Hereditarily Non Uniformly Perfect Sets

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.


Introduction and results
Various types of non-smooth sets arise naturally in many mathematical settings. Julia sets, attractor sets generated from iterated function systems, bifurcation sets in parameter spaces such as the boundary of the Mandelbrot set, and Kakeya sets are all examples of non-smooth sets which are studied intensely. Classical analysis has given way to fractal geometry for the purposes of studying such "pathological" sets. Many tools such as Hausdorff measure and Hausdorff dimension, logarithmic capacity, porosity, Lebesgue measure, and uniform perfectness have been utilized to discern certain fundamental "thickness/thinness" properties of such pathological sets. In this paper we compare how the above tools and properties relate to each other with regard to compact sets in the complex plane C. In particular, we are interested in how the thinness properties with respect to each of these notions relate. We will consider compact sets E ⊂ C and study the following conditions: dim H E = 0, Cap E = 0, m 2 (E) = 0, E is porous. However, these four properties are hereditary properties of thinness in the sense that if E satisfies one of these properties, then all subsets of E also satisfy the same property. Uniform perfectness of a (compact) set E (see definition below) is a property quantifying a uniform thickness near each point of E. To get at a compatible notion of thinness we need to require more than just that the set fails to be uniformly perfect. For example, a set E = F ∪ {z 0 }, where F is uniformly perfect and z 0 / ∈ F , fails to be uniformly perfect (since z 0 is isolated), yet E is "thick" near all of the points of F . Thus to capture the correct idea of being "thin", as a counterpart to the uniformly perfect notion of "thick", we offer the following.
where the infimum is taken over countable covers of E by sets B k such that each d(B k ) < ǫ. For our purposes we will only use dimension functions of the form h(t) = t α (with α > 0) and we shall employ the notation L α (E) = L h (E) and H α (E) = H h (E). We note that with some effort one can show that L h (E) = 0 if and only if H h (E) = 0. A simple calculation (see [4], Ch. 2) shows that if Also found in [4] are the facts that ( 2.2. Uniform perfectness. The notion of uniform perfectness was first introduced by Beardon and Pommerenke in [2]. A set is called perfect if it has no isolated points, whereas the notion of uniform perfectness is a quantified version of perfectness. Uniformly perfect sets can be equivalently defined in many different ways and be defined for sets in R n , but we are mainly concerned with sets in the complex plane C and so we define uniformly perfect sets as follows. Definition 2.1. For w ∈ C and r, R > 0 the true annulus A = Ann(w; r, R) = {z : r < |z − w| < R} is said to separate a set F ⊂ C if F intersects both components of C \ A and F ∩ A = ∅.  Thus a set is uniformly perfect if there is a uniform bound on how "fat" (large modulus) an annulus can be and still separate the set. Uniform perfectness, in a sense, measures how "thick" a set is near each of its points and is related in spirit to many other notions of thickness such as 1 More generally, a compact subset F ⊂ R n is uniformly perfect if there exists a constant c > 0 such that F ∩ Ann(a; cr, r) = ∅ for any a ∈ F and 0 < r < d(F ). Here, of course, the Euclidean metric in R n is used to define the annular region Ann(a; cr, r).
Hausdorff content and dimension, logarithmic capacity and density, Hölder regularity, and positive injectivity radius for Riemann surfaces. For an excellent survey of uniform perfectness and how it relates to these and other such notions see Pommerenke [8] and Sugawa [10]. In particular, we note that uniformly perfect sets are necessarily uncountable.
2.3. Logarithmic capacity. The logarithmic capacity of a compact set E ⊂ C can also be defined in many different, yet equivalent, ways (see [6], Ch. 1). For example, one may define it in terms of the asymptotic behavior of Green's function defined on C \ E with pole at ∞ or in terms of the infimum of an energy integral. We shall employ the following definition given via the transfinite diameter. Set P n = max j<k |z j − z k | where z j ∈ E for j = 1, . . . , n. The sequence D n = P 2/n(n−1) n is non-increasing (see [1], p. 23) and the logarithmic capacity is defined by Cap E = lim n→∞ D n .
We note that capacity is monotone, i.e., Cap F ≤ Cap E whenever F ⊆ E, and Cap C = 0 when C is countable. The countability result follow from the obvious fact that a finite set has zero capacity and that for Borel sets to be porous if there exists a constant c > 0 such that for any point a ∈ E and radius r > 0 there exists a ball B(b, cr) ⊂ B(a, r) such that B(b, cr) ∩ E = ∅. Table 1 Here we present the proofs for the statements made in Table 1. Many of these results are known (being based on classical theorems), but since the arguments are short we include them here for completeness. We also note that most results apply to R n as well, but since that is not the focus of this paper so we do not detail arguments for such.
