# Bernstein sets of large cardinality

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 $|X|=\mathfrak c^+$?

• Taras, could you suggest a reference for the result you mention in paragraph 3? Jun 26 '16 at 22:12
• 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 $\mathcal C=\{C_\alpha\}_{\alpha\in 2^\kappa}$. Jun 27 '16 at 4:46
• 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 $|Y|=\kappa^\omega$. Jun 27 '16 at 4:50
• 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 $C$. Jun 27 '16 at 4:54

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.
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).
• 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.) Jun 27 '16 at 16:05