10 Typical compact sets

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1 Tel Aviv University, 2013 Measure and category Typical compact sets 10a Covering, packing, volume, and dimension b A space of compact sets c Dimensions of typical sets d Besicovitch sets Hints to exercises Index a Covering, packing, volume, and dimension Covering numbers N ε (X) and packing numbers M ε (X) (natural numbers or ) are defined for ε > 0 and a metric space X by 1 N ε (X) = inf{ A : x X a A ρ(x, a) < ε}, M ε (X) = sup{ B : b 1, b 2 B (b 1 b 2 = ρ(b 1, b 2 ) ε)} ; here A, B run over all finite subsets of X, and... is the number of elements. 10a1 Lemma. M 2ε (X) N ε (X) M ε (X). Proof. M 2ε (X) N ε (X): we have a one-to-one map B A, since ρ(b 1, a 1 ) < ε and ρ(b 2, a 2 ) < ε and b 1 b 2 imply a 1 a 2. N ε (X) M ε (X): if M ε (X) <, we take a maximal B and note that it is a possible A. 10a2 Exercise. The following three conditions on a metric space X are equivalent: (a) ε > 0 N ε (X) < ; (b) ε > 0 M ε (X) < ; (c) every sequence (of points of X) has a Cauchy subsequence. 10a3 Corollary. The following three conditions on a complete metric space X are equivalent: (a) ε > 0 N ε (X) < ; (b) ε > 0 M ε (X) < ; (c) X is compact. ε. 1 Not equivalently, but equally well, one may use ρ(x, a) ε in concert with ρ(b 1, b 2 ) >

2 Tel Aviv University, 2013 Measure and category 84 Interestingly, it appears that (a) and (b) do not depend on the choice of a complete metric on a (completely) metrizable space. (However, (0, 1) is homeomorphic to R... ) A subset E of a metric space X is itself a metric space; thus, N ε (E), M ε (E) are well-defined, and 10a1 10a3 apply. Compact subsets of R n are a notable special case. For a cube E R n it is easy to see that N ε (E) C/ε n and M ε (E) c/ε n for some c, C (0, ) (dependent on n but not E). Using the inequality M 2ε (E) N ε (E) we get N ε (E) M ε (E) 1 ε n ; here α β means that cα β Cα for some c, C (0, ). The same holds for every bounded set E R n with nonempty interior (in particular, a ball). For such E we get log N ε (E) log 1/ε n, log M ε (E) log 1/ε n as ε 0 +. Accordingly, one defines the lower and upper Minkowski(-Bouligand) dimension 1 dim M (E) = lim inf log N ε (E) log 1/ε, dim log N ε (E) M(E) = lim sup log 1/ε, and if these are equal, the Minkowski dimension dim M (E) is equal to both. (Equivalently, M ε (E) may be used.) Now we turn to a bounded set E R n and Lebesgue measure of its closed ε-neighborhood E +ε. 10a4 Lemma. For all ε > 0, M 2ε (E) m(e +ε) C n ε n 2 n N ε (E), where C n is the volume of the n-dimensional unit ball. Proof. First, N ε (E) = A, E A +ε, thus E +ε A +2ε and m(e +ε ) C n (2ε) n A. Second, M 2ε (E) = B, B E, thus m(e +ε ) C n ε n B. 1 Also box (counting) dimension, and (for the upper dimension) entropy dimension, Kolmogorov dimension, Kolmogorov capacity. By the way, the well-known Hausdorff dimension never exceeds the lower Minkowski dimension.

