A question by Furstenberg

September 25, 2009

Yesterday I was talking about some properties of different dimensions with Furstenburg. Somehow I mentioned Kekaya, and he told me about the following question he has been longing to solve (which is amazingly many similarities to Kekaya):

For set S \in \mathbb{R}^2, if \exists \delta>0 s.t. for all direction \theta, \exists line l with direction \theta s.t. \dim (l \cap S) > \delta . Does it follow that \dim(A) \geq 1 ?

Note that a stronger conjecture would be \dim(A) is at least 1+\delta which when taking \delta = 1 would give a generalization of the 2-dimensional Kekaya. (i.e. instead of having to have a line segment, we only require a 1-dimension set in each direction)

Reviewing the proofs of the 2-dimsional Kekaya, I found they all rely on the fact that the line segment is connected…Hence it might be interesting to even find an answer to the following question:

If A \subseteq \mathbb{R}^2 contains a measure 1 set in every direction, does it follow that \dim(A)=2?

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