Random intersections of thick Cantor sets Roger L. Kraft roger@calumet.purdue.edu Kraft_Roger_L 1$, then it is well known that the difference set $C_{1}-C_{2}$ is a disjoint union of closed intervals. B.~Williams showed that for some $t\in \interior (C_{1}-C_{2})$, it may be that $C_{1}\cap (C_{2}+t)$ is as small as a single point. However, the author previously showed that generically, the other extreme is true; $C_{1}\cap (C_{2}+t)$ contains a Cantor set for all $t$ in a generic subset of $C_{1}-C_{2}$. This paper shows that small intersections of thick Cantor sets are also rare in the sense of Lebesgue measure; if $\tau _{1}\tau _{2}>1$, then $C_{1}\cap (C_{2}+t)$ contains a Cantor set for almost all $t$ in $C_{1}-C_{2}$.]]> , $C_{2}$ be Cantor sets embedded in the real line, and let $\tau _{1}$, $\tau _{2}$ be their respective thicknesses. If $\tau _{1}\tau _{2}>1$, then it is well known that the difference set $C_{1}-C_{2}$ is a disjoint union of closed intervals. B. Williams showed that for some $t\in \operatorname{int}(C_{1}-C_{2})$, it may be that $C_{1}\cap (C_{2}+t)$ is as small as a single point. However, the author previously showed that generically, the other extreme is true; $C_{1}\cap (C_{2}+t)$ contains a Cantor set for all $t$ in a generic subset of $C_{1}-C_{2}$. This paper shows that small intersections of thick Cantor sets are also rare in the sense of Lebesgue measure; if $\tau _{1}\tau _{2}>1$, then $C_{1}\cap (C_{2}+t)$ contains a Cantor set for almost all $t$ in $C_{1}-C_{2}$.

]]> Let be Cantor sets embedded in the real line, and let be their respective thicknesses. If then it is well known that the difference set is a disjoint union of closed intervals. B. Williams showed that for some it may be that is as small as a single point. However, the author previously showed that generically, the other extreme is true; contains a Cantor set for all in a generic subset of . This paper shows that small intersections of thick Cantor sets are also rare in the sense of Lebesgue measure; if then contains a Cantor set for almost all in . GH GH J. Guckenheimer, P. Holmes, Nonlinear Oscillations, Dynamical Systems, and Bifurcation of Vector Fields , Springer-Verlag, New York, 1983. 93e:58046 HKY HKY B.R. Hunt, I. Kan, J. A. Yorke, When Cantor sets intersect thickly , Trans. Amer. Math Soc. 339 (2) (1993), 869--888. 94f:28010 KP KP R. Kenyon, Y. Peres, Intersecting random translates of invariant Cantor sets , Invent. Math. 104 (3) (1991), 601--629. 92g:28018 K1 K1 R. L. Kraft, Intersections of Thick Cantor Sets , Mem. Amer. Math. Soc. 97 (468) (1992). 92i:28010 K2 K2 , One point intersections of middle- Cantor sets , Ergodic Theory Dynam. Systems 14 (3) (1994), 537--549. 95i:54050 K3 K3 , What's the difference between Cantor sets , Amer. Math. Monthly 101 (7) (1994), 640--650. 95f:04006 K4 K4 , A golden Cantor set , Amer. Math. Monthly 105 (8) (1998). 99:01 KKY KKY I. Kan, H. Koak, J. Yorke, Antimonotonicity: concurrent creation and annihilation of periodic orbits , Ann. of Math. (2) 136 (2) (1992), 219--252. 94c:58135 MO MO P. Mendes, F. Oliveira, On the topological structure of the arithmetic sum of two Cantor sets , Nonlinearity 7 (2) (1994), 329--343. 95j:58123 N1 N1 S. E. Newhouse, The abundance of wild hyperbolic sets and non--smooth stable sets for diffeomorphisms , Publ. Math. IHES 50 (1979), 101--151. 82e:58067 N2 N2 , Lectures on dynamical systems , Dynamical Systems, C. I. M. E. Lectures, Bressanone, Italy, June, 1978, Progress in Mathematics, No. 8, Birkhauser, Boston, 1980, pp. 1--114. 81m:58028 PT1 PT1 J. Palis, F. Takens, Hyperbolicity and the creation of homoclinic orbits , Ann. of Math. (2) 125 (2) (1987), 337--374. 89b:58118 PT2 PT2 , Hyperbolicity and sensitive chaotic dynamics at homoclinic bifurcations: Fractal dimensions and infinitely many attractors , Cambridge Studies in Advanced Mathematics, 35, Cambridge University Press, Cambridge, 1993. 94h:58129 PS PS Y. Peres, B. Solomyak, Self-similar measures and intersections of Cantor sets , Trans. Amer. Math Soc. 350 (10) (1998), 4065--4087. 98m:26009 R R D. Ruelle, Elements of Differentiable Dynamics and Bifurcation Theory , Academic Press, New York, 1989. 90f:58048 S S A. Sannami, An example of a regular Cantor set whose difference set is a Cantor set with positive measure , Hokkaido Math. J. 21 (1) (1992), 7--24. 93c:58116 W W R. F. Williams, How big is the intersection of two thick Cantor sets , Continuum Theory and Dynamical Systems (M. Brown, ed.), Proc. Joint Summer Research Conference on Continua and Dynamics (Arcata, California, 1989), Amer. Math. Soc., Providence, R.I., 1991. 92f:58116 WZ WZ R. Wheeden, A. Zygmund, Measure and Integral: An Introduction to Real Analysis , Marcel Dekker, Inc., New York, 1977. 58:11295

