As the title suggests, I'm interested in a subset of the real line which is neither meagre nor comeagre in any interval. Does anyone have an example?
Added. See the comments for some discussion about Bernstein sets, which may provide a solution. I don't know anything about these sets, so a detailed answer or reference in that connection would certainly be welcome.
Here are some thoughts of mine and of one of my classmates regarding the problem. My original idea was as follows: Let $\mathcal V$ be a Vitali set (i.e., $\mathcal V$ contains exactly one element from each additive coset of $\mathbb Q$), choose dense subset $S$ of $\mathbb Q$ whose complement $\mathbb Q \setminus S$ in $\mathbb Q$ is also dense, and let $E = \mathcal{V} + S$. Then $E^c = \mathcal{V} + \mathbb{Q}\setminus S$, and it seems like $E$ should have the required property, but I have been unable to prove that this is so.
My classmate had another idea. He worked in $[-1,1]\setminus\{0\}$, seeking a set $E$ with the following properties.
- The complement of $E$ is obtained from $E$ via reflection through the origin, that is, $E^c = -E$.
- If $E \cap I$ is the intersection of $E$ with some dyadic subinterval $I \subset [-1,1]$ of the $k$th generation, then $E$ is similar to this intersection, in the sense that $E = 2^k(E\cap I) + a$ for some real number $a$; that is, the intersection of $E$ with any dyadic subinterval is a scaled translate of $E$.
If I'm not mistaken, it is (relatively) straightforward to verify that any set $E$ satisfying 1. and 2. provides an example. As far as the existence of such a set is concerned, my classmate made an argument, but it was fairly complicated, and I cannot recall it well enough off the top of my head to verify its correctness.