9
$\begingroup$

For a compactification $c\mathbb N$ of $\mathbb N$ let $\mathcal H(c\mathbb N,\mathbb N)$ be the group of homeomorphisms $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$. The group $\mathcal H(c\mathbb N,\mathbb N)$ determines the subgroup $$S_{\mathbb N,c\mathbb N}:=\{h{\restriction}\mathbb N:h\in\mathcal H(c\mathbb N,\mathbb N)\}$$ in the permutation group $S_{\mathbb N}$ of $\mathbb N$.

Observe that for the one-point compactification $\alpha\mathbb N$ of $\mathbb N$ the group $S_{\mathbb N,\alpha\mathbb N}$ coincides with the whole group $S_{\mathbb N}$ whereas for the Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ the group $S_{\mathbb N,\beta\mathbb N}$ coincides with the group $S_{<\mathbb N}$ of finitely supported permutations of $\mathbb N$ (i.e., permutations that move only finitely many points of $\mathbb N$).

For any compactification $c\mathbb N$ of $\mathbb N$ we have $$S_{<\mathbb N}=S_{\mathbb N,\beta\mathbb N}\subset S_{\mathbb N,c\mathbb N}\subset S_{\mathbb N,\alpha\mathbb N}=S_{\mathbb N}$$so $S_{\mathbb N,c\mathbb N}$ is intermediate between $S_{\mathbb N,\beta\mathbb N}$ and $S_{\mathbb N,\alpha\mathbb N}$.

It is easy to prove that the one-point compactification of $\mathbb N$ is the unique compactification $c\mathbb N$ of $\mathbb N$ with $S_{\mathbb N,c\mathbb N}=S_{\mathbb N}$.

Question. Is the Stone-Cech compactification a unique compactification of $\mathbb N$ with $S_{\mathbb N,c\mathbb N}=S_{<\mathbb N}$?

Remark. Negative answer to this question will imply negative answer to this MO-problem.

$\endgroup$

1 Answer 1

4
$\begingroup$

Analyzing the answer of @James Hanson to my preceding question, I realized that this question also has a simple negative answer: the quotient space $c\mathbb N:=\beta\mathbb N/\{p,q\}$ of $\beta\mathbb N$ by any doubleton $\{p,q\}\subset\beta\mathbb N\setminus\mathbb N$ is not homeomorphic to $\beta\mathbb N$ but has the smallest possible permutation group $S_{\mathbb N,c\mathbb N}=S_{<\mathbb N}$. This compactification $c\mathbb N$ also is not soft (according to this definition).

$\endgroup$

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.