[GH]
J. Guckenheimer, P. Holmes, Nonlinear Oscillations, Dynamical Systems, and Bifurcation of Vector Fields, Springer-Verlag, New York, 1983. MR 93e:58046

[HKY]
B.R. Hunt, I. Kan, J. A. Yorke, When Cantor sets intersect thickly, Trans. Amer. Math Soc. 339 (2) (1993), 869-888. MR 94f:28010

[KP]
R. Kenyon, Y. Peres, Intersecting random translates of invariant Cantor sets, Invent. Math. 104 (3) (1991), 601-629. MR 92g:28018

[K1]
R. L. Kraft, Intersections of Thick Cantor Sets, Mem. Amer. Math. Soc. 97 (468) (1992). MR 92i:28010

[K2]
-, One point intersections of middle-$\alpha $ Cantor sets, Ergodic Theory Dynam. Systems 14 (3) (1994), 537-549. MR 95i:54050

[K3]
-, What's the difference between Cantor sets, Amer. Math. Monthly 101 (7) (1994), 640-650. MR 95f:04006

[K4]
-, A golden Cantor set, Amer. Math. Monthly 105 (8) (1998). CMP 99:01

[KKY]
I. Kan, H. Ko[??]ak, J. Yorke, Antimonotonicity: concurrent creation and annihilation of periodic orbits, Ann. of Math. (2) 136 (2) (1992), 219-252. MR 94c:58135

[MO]
P. Mendes, F. Oliveira, On the topological structure of the arithmetic sum of two Cantor sets, Nonlinearity 7 (2) (1994), 329-343. MR 95j:58123

[N1]
S. E. Newhouse, The abundance of wild hyperbolic sets and non-smooth stable sets for diffeomorphisms, Publ. Math. IHES 50 (1979), 101-151. MR 82e:58067

[N2]
-, Lectures on dynamical systems, Dynamical Systems, C. I. M. E. Lectures, Bressanone, Italy, June, 1978, Progress in Mathematics, No. 8, Birkhäuser, Boston, 1980, pp. 1-114. MR 81m:58028

[PT1]
J. Palis, F. Takens, Hyperbolicity and the creation of homoclinic orbits, Ann. of Math. (2) 125 (2) (1987), 337-374. MR 89b:58118

[PT2]
-, Hyperbolicity and sensitive chaotic dynamics at homoclinic bifurcations: Fractal dimensions and infinitely many attractors, Cambridge Studies in Advanced Mathematics, 35, Cambridge University Press, Cambridge, 1993. MR 94h:58129

[PS]
Y. Peres, B. Solomyak, Self-similar measures and intersections of Cantor sets, Trans. Amer. Math Soc. 350 (10) (1998), 4065-4087. MR 98m:26009