3 Tel Aviv University, 2013 Measure and category 85 10a5 Corollary. If dim M (E) exists, then log m(e +ε ) n dim M (E) as ε 0+ ; and in every case, n dim M (E) = lim inf log m(e +ε ) lim sup log m(e +ε ) = n dim M (E). 10b A space of compact sets Given a metric space X, we denote by K(X) the set of all nonempty compact subsets of X. For each K K(X) we introduce its distance function d K : X [0, ) by d K (x) = dist(x, K) = min ρ(x, y). y K Note that d K (x) d K (y) ρ(x, y); K = {x : d K (x) = 0}; and K 1 K 2 d K1 d K2. Also, d K1 K 2 = min(d K1, d K2 ), while the evident inequality d K1 K 2 max(d K1, d K2 ) is generally strict (even if K 1 K 2 ). We endow K(X) with the Hausdorff metric d H, d H (K 1, K 2 ) = d K1 d K2 = sup d K1 (x) d K2 (x). x X 10b1 Exercise. (a) sup x X ( dk1 (x) d K2 (x) ) = max x K2 d K1 (x); (b) d H (K 1, K 2 ) = max ( max x K2 d K1 (x), max x K1 d K2 (x) ). Denoting K +ε = {x : d K (x) ε} we have d H (K 1, K 2 ) = min{ε : K 1 (K 2 ) +ε K 2 (K 1 ) +ε }. Here is a metric-free description 1 of the topology on K(X). 10b2 Exercise. The following two conditions on K, K 1, K 2, K(X) are equivalent: (a) K n K; (b) for every open U X, (b1) if K U then K n U for all n large enough; (b2) if K U then K n U for all n large enough. 1 So-called Vietoris topology on K(X).

4 Tel Aviv University, 2013 Measure and category 86 If two metrics on X are equivalent then the two corresponding Hausdorff metrics on K(X) are equivalent. Thus, a metrizable space K(X) is welldefined for every metrizable space X. 10b3 Exercise. If X is separable then K(X) is separable. We return to a metric space X. 10b4 Exercise. Let K 1, K 2, K(X), K 1 K 2..., then d Kn where K = n K n. d K 10b5 Exercise. Let K 1, K 2, K(X) be such that the set K = Cl(K 1 K 2... ) is compact; then (a) d K = inf n d Kn ; (b) lim inf n d Kn = d K where K = n Cl(K n K n+1... ) is the socalled topological upper limit of K n ; (c) K lim sup n K n, and this inequality is generally strict. 10b6 Exercise. Let K 1, K 2, K(X) and ε n 0. If X is complete then the set (K n ) +εn is compact. 10b7 Proposition. If X is complete then K(X) is complete. n Proof. Given a Cauchy sequence K 1, K 2, K(X), we take ε n 0 such that d H (K n, K n+k ) ε n, then K n+k (K n ) +εn, therefore K n (K 1 K n ) +εn. n n The latter is compact by 10b6. Thus, Cl( n K n ) is compact. By 10b5, lim inf n d Kn = d K. However, d Kn are a Cauchy sequence; thus d Kn d K 0, that is, K n K. 10b8 Corollary. (a) If X is completely metrizable then K(X) is completely metrizable; (b) if X is Polish then K(X) is Polish.

5 Tel Aviv University, 2013 Measure and category 87 10c Dimensions of typical sets 10c1 Theorem. 1 Quasi all K K(R n ) satisfy dim M (K) = 0, dim M (K) = n. This fact should be another manifestation of the phenomenon seen in 1e1(b). I wonder, is there a general theorem that implies both as special cases? 10c2 Corollary. Quasi all K K(R n ) are null sets, and (therefore) nowhere dense. Strangely, small sets are the majority... In order to avoid the question of continuity of m(k +ε ) in K (for a fixed ε) we introduce f ε (K) = 1 ε ε 0 m(k +a ) da = R n ( 1 1 ε d K) + dm. If K m K then d Km d K uniformly, thus f ε (K m ) f ε (K) (since the relevant part of R n is bounded). It means that f ε : K(R n ) (0, ) is continuous. By monotonicity, 1 2 m(k +ε/2) 1 ε ε therefore ( ( 1 ) ) log m(k +ε/2 ) 1 + O log 1/ε log(ε/2) In combination with 10a5 it gives 0 m(k +a ) da m(k +ε ), log f ε(k) log m(k +ε). n dim M (K) as ε 0+ if dim M (K) exists; and in every case, (10c3) n dim M (K) = lim inf lim sup = n dim M (K). 1 P.M. Gruber (1989) Dimension and structure of typical compact sets, continua and curves, Monatshefte für Mathematik 108,

