# Is each Parovichenko compact space homeomorphic to the remainder of a soft compactification of $\mathbb N$?

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?

• Is there a characterization of non-(either soft or not) Parovichenko compacts? – მამუკა ჯიბლაძე Sep 1 '18 at 5:17
• @მამუკაჯიბლაძე I added (to my question) some known information about (soft) Parovichenko spaces. – Taras Banakh Sep 1 '18 at 5:56
• For what it's worth: every compactification with the ordinal $\omega_1+1$ as its remainder is soft. – KP Hart Sep 20 '18 at 19:49

## 1 Answer

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 $$\omega_1+1+\omega_1^\ast$$.

• 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 $\mathfrak t=\mathfrak c$? – Taras Banakh Nov 9 '18 at 17:30
• 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. – KP Hart Nov 9 '18 at 22:37
• There is now a consistent counterexample, see the answer. – KP Hart Nov 12 '18 at 22:47
• Thank you. But It would be helpful to remind what does the principle (NT) says exactly. – Taras Banakh Nov 12 '18 at 23:27
• I have added the formulation to the note. – KP Hart Nov 13 '18 at 13:01