[R]
D. Ruelle, Elements of Differentiable Dynamics and Bifurcation Theory, Academic Press, New York, 1989. MR 90f:58048

[S]
A. Sannami, An example of a regular Cantor set whose difference set is a Cantor set with positive measure, Hokkaido Math. J. 21 (1) (1992), 7-24. MR 93c:58116

[W]
R. F. Williams, How big is the intersection of two thick Cantor sets?, Continuum Theory and Dynamical Systems (M. Brown, ed.), Proc. Joint Summer Research Conference on Continua and Dynamics (Arcata, California, 1989), Amer. Math. Soc., Providence, R.I., 1991. MR 92f:58116

[WZ]
R. Wheeden, A. Zygmund, Measure and Integral: An Introduction to Real Analysis, Marcel Dekker, Inc., New York, 1977. MR 58:11295
]]> 1999 American Mathematical Society Transactions of the American Mathematical Society Trans. Amer. Math. Soc. tran 352 03 19990920 October 14, 1997 September 20, 1999 Cantor sets difference sets thickness 1315 1315-1328 14 S0002-9947-99-02464-2 10.1090/S0002-9947-99-02464-2 0002-9947 1088-6850 English 28A80 58F99 1991 http://www.ams.org/jourcgi/jour-getitem?pii=S0002-9947-99-02464-2 If C 1 , C 2 are Cantor sets embedded in the real line, then their difference set is equation C 1-C 2 ,x-yxC 1 and yC 2 , . equation The difference set has another, more dynamical, definition as equation C 1-C 2 , t C 1(C 2 t) , , equation where C 2 t , x t xC 2 , is the translation of C 2 by the amount t . There are two reasons to say that the second definition is dynamical. First, it gives a dynamic way of visualizing the difference set; if we think of C 1 as being fixed in the real line and think of C 2 as sliding across C 1 with unit speed, then C 1-C 2 can be thought of as giving those times when the moving copy of C 2 intersects C 1 . Second, it has become a tool for studying dynamical systems. One Cantor set sliding over another one comes up in various studies of homoclinic phenomena, such as infinitely many sinks, N1 , antimonotonicity, KKY , and -explosions, PT1 ; for an elementary explanation of this, see pp. 331--342 GH or pp. 110--115 R. This has led to a number of problems and results of the following form: Given conditions on the sizes of C 1 and C 2 , what can be said of the sizes of either C 1-C 2 , or C 1(C 2 t) for tC 1-C 2 . A wide variety of notions of size have been used, such as cardinality, topology, measure, Hausdorff dimension, limit capacity, and thickness; see for example HKY , KP , MO , PT2 , PS , , and . In this paper we will be concerned with the thickness of C 1 and C 2 , and our conclusion will be about the topology of C 1(C 2 t) for almost every tC 1-C 2 . It is not hard to show that the difference set of two Cantor sets C 1 , C 2 is always a compact, perfect set. So the simplest structure that we can expect C 1-C 2 to have is the disjoint union of closed intervals. There is a condition we can put on C 1 and C 2 that will guarantee this; if 1 , 2 are the thicknesses of C 1 , C 2 , and if 1 2 1 , then C 1-C 2 is a disjoint union of closed intervals. What about the size of C 1(C 2 t) for tC 1-C 2 In it was shown that even when 1 2 1 , it is possible that C 1(C 2 t) can be as small as a single point for some t(C 1-C 2) . But in Chapter 3 K1 , it was shown that this is exceptional, at least in the sense of category, and that in fact the other extreme is the case; if 1 2 1 , then C 1(C 2 t) contains a Cantor set for all t in a generic subset of C 1-C 2 . Our main result in this paper is to prove a similar result for Lebesgue measure. thm Let C 1 , C 2 be Cantor sets embedded in the real line and let 1 , 2 be their respective thicknesses. If 1 2 1 , then C 1(C 2 t) contains a Cantor set for almost all tC 1-C 2 . thm It is worth mentioning here that, in , HKY , and K1 , conditions are given on 1 and 2 so that C 1(C 2 t) contains a Cantor set for all t(C 1-C 2) . Before proving Theorem 1, let us look at the definition of thickness and see how it is used. If C is a Cantor set embedded in the real line, then the complement of C is a