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 (see $\S$3.10),

Problem 1. Is each Parovichenko compact space soft Parovichenko?

Remark 2. The Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ is soft, but there are simple examples 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 $<\mathfrak p$).

Problem 2. Is each (Frechet-Urysohn) sequential Parovichenko space soft Parovichenko?

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.

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?

  • $\begingroup$ Is there a characterization of non-(either soft or not) Parovichenko compacts? $\endgroup$ Sep 1, 2018 at 5:17
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    $\begingroup$ @მამუკაჯიბლაძე I added (to my question) some known information about (soft) Parovichenko spaces. $\endgroup$ Sep 1, 2018 at 5:56
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    $\begingroup$ For what it's worth: every compactification with the ordinal $\omega_1+1$ as its remainder is soft. $\endgroup$
    – KP Hart
    Sep 20, 2018 at 19:49
  • $\begingroup$ I read a definition of "Parovichenko space" as: a Stone space $X$ with weight $\mathbf{c}$, no isolated point and in which every nonempty countable intersection of open subsets has nonempty interior. (Under CH this characterizes $\beta\mathbf{N}\smallsetminus\mathbf{N}$ up to homeomorphism.) Is this related? $\endgroup$
    – YCor
    Aug 18, 2019 at 10:39
  • $\begingroup$ @YCor No, "your" Parovichenko space is different than "mine". $\endgroup$ Aug 21, 2019 at 11:22

1 Answer 1


Here is a partial answer: the Continuum Hypothesis implies that all Parovichenko spaces are soft-Parovichenko; 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 $\omega_1+1+\omega_1^\ast$.

  • $\begingroup$ Thank you for this partial solution. Indeed, the construction written in the pdf-file is rather complicated. Is it indeed requires the full strenth of CH, or something weaker like $\mathfrak t=\mathfrak c$? $\endgroup$ Nov 9, 2018 at 17:30
  • $\begingroup$ All I can say about CH is that it appears to be needed in this proof: it needs countable initial segments and is the density of $X$ is equal to continuum then you expect to have to take care of continuum many pairs. The proof can be simplified a bit and I will do that later. $\endgroup$
    – KP Hart
    Nov 9, 2018 at 22:37
  • $\begingroup$ There is now a consistent counterexample, see the answer. $\endgroup$
    – KP Hart
    Nov 12, 2018 at 22:47
  • $\begingroup$ Thank you. But It would be helpful to remind what does the principle (NT) says exactly. $\endgroup$ Nov 12, 2018 at 23:27
  • $\begingroup$ I have added the formulation to the note. $\endgroup$
    – KP Hart
    Nov 13, 2018 at 13:01

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