(2) Example 4.1 below gives a compact set I which is HNUP yet dim H I > 0.
verifies the assertion. The middle third Cantor set is a specific example which is known to have Hausdorff dimension log 2/ log 3.
(5) Example 4.2 below gives a compact set I such that dim H I = 0 yet Cap I > 0.
(6) Example 4.1 below gives a compact set I which is HNUP yet Cap I > 0.
(7) Any compact interval of length L is known to have capacity L/4 (see [11], p. 84), yet has zero two dimensional measure.
(9) Since Hausdorff dimension is monotone, the claim follows from the known fact that uniformly perfect sets necessarily have positive Hausdorff dimension. In fact, Järvi and Vuorinen (see [7], p. 522) have shown that a compact set E ⊂ C is uniformly perfect if and only if there exist constants C > 0 and α > 0 such that L α (E ∩ B(a, r)) ≥ Cr α for all a ∈ E and 0 < r < d(E)/2.
(10) Since uniformly perfect sets necessarily have positive capacity, the claim follows from the monotonicity of capacity. In particular, Pommerenke [8] has shown that a compact set E ⊂ C is uniformly perfect if and only if there exists c > 0 such that Cap (B(z, r) ∩ E) ≥ cr for all z ∈ E and 0 < r < d(E).
(11) Any compact interval has zero two dimensional measure and is trivially uniformly perfect since it is connected (and therefore cannot be separated by any annulus).
(12) Any compact interval of the real line is trivially uniformly perfect since it is connected, yet is also porous with constant c = 1/2 as in (4). [11], p. 58) and so the assertion follows.
(15) Conjecture 4.1 given later in this paper is that the answer is yes.
(16) Given a point a in a porous set E and radius r > 0, we must have Thus we must have m 2 (E) = 0. In fact, a stronger result holds (see [4], Proposition 3.12) which shows that a porous set E ⊂ C must have dim H E < 2.
(20) For a ∈ C and r > 0, let C(a, r) = {z : |z − a| = r}. Consider the set E = ∪ ∞ n=1 C(0, 1/n). The ball B( ) is a ball of largest radius which is contained in both C\E and B(0, 1/n). Since the ratio of the radii 1/2n(n+1) 1/n = 1 2(n+1) → 0 we see that E fails to be porous at the origin. Thus the set E is not porous, yet clearly m 2 (E) = 0.
(17-19) Replace each circle in (20) by a "discrete circle" of 100,000 points equally 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 also that e k+1 = 1 ≤ e k and thus e k strictly decreases to 0. Note that because the gaps e k are decreasing the distance between any basic subintervals of I k must be separated by a distance e k , whether or not these basic subintervals come from the same ("parent") basic interval from I k−1 .
Define set Since by (4.1) the basic intervals in I k are "equally spaced", Example 4.6 (and the discussion on p. 59) of [4] yields that

Remark 4.1. Formula (4.3) is based on the existence of a mass distribution (measure) as constructed
in the discussion prior to Proposition 1.7 of [4]. The construction, however, is known not to work in the full generality as stated in [4], but it does work in the situation we present here since we are in the nice situation that each basic interval is compact. See [5] by N. Falkner, a Mathematical Review that describes how the general construction of the measure can fail and also states some sufficient conditions to ensure success.
Key to our results is part (1) of the following dichotomy.  is an annulus which separates I. We show that R/r ≤ M . Since C separates I, we must have that C separates I k for large k (in particular, whenever A k < R − r any basic interval of I k meeting C would have one of its endpoints in C thus showing that C does not separate I since such an endpoint is also in I). Let n be the smallest positive integer such that C separates I n . Considering C ∩ R and the fact that C separates I n but does not separate I n−1 , we see that R − r ≤ e n . Since B(z, r) contains some basic interval of I n (of length A n ), we have 2r ≥ A n . Hence using (4.1) we see R r ≤ r+en r ≤ 1 + en Remark 4.2. The artificial-looking condition 0 < a ≤ 1/(m + 1) in Theorem 4.1 was chosen to simplify the related proofs (by, in particular, to forcing the gap sizes to be decreasing), but it is worth noting that it can be relaxed to 0 < a < 1/m though we shall omit its more involved proof. . Define a k = a n if k = 2 n for n = 1, 2, . . . and a k = a otherwise. Then the set I given in (4.2) satisfies the following: We note that if we select instead a k = a for all k then by (4.3) dim H I would still equal log m − log a and by Theorem 4.1(2) I would be uniformly perfect. Thus by reducing the total length of basic intervals in the k = 2 n step by a very small factor (a n ) times the total length of the basic intervals in the k = 2 n − 1 step, but doing so only sparingly (for k = 2 n ), results in thinning out the set in a HNUP sense, but does not thin out the set in a Hausdorff dimension sense.
proof of (a) in Example 4.1. Let k be a positive integer and let n 0 be the integer such that 2 n0 ≤ k < 2 n0+1 . Thus we have log k proof of (b) in Example 4.1. This follows from (1) in Table 1.
proof of (c) in Example 4.1. This follows from Theorem 4.1(1) since a 2 n = a n → 0.