6 Tel Aviv University, 2013 Measure and category 88 Proof of Theorem 10c1. Finite K are dense in K(R n ) and are of Minkowski dimension 0. Thus, n (as ε 0) on a dense set. By 5b9, lim sup n (thus, = n) for quasi all K K(R n ). By (10c3), dim M (K) = 0 for quasi all K. On the other hand, sets of Minkowski dimension n are also dense (try finite unions of balls; or even a ball plus a finite set). Thus, on a dense set. By 5b9 (again), lim inf 0 (as ε 0+) 0 (thus, = 0) for quasi all K K(R n ). By (10c3), dim M (K) = n for quasi all K. 10c4 Exercise. Quasi all K K(R n ) satisfy where U runs over all open sets. U ( K U = dim M (K U) = n ), 10c5 Corollary. Quasi all K K(R n ) are perfect sets. By Theorem 10c1, quasi all K K(R n ) satisfy α > 0 On the other hand, lim inf (10c6) m(k +ε ) 0, m(k +ε ) m(k +ε ) = 0, lim sup =. ε n α ε α m(k +ε ) ε n (as ε 0+) since m(k) = 0, and K is infinite (and m(e +ε) C ε n n E for finite E). A more detailed analysis leads to a stronger result. We recall the anti-egorov phenomenon discussed in Sect. 7a (recall also 9b2) and give it a name. 1 1 Not a standard terminology.

7 Tel Aviv University, 2013 Measure and category 89 10c7 Definition. Let X be a completely metrizable space, and f 1, f 2, : X (0, ). We say that quasi everywhere f n 0 with nothing to spare if {x : f n (x) 0} is comeager, but {x : A n f n (x) 0} is meager whenever A n. Similarly, we say that quasi everywhere f n with nothing to spare if {x : f n (x) } is comeager, but {x : a n f n (x) } is meager whenever a n 0. (And the same for ε 0+ instead of n ). 10c8 Theorem. For quasi all K K(R n ), m(k +ε ) 0 (as ε 0+) with nothing to spare, m(k +ε ) ε n (as ε 0+) with nothing to spare. Proof. Convergence is already established, see (10c6); nothing to spare will be proved. Using 10a4 (and 10a1) we reformulate it equivalently as follows: N ε (K) (as ε 0+) with nothing to spare, ε n N ε (K) 0 (as ε 0+) with nothing to spare. The first relation. Let a(ε) 0; we have to prove that {K : a(ε)n ε (K) } is meager. It is sufficient to prove for arbitrary ε 0 that the set S = {K : ε ε 0 a(ε)n ε (K) 1} is nowhere dense. We ll prove that a finite K 0 K(R n ) cannot belong to the closure of S; this is sufficient, since these K 0 are dense in K(R n ). We take ε ε 0 such that a(ε) K 0 < 1. Every K such that d H (K 0, K) ε satisfies N ε (K) K 0, thus, a(ε)n ε (K) < 1. We see that S misses the ε-neighborhood of K 0. The second relation. Let A(ε) ; we have to prove that {K : A(ε)ε n N ε (K) 0} is meager. It is sufficient to prove for arbitrary ε 0 that the set S = {K : ε ε 0 A(ε)ε n N ε (K) 1} is nowhere dense. We ll prove that K 0 K(R n ) with nonempty interior cannot belong to the closure of S; this is sufficient, since these K 0 are dense in K(R n ). There exists c > 0 such that ε n N ε (K 0 ) c for all ε. We take ε ε 0 such that A(ε)c > 2 n. Every K such that d H (K 0, K) ε satisfies N ε (K) N 2ε (K 0 ), thus, A(ε)ε n N ε (K) > 1. We see that S misses the ε-neighborhood of K 0.