disjoint union of open intervals. We call the components of the complement of C the gaps of C . Let U n n 1 be an ordering of the bounded gaps of C by decreasing length, so U n 1 U n , where U denotes the Lebesgue measure of U . Let I 1 denote the smallest closed interval containing C . For n 1 , let I n I 1 ( i 1 n-1 U i) . Note that I n has n components. Let A n denote the component of I n that contains U n . Let L n and R n denote the left and right components of A nU n . Then the thickness of C is defined by equation ( C ) n L n U n , R n U n . equation This definition of thickness is from ; in both and pp. 15--16 K1 it is shown that (i) this definition does not depend on the choice of an ordering for the gaps of C in the case when U n 1 U n for some n , and (ii) this definition is equivalent to the usual definition of thickness (e.g., pp. 99--100 N2 ). Thickness gives us a way of measuring the size of Cantor sets embedded in the real line. The larger the thickness, the bigger'' the Cantor set. So for example, as a consequence of the next lemma the condition 1 2 1 implies that C 1 and C 2 are big enough that their difference set is large in the sense that C 1-C 2 is a disjoint union of closed intervals. lem 1 lem lem:2 Let C 1 , C 2 be Cantor sets embedded in the real line, with thicknesses 1 , 2 . If 1 2 1 and neither C 1 nor C 2 is contained in a gap of the other, then C 1C 2 . lem This lemma is often referred to as the Gap Lemma, p. 63 PT2 . There is a slightly stronger version of the Gap Lemma that uses the notion of an overlapped point in the intersection of two Cantor sets. This is a simple, but useful, definition from pp. 17--18 K1 . Suppose that xC 1C 2 . Let U n n 1 and V n n 1 denote the bounded gaps, and let I 1 , J 1 denote the convex hulls, of C 1 and C 2 . Let A n and B n denote the components of I 1( i 1 n-1 U i) and J 1( i 1 n-1 V i) , respectively, that contain x . Then x is an overlapped point from C 1C 2 if A nB n has nonempty interior for all n . To put this another way, if xC 1C 2 , then x is not an overlapped point if and only if there is an n such that A nB n x , i.e., A n and B n look like the following picture. equation .5 A n .7 x 1.2 1.2 .8 1.5 B n equation Now we can state the slightly stronger version of the Gap Lemma. lem lem:3 Let C 1 , C 2 be Cantor sets embedded in the real line, with thicknesses 1 , 2 . If 1 2 1 and neither C 1 nor C 2 is contained in the closure of a gap of the other, then C 1C 2 contains an overlapped point. lem This version of the Gap Lemma implies that C 1-C 2 is a disjoint union of closed intervals, and that C 1(C 2 t) contains an overlapped point for all t( C 1-C 2 ) . It is not hard to see that C 1(C 2 t) contains only non-overlapped points when t is a boundary point of C 1-C 2 . We say that Cantor sets C 1 and C 2 are interweaved if neither C 1 nor C 2 is contained in the closure of a gap of the other. Here is a sketch of the proof of the Gap Lemma. Let U n n 1 and V n n 1 denote the bounded gaps, and let I 1 , J 1 denote the convex hulls, of C 1 and C 2 , respectively. The key idea is that, since 1 2 1 , we cannot have the following picture of I 1U 1 and J 1V 1 . equation .5 L 1 1.0 U 1 .8 R 1 1.2 .9 1.0 1.2857 .8 1.1 .7 1.5 L 1 .9 V 1 1.0 R 1 equation So it must be that the intersection of I 1U 1 and J 1V 1 has nonempty interior. A careful induction argument, based on the above idea, gives that the intersection of I 1( i 1 n U i) and J 1( i 1 n V i) has nonempty interior for all n 1 ; this implies that C 1C 2 contains an overlapped point. Notice that if the hypothesis 1 2 1 is replaced with 1 21 , then we can still conclude that C 1C 2 , but we cannot conclude that C 1C 2 contains an overlapped point. If C is a Cantor set embedded in the real line, then the components of each I n are called the bridges of C ; if B is any bridge of C , then BC is called a segment of C . Clearly any segment of C is also a Cantor set. As a consequence of the definition of thickness, we