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A metrizable space $X$ will be called a generalized Bernstein set if every closed completely metrizable subspace $C$ of $X$ has cardinality $|C|<|X|$.

It is well-known that the real line contains a (generalized) Bernstein set of cardinality $\mathfrak c$.

Moreover, for every cardinal $\kappa$ with $\kappa^\omega=2^\kappa$ there exist a generalized Bernsten set $X$ of cardinality $|X|=2^\kappa=\kappa^\omega$. In particular, there exists a generalized Bernstein set of cardinality $\beth_{\omega+1}$.

Question. What are possible cardinalities of generalized Bernstein sets? For example, is there a generalized Bernstein set $X$ of cardinality $|X|=\mathfrak c^+$?

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  • $\begingroup$ Taras, could you suggest a reference for the result you mention in paragraph 3? $\endgroup$
    – Andrés E. Caicedo
    Commented Jun 26, 2016 at 22:12
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    $\begingroup$ I do not know such a reference, but know a proof: Given a cardinal $\kappa$ with $\kappa^\omega=2^\kappa$, endow $\kappa$ with the discrete $\{0,1\}$-valued metric and consider the complete metric space $X=\kappa^\omega$ of weight $\kappa$. This space contains at most $2^\kappa$ $G_\delta$-subsets, so we can enumerate the family $\mathcal C$ of completely metrizable subsets of cardinality $\kappa^\omega$ as $\mathcal C=\{C_\alpha\}_{\alpha\in 2^\kappa}$. $\endgroup$
    – Taras Banakh
    Commented Jun 27, 2016 at 4:46
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    $\begingroup$ Then by transfinite induction for every ordinal $\alpha<2^\kappa$ choose two distinct points $x_\alpha,y_\alpha$ in the set $C_\alpha\setminus\{x_\beta,y_\beta\}_{\beta<\alpha}$. The choice of $x_\alpha,y_\alpha$ is always possible as $|C_\alpha|=\kappa^\omega=2^\kappa>|\alpha+\alpha|$. Then $Y=\{y_\alpha\}_{\alpha<2^\kappa}$ is a generalied Bernstein set of cardinality $|Y|=\kappa^\omega$. $\endgroup$
    – Taras Banakh
    Commented Jun 27, 2016 at 4:50
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    $\begingroup$ To show that $Y$ is a generalized Bernstein set, assume that $Y$ contains a completely metrizable set $C$ of cardinality $|C|=|Y|=\kappa^\omega$. Then $C=C_\alpha$ for some $\alpha<2^\kappa$ and then $x_\alpha\in C\setminus Y$, which means that $C\not\subset Y$. But this contradicts the choice of $C$. $\endgroup$
    – Taras Banakh
    Commented Jun 27, 2016 at 4:54

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After thinking a night on this question and waking up, I realized that the answer is almost trivial: there are restrictions on possible cardinalities of generalized Bernstein set.

Any metrizable space $X$ of density $\kappa$ has cardinality $|X|\le\kappa^\omega$ and contains a discrete (and hence completely metrizable) subspace $D$ of cardinality $|D|=\kappa$. If $\kappa=\kappa^\omega$ (which is the case for the cardinal $\mathfrak c^+$), then $X$ cannot be a generalized Bernstein set.

On the other hand, there is also a less trivial fact involving (even more) generalized Bernstein sets.

Let $\mathfrak c^{+0}=\mathfrak c$ and $\mathfrak c^{+(n+1)}=(\mathfrak c^{+n})^+$ for $n\in\omega$. So, $\mathfrak c^{+n}$ is the $n$th successor cardinal of $\mathfrak c$. Let also $\mathfrak c^{+\omega}=\sup_{n\in\omega}\mathfrak c^{+n}$.

By induction it can be shown that each metrizable space $X$ of cardinality $|X|<\mathfrak c^{+\omega}$ contains a subset $B\subset X$ of cardinality $|B|=|X|$ such that for every uncountable Polish subspace $P\subset X$ both sets $P\cap B$ and $P\setminus B$ are not empty. This implies that $B$ contains no uncountable Polish subspaces, so $B$ can be considered as a generalized Bernstein set (in a somewhat different sense).

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    $\begingroup$ Concerning your last paragraph: Bill Weiss has shown that under certain set-theoretic assumptions ("there are no measurable cardinals" will do), every Hausdorff space contains a subset $B$ with the property you describe. (See his article "Partitioning topological spaces" in Mathematics of Ramsey Theory -- unfortunately, I do not know of a free online version.) On the other hand, using a supercompact cardinal Shelah has found a model with a space $X$ admitting no such subspace. (See Sh 460 and search it for "Cantor discontinuum problem" -- it's somewhere near the end.) $\endgroup$
    – Will Brian
    Commented Jun 27, 2016 at 16:05
  • $\begingroup$ Thank you, Will. I looked at Sh460 (shelah.logic.at/files/460.pdf) and it gives a consistency of the existence of the Berstein set in any topological space. On the other hand, the paper Sh668 (arxiv.org/pdf/math/9906025v1.pdf) proves the consistency of the existence of a compact Hausdorff space without a Bernstein set. So, what I have asked appeared to be just another bicycle :( $\endgroup$
    – Taras Banakh
    Commented Jun 27, 2016 at 17:18

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