Abstract
We introduce the concept of hereditarily non uniformly perfect sets, compact sets for which no compact subset is uniformly perfect, and compare them with the following: Hausdorff dimension zero sets, logarithmic capacity zero sets, Lebesgue 2-dimensional measure zero sets, and porous sets. In particular, we give an example of a compact set in the plane of Hausdorff dimension 2 (and positive logarithmic capacity) which is hereditarily non uniformly perfect.
Highlights
Introduction and resultsVarious types of non-smooth sets arise naturally in many mathematical settings
The effect of reducing the total length of the basic intervals in step 2n by a very small factor of an times those in the previous step 2n − 1, but only doing this sparingly at these times 2n, is to ensure that these sets are thin in the sense of being hereditarily non uniformly perfect (HNUP), but not thin in the sense of Hausdorff dimension as dimH I > 0
We see that by Theorem 4.1(1), Iais HNUP whenever lim inf ak = 0, which as we demonstrate occurs for λm-a.a. a = (a1, a2, . . . ) ∈ X
Summary
(17-19) Replace each circle in (20) by a “discrete circle” of 100,000 points spaced on the given circle to obtain a set that is not porous (at the origin). Since this set is countable it must have Hausdorff dimension zero, logarithmic capacity zero, and be HNUP. Note that by Theorem 1.5 in [10] for each lattice Γ with cusp at infinity its Diophantine set D(Γ) ⊂ Rn is σ-porous and has n-dimensional Lebesgue measure zero. When Γ is a nonuniform Kleinian lattice of Mobius maps, its Diophantine set D(Γ) ⊂ R2 is absolutely winning Since such a set D(Γ) has zero m2 measure, any positive m2-measure compact subset E of well approximable points W (Γ) ⊂ R2 must be HNUP. Because the gaps ek are decreasing, the distance between any basic subintervals of Ik must be separated by a distance at least ek, whether or not these basic subintervals come from the same (“parent”) basic interval from Ik−1
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