8 Tel Aviv University, 2013 Measure and category 90 In this sense, ε n m(k +ε ) 1 and 1 N ε (K) 1 ε n (as ε +0) with nothing to spare. 10d Besicovitch sets By Besicovitch sets we mean planar compact sets that contain unit line segments in every direction. We want to minimize the area of such set. It was conjectured that the deltoid is optimal, of area π/8. Amazingly, the minimal area is zero! 1 10d1 Exercise. The set B of all Besicovitch sets is a closed subset of K(R 2 ). Deltoid. Its boundary is { 1 2 eit e 2it : 0 t 2π}. Thus we may talk about typical Besicovitch sets. 10d2 Theorem. 2 Quasi all Besicovitch sets are null sets. A spectacular manifestation of the tendency small sets are the majority! On K(R n ), Lebesgue measure K m(k) is an upper semicontinuous function. Moreover, K µ(k) is upper semicontinuous on K(X) for every locally finite measure µ on X. Proof: K +ε K, therefore µ(k +ε ) µ(k); if µ(k) < a then µ(k +ε ) < a for a small ε, and µ(k 1 ) < a whenever d H (K 1, K) ε. Thus, in order to prove Theorem 10d2 it is sufficient to prove that {K B : m(k) < ε} is dense in B for all ε. We divide the set of all directions in two subsets: these closer to the x axis, and to y axis. We have B = B 1 B 2 where B 1 consists of sets that contain unit line segment in every direction closer to the x axis, and B 2 to y. It is sufficient to prove that {K B 1 : m(k) < ε} is dense in B 1, since if K 1 B 1, K 2 B 2, m(k 1 ) < ε, m(k 2 ) < ε, then K 1 K 2 B, m(k 1 K 2 ) < 2ε and d H (K 1 K 2, K) max ( d H (K 1, K), d H (K 2, K) ) (think, why). We ll prove a stronger claim: quasi all sets of B 1 are null sets. 1 Some authors define Besicovitch sets as null sets with that property. 2 T.W. Körner (2003) Besicovitch via Baire, Studia Math. 158, See also Sect. 4.5 in book: E.M. Stein and R. Shakarchi, Functional analysis, Princeton 2011.

9 Tel Aviv University, 2013 Measure and category 91 To this end it is sufficient to prove that (10d3) m ( K ([a, a + ε] R) ) < Aε 2 for quasi all K B 1, whenever a R, ε > 0; here A is some absolute constant. Given a and ε, we have an open set of such K B 1, and we ll prove that this open set is dense in B 1. Given K B 1 and N, we seek K 1 B 1 satisfying (10d3) and d H (K 1, K) = O(1/N). 1 We take directions d 1,..., d N, closer to the x axis, that are O(1/N)- dense among all such directions. For each i = 1,..., N we choose a unit line segment S i K in the direction d k. Rotating S i by angles ±O(1/N) around one of its points (specified below) we get S i K(R 2 ) such that the set S = S 1 S N belongs to B 1 and S K +O(1/N). We choose the center of the rotation as the point of S i most close to the line {a + ε 2 } R (be it on the line or not). Then m( S i ) = O(ε 2 /N) (think, why), thus m( S) = O(ε 2 ). It remains to take K 1 = S K 0 where K 0 K is a compact null set (even finite, if you like) such that K A +O(1/N) ; indeed, then K (K 1 ) +O(1/N) and K 1 K +O(1/N). Hints to exercises 10a2: (c)= (a): if no A is finite then some B is infinite; (a)= (c): if A is finite then some ε-ball contains x n for infinitely many n. 10b3: try finite (compact) sets. 10b6: use 10a3. 10c4: use a countable basis. 10d1: K n K. Index Besicovitch set, 90 Hausdorff metric, 85 Minkowski dimension, 84 nothing to spare, 89 d H, 85 d K, 85 K, 86 K(X), 85, 86 M ε, 83, 84 N ε, 83, 84 1 All O(1/N) are absolute (I mean, with absolute constants).