have the following simple lemma, p. 16 K1 , which will allow us to apply the Gap Lemma locally.'' lem lem:4 Let C be a Cantor set embedded in the real line with thicknesses . If C' is any segment of C , then the thickness of C' is greater than or equal to . lem The main result we need in order to prove Theorem 1 is the following lemma, which at first glance seems to be only slightly stronger than the Gap Lemma. lem lem:5 Let C 1 , C 2 be Cantor sets embedded in the real line, with thicknesses 1 , 2 . If 1 2 1 , then C 1(C 2 t) contains at least two overlapped points for almost all tC 1-C 2 . lem Before proving this lemma, let us see how it is used to prove Theorem 1. proof Proof of Theorem 1 Let C 1,n n 1 be any ordering of all the segments of C 1 , and let C 2,n n 1 be any ordering of all the segments of C 2 . Then, by Lemmas 4 and 5, for any i and j there is a set E ij C 1,i -C 2,j of measure zero, such that C 1,i (C 2,j t) contains at least two overlapped points for all t(C 1,i -C 2,j )E ij . Let E i,j E ij . So then E has measure zero, and EC 1-C 2 . Using terminology from p. 20 K1 , if t(C 1-C 2)E , then C 1(C 2 t) has no isolated overlapped points . An overlapped point is isolated if there is a neighborhood of it which contains no other overlapped points. In pp. 20--21 K1 it is shown that if the intersection of two Cantor sets does not contain isolated overlapped points, then the intersection must contain a Cantor set. But here we will sketch a proof that if t(C 1-C 2)E , then C 1(C 2 t) contains a Cantor set. Let t(C 1-C 2) E . Then C 1(C 2 t) contains at least two overlapped points, so let x , y be distinct overlapped points in C 1(C 2 t) . Choose integers i 1 , j 1 large enough so that x and y are in distinct components of I i 1 and J j 1 t . Let K 1 K 1,1 K 1,2 denote the two components of I i 1 that contain x and y , and let L 1 L 1,1 L 1,2 denote the two components of J j 1 t that contain x and y . Since t(C 1-C 2) E , K 1,1 L 1,1 contains at least two overlapped points from C 1(C 2 t) , and so does K 1,2 L 1,2 . Now choose integers i 2 i 1 and j 2 j 1 large enough so that these four overlapped points are in distinct components of I i 2 and J j 2 t , and let K 2 1 4 K 2, and L 2 1 4 L 2, denote these components. In general, suppose we are given integers i n and j n , and 2 n distinct components K n 1 2 n K n, from I i n , and 2 n distinct components L n 1 2 n L n, from J j n t , such that each of K n, L n, contains an overlapped point from C 1(C 2 t) . Then, since t(C 1-C 2) E , each of K n, L n, actually contains two overlapped points from C 1(C 2 t) . So we can choose integers i n 1 i n and j n 1 j n large enough so that these 2 n 1 overlapped points are contained in 2 n 1 distinct components K n 1 1 2 n 1 K n 1, from I i (n 1) , and L n 1 1 2 n 1 L n 1, from J j (n 1) t . So for every n1 , the set K nL n has 2 n components, (K nL n)(I i n J j n ) , and (K n 1 L n 1 )(K nL n) . Finally, the set n 1 (K nL n) is a Cantor set contained in C 1(C 2 t) . proof Now we shall begin working on the proof of Lemma 5. For Cantor sets C 1 , C 2 with thicknesses 1 , 2 , and 1 2 1 , let equation , t C 1(C 2 t) contains exactly one overlapped point , , equation and equation , t C 1(C 2 t) contains two or more overlapped points , . equation Notice that , and C 1-C 2 up to a set of measure zero (in fact (C 1-C 2)() is a countable set). To prove Lemma 5, we need to show that has measure zero. To do this, it helps to make a distinction between three kinds of overlapped points. Suppose that xC 1C 2 is an overlapped point. Let A n and B n denote the components of I 1( i 1 n-1 U i) and J 1( i 1 n-1 V i) , that contain x (where U n n 1 and V n n 1 denote the bounded gaps, and I 1 , J 1 denote the convex hulls, of C 1 , C 2 ). Then x is an overlapped point of the first, second, or third kind , respectively, if one of the following three conditions holds, respectively; enumerate x(A n) and x(B n) for all n , x(A n) for all n and there is