# Is each compactification of $\mathbb N$ soft?

Definition. A compactification $c\mathbb N$ of the countable discrete space $\mathbb N$ is defined to be soft if for any disjoint sets $A,B\subset\mathbb N\subset c\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there exists a homeomorphism $h$ of $c\mathbb N$ such that $h(A)\cap B$ is infinite and $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$.

Problem. Is each compactification of a countable discrete space soft?

Remark. The Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ is soft as no disjoint sets $A,B\subset\mathbb N$ have $\bar A\cap\bar B\ne\emptyset$. On the other hand, a compactification $c\mathbb N$ is soft if the space $c\mathbb N$ is Frechet-Urysohn or has sequential square $c\mathbb N\times c\mathbb N$. So, a counterexample if exists should be rather exotic.

• If you take a copy of $\beta \mathbb{N}$ and a copy of the one point compactification of $\mathbb{N}$ and glue the limit point to some arbitrary point in $\beta \mathbb{N} \setminus \mathbb{N}$ doesn't that fail to be soft? – James Hanson Aug 30 '18 at 20:30
• @JamesHanson You are right! So easy! Please write down your comment as an answer and I will accept it as a solution in order to close this question. – Taras Banakh Aug 30 '18 at 20:35
• I hope you did not mean "close" in the MO technical sense :D – მამუკა ჯიბლაძე Sep 1 '18 at 5:15

Let $A=\{0,2,4,\dots\}$ be the even numbers and let $B=\{1,3,5,\dots\}$ be the odd numbers. Topologize $A\cup \beta B$ so that $A$ is a sequence limiting to a unique point in $\beta B \setminus B$. This is a compactification of $\mathbb{N}$ that fails to be soft, since any homomorphism of the required form would give a non-trivial sequence in $\beta B$ limiting to a unique point, but no such sequence exists.