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



Bernstein sets of large cardinality
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).


