*$ is a $P$-set. The same paper also answers the other bonus question: every $P$-set is the fixed-point set of an involution. Apply this to a $P$-point.

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  • $\begingroup$ Thank you very much! this indeed solves all the questions I addressed, and is indeed a prior reference. $\endgroup$
    – YCor
    Commented May 6, 2021 at 11:20

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