6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Is there a characterization of non-(either soft or not) Parovichenko compacts? $\endgroup$ – მამუკა ჯიბლაძე Sep 1 '18 at 5:17
  • 1
    $\begingroup$ @მამუკაჯიბლაძე I added (to my question) some known information about (soft) Parovichenko spaces. $\endgroup$ – Taras Banakh Sep 1 '18 at 5:56
  • 1
    $\begingroup$ For what it's worth: every compactification with the ordinal $\omega_1+1$ as its remainder is soft. $\endgroup$ – KP Hart Sep 20 '18 at 19:49
6
$\begingroup$

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

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.