**Definition 1.** A compactification $c\mathbb N$ of the discrete space $\mathbb N$ is called *soft* if for any disjoint sets $A,B\subset\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there exists a homeomorphism $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$ and the set $\{x\in A:h(x)\in B\}$ is infinite.


**Definition 2.** A compact Hausdorff space $X$ is called *Parovichenko* (resp. *soft Parovichenko*) if $X$ is homeomorphic to the remainder $c\mathbb N\setminus\mathbb N$ of some (soft) compactification $c\mathbb N$ of $\mathbb N$?


**Remark 1.** By a classical Parovichenko Theorem, each compact Hausdorff space of weight $\le\aleph\_1$ is Parovichenko. Hence, under CH a compact Hausdorff space is Parovichenko if and only if it has weight $\le\mathfrak c$. By a result of Przymusinski, each perfectly normal compact space is Parovichenko. On the other hand, Bell constructed an consistent example of a first-countable compact Hausdorff space, which is not Parovichenko. More information and references on Parovichenko spaces can be found in [this survey of Hart and van Mill](https://fa.its.tudelft.nl/~hart/37/publications/the_papers/open_problems_on_betaN.pdf) (see $\S$3.10), 



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> **Problem 1.** Is each Parovichenko compact space soft Parovichenko?
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**Remark 2.** The Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ is soft, but there are [simple examples](https://mathoverflow.net/questions/309458/is-each-compactification-of-mathbb-n-soft) of compactifications which are not soft. A compactification $c\mathbb N$ of $\mathbb N$ is soft if for any disjoint sets $A,B\subset\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there are sequences $\{a\_n\}\_{n\in\omega}\subset A$ and $\{b\_n\}\_{n\in\omega}\subset B$ that converge to the same point $x\in\bar A\cap\bar B$. This implies that a compactification $c\mathbb N$ is soft if the space $c\mathbb N$ is Frechet-Urysohn or has sequential square. This also implies that *each first-countable Parovichenko space is soft Parovichenko* (more generally, *a Parovichenko space $X$ is soft Parovichenko if each point $x\in X$ has a neighborhood base of cardinality $<\mathfrak p$*).



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> **Problem 2.** Is each (Frechet-Urysohn) sequential Parovichenko space soft Parovichenko? 
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The following concrete version of Problem 1 describes an example of a Parovichenko space for which we do not know if it is soft Parovichenko.



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> **Problem 3.** Let $X$ be a compact space that can be written as the union $X=A\cup B$ where $A$ is homeomorphic to $\beta\mathbb N\setminus\mathbb N$, $B$ is homeomorphic to the Cantor cube $\{0,1\}^\omega$ and $A\cap B\ne\emptyset$. Is the space $X$ soft Parovichenko?
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Is each Parovichenko compact space homeomorphic to the remainder of a soft compactification of $\mathbb N$?
Here is a partial answer: the [Continuum Hypothesis implies that all Parovichenko spaces are soft-Parovichenko](http://fa.its.tudelft.nl/~hart/37/publications/the_papers/soft-compactifications.pdf); the proof is a bit long, so I put it in a PDF-file on my website.


Also, I retract my claim in the comments that all compactifications with $\omega\_1+1$ as a remainder are soft. It is true, in ZFC, that $\omega\_1+1$ is soft-Parovichenko but "all compactifications with remainder $\omega\_1+1$ are soft" is equivalent to $\mathfrak{t}>\omega\_1$.


Added 2018-11-12: The note linked to above now contains a, consistent, example of a Parovichenko space that is not soft-Parovichenko. The example is the ordered space $\omega\_1+1+\omega\_1^\ast$.


