I'm looking for concrete examples of a countable Bernstein set and an uncountable Bernstein set. I haven't been able to find or construct any specific examples so far.
Asked
Active
Viewed 87 times
1
-
3I'm fairly certain that Bernstein sets are not Lebesgue-measureable – John White Aug 14 '20 at 20:36
-
1Do you know the usual construction by transfinite induction? It's not exactly explicit, but it is the best you can get by Brian's answer – Alessandro Codenotti Aug 15 '20 at 07:13
-
1In this answer I went through the construction mentioned by Alessandro Codenotti. – Brian M. Scott Aug 15 '20 at 16:34
1 Answers
5
Bernstein sets are non-measurable, so there are no concrete examples: assuming the existence of an inaccessible cardinal, it is consistent with $\mathsf{ZF}$ that all subsets of the reals are measurable.
All Borel sets are measurable, so there cannot be a countable Bernstein set.

Brian M. Scott
- 616,228