A Tychonoff space $X$ is defined to have countable separation if some (equivalently, any) compactification $bX$ of $X$ contains a countable family $\mathcal U$ of open sets such that for any points $x\in X$ and $y\in bX\setminus X$ there is a set $U\in\mathcal U$ containing exactly one point of the doubleton $\{x,y\}$. The class of spaces with countable separation is very rich: for every compact Hausdorff space $K$ the family of subspaces of $K$ having countable separation is a $\sigma$-algebra, closed under the Suslin $A$-operation. This notion was indroduced by Kenderov, Kortezov and Moors in 2001 (at least).

It can be shown that each metrizable space of density at most continuum has countable separation. On the other hand, by tranfinite induction it is possible to construct a metrizable space of density $\beth_\omega$ without countable separation (this follows from the fact that each Tychonoff space $(X,\tau)$ of cardinality $|X|\ge|\tau^\omega|>\mathfrak c$ contains a subspace $Y$ without countable separation).

Let us recall that $\beth_\omega=\sup_{n\in\omega}\beth_n$ where $\beth_0=\omega$ and $\beth_{n+1}=2^{\beth_n}$, so $\beth_\omega$ is rather large. The key property of the cardinal $\kappa=\beth_\omega$ in this context is that $|\kappa^\omega|=2^\kappa>\mathfrak c$.

Question: What is the smallest density of a metrizable space without countable separation? Can it be equal to $\mathfrak c^+$? Or it is always $\ge \aleph_\omega$?

It can be shown that every (complete) metric space $X$ of cardinality $|X|=\mathfrak c^+$ contains a subset $B\subset X$ such that for every uncountable Polish subspace $P$ of $X$ both sets $P\cap B$ and $P\setminus B$ are not empty. Such set $B$ will be called a Bernstein set in $X$.

Question. Can a Bernstein set $B$ in the Hilbert space $\ell_2(\mathfrak c^+)$ of density $\mathfrak c^+$ have countable separation?

A negative answer to the last question would follows from the affirmative answer to this problem.

  • $\begingroup$ Wouldn't it be equivalent to ask about the discrete spaces? $\endgroup$ – Włodzimierz Holsztyński Jun 26 '16 at 7:18
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    $\begingroup$ No, discrete spaces are Cech-complete and hence have countable separation. So, a counterexample should be very "non-Borel". $\endgroup$ – Taras Banakh Jun 26 '16 at 11:20
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    $\begingroup$ In fact, for any compact Hausdorff space $K$ the algebra $\mathcal A$ of subsets of $K$ with countable separation is closed under some quite "fantastic" operations. Namely, for any sequence $(A_n)_{n\in\omega}\in\mathcal A^\omega$ and any subset $\Omega\subset\{-1,1\}^\omega$ the set $\Omega(A_n)_{n\in\omega}=\bigcup_{f\in \Omega}\bigcap_{n\in\omega}f(n)\cdot A_n$ belongs to the algebra $\mathcal A$. Here $1\cdot A_n=A_n$ and $(-1)\cdot A_n=X\setminus A_n$. The operation $\Omega(A_n)_{n\in\omega}$ can be considered as a quite far generalization of the Suslin $A$-operation! $\endgroup$ – Taras Banakh Jun 26 '16 at 11:23
  • $\begingroup$ By the way, does anybody know a good reference for such general operations $\Omega(A_n)_{n\in\omega}$? And for algebras of sets, closed under such operations? $\endgroup$ – Taras Banakh Jun 26 '16 at 11:26
  • $\begingroup$ Indeed, Taras, your operations are truly fantastic. Wouldn't logicians have a lot to say about them? *** BTW, thank you dor answering my (naive) question about the discrete spaces. $\endgroup$ – Włodzimierz Holsztyński Jun 27 '16 at 5:55

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