The following result shows that we have lots and lots of examples where Iā is HNUP and has positive Hausdorff dimension. Proof. Considering the i.i.d. random variables a n we note that the strong law of large numbers (or the ergodicity of the shift map on X with respect toλ m ) applied to log a n shows that for λ m -a.a. a = (a 1 , a 2 , . . . ) ∈ X we have 1 k log(a 1 . . . a k ) = 1 k k j=1 log a j → E(log a 1 ) = (m+1) Also, we see that by Theorem 4.1(1), Iā is HNUP whenever lim inf a k = 0, which as we now demonstrate occurs for λ m -a.a. ā = (a 1 , a 2 , . . . ) ∈ X. For each positive integer n, let D n be the set ofā such that a j > 1 n for all j and noteλ m (D n ) = 0. Sinceλ m is invariant under the shift map σ, we have that for any positive integer p, proof of (b) in Example 4.2. Let E k be the set of 2 k+1 endpoints of the 2 k basic intervals of I k . For disctinct points z, z ′ ∈ I, let µ(z, z ′ ) = max{k : z and z ′ lie in the same basic interval of I k }. Thus for z, z ′ ∈ I, we see that µ(z, z ′ ) = 0 implies that e 1 ≤ |z − z ′ | ≤ A 0 , and, in general, since the gaps Given z ∈ E k and ℓ ∈ {0, . . . , k} there are exactly 2 k−ℓ points z ′ ∈ E k with µ(z, z ′ ) = ℓ and so for fixed z ∈ E k we have Thus , and so We show Cap I > 0 by using (4.1) to assert e k+1 ≥ BA k and then computing proof of (c) in Example 4.2. This follows from Theorem 4.1(1) since a k = a k → 0.  a ∈ E. Hence if m 2 (E) > 0 it seems that for a choice of a and r the set B(a, r) ∩ E must be so "thick" as to contain a uniformly perfect subset. However, carefully extracting such a subset has eluded the authors.
Remark 4.6. We note that the 1-dimensional version of Conjecture 4.1 (i.e., if E ⊂ R is HNUP, then it has one-dimensional measure zero) seems like it would be equivalent in the sense that resolution of one version would, through the same techniques, lead to a resolution of the other.  Proof. Select a subsequence a kn such that a kn → 0. Consider a point (x, y) ∈ E. Now consider a basic interval J x of I k (in the construction of Iā) containing x and a basic interval J y of I k containing y. As before we let x Jx denotes the midpoint of J x , x Jy denotes the midpoint of J y , r k = A k /2, and R k = r k + e k . Hence we see that is a (square shaped) annulus that separates E. The picture quickly shows that S contains the true annulus C = Ann((x Jx , y Jy ); √ 2r k , R k ) which also separates E.
Since as shown in the proof of Theorem 4.1(1), R kn /r kn → +∞ (with R kn → 0), we see that for each point (x, y) ∈ E there exists an annulus C of arbitrarily large modulus with arbitrarily small outer radius which separates E and has (x, y) in the bounded component of the complement of C.
Hence, (x, y) can never be a point in a uniformly perfect subset of E and thus no subset of E can be uniformly perfect.
proof of Theorem 1.3. Choose 0 < ρ n < 1 such that ρ n strictly decreases to 0. Consider the annuli B n = Ann(0; ρ n+1 , ρ n ). For each m = 2, 3, . . . let E ′ m be a translated and scaled down copy of E m such that E ′ m ⊂ B 2m . Note that each E ′ m is HNUP because each E m is HNUP and uniformly perfect sets remain uniformly perfect after translation and scaling. of Proposition 4.1 worked because of the high degree of uniformity in the sets W m , that is, given (x, y) ∈ W m × W m , for each basic set J x containing x there was a basic set J y containing y of the (exact) same size and both J x and J y had the exact same gap e n around them. We do not have such a uniform structure in W × W . It would be interesting, however, to be able to settle the following related question.
Open Question: Must E × E ′ be a HNUP subset of C when each E and E ′ are HNUP subsets of R?