an n such that x is an endpoint of B n , or x(B n) for all n and there is an n such that x is an endpoint of A n , there is an n such that x is an endpoint of both A n and B n , and A nB n x . enumerate Figure 1 gives an idea of what the three different kinds of overlapped points look like with respect to the bridges A n and B n . For specific examples of Cantor sets whose intersection contains a single overlapped point of either the first or third kind, see K3 and K4 . figure t 1.45 x 1.5truein .5 3 1.5truein 3 1.45 x 1.5truein .5 3 1.5 1.5 1.45 x 1.5 2 1.5 1.5 Overlapped points of the first, second, and third kind. figure If t , then C 1(C 2 t) contains only one overlapped point; so we can partition into three subsets according to whether C 1(C 2 t) contains an overlapped point of the first, second or third kind. There are only a countable number of tC 1-C 2 for which C 1(C 2 t) can have an overlapped point of the third kind (since there are only a countable number of endpoints'' in C 1 or C 2 ), so the part of for which C 1(C 2 t) contains an overlapped point of the third kind has measure zero. So we need to concentrate on the part of for which C 1(C 2 t) contains an overlapped point of the first or second kind. Define equation ' , t the overlapped point in C 1(C 2 t) is not of the third kind . equation We need to show that ' has measure zero. Our proof is by contradiction; we assume that ' has positive measure, but then show that no point of ' is a density point. The main part of the proof is the next lemma; it gives a lower bound on the density of in a neighborhood of any point t' . We need two more definitions. Let us say that two bounded, closed, intervals are linked if each one contains exactly one boundary point of the other; see pp. 63--64 PT2 . We say that two Cantor sets embedded in the real line are linked Cantor sets if their convex hulls are linked. Notice that linked Cantor sets are interweaved. lem lem:6 Let C 1 , C 2 be linked Cantor sets, with thicknesses 1 , 2 , such that C 1C 2 contains a single overlapped point which is of the first or second kind. If 1 2 1 , then there is a constant ( 1, 2) 0 , which only depends on 1 and 2 , and a neighborhood (a,b) of 0 , such that equation (a,b) b-a . equation lem proof Let I , J denote the convex hulls of C 1 , C 2 . We are assuming that I and J are linked so they are positioned, relative to each other, something like the following. equation 1.4 I .75 2.2 2.6 .25 .8 .7 J equation Let U denote the longest gap of C 1 which intersects with J , and let V denote the longest gap of C 2 which intersects with I . Now we make the following claim: Either the closure of U contains an endpoint of J , or the closure of V contains an endpoint of I . To prove the claim, suppose it is not true; suppose that the closure of U does not contain an endpoint of J , and the closure of V does not contain an endpoint of I . So U and V might be positioned, relative to each other, something like the following picture. equation 1.4 .75 U .45 .75 1.2 .5 .5 .9 .4 1.3 .25 .8 V .7 equation But then C 1 and C 2 have (at least) two pairs of linked segments, so by Lemma 4 and the Gap Lemma, C 1C 2 contains at least two overlapped points, which is a contradiction, which proves the claim. proof Now we have two cases to consider. The first case is when both the closure of U contains an endpoint of J , and the closure of V contains an endpoint of I . The second case is when either the closure of U does not contain an endpoint of J , or the closure of V does not contain an endpoint of I . proof Case 1 In this case IU and JV are positioned, relative to each other, as in the following picture. equation .75 a 0 .75 A .7 a 1 .25 U .45 R .75 1.5 .5 .5 .6 .6 1.4 .25 L .6 V .35 b 0 .65 B .75 b 1 equation Notice that we have two linked bridges, which are denoted by A and B (the intervals A and B cannot have a common endpoint, since it would have to be either an overlapped point of the third kind or a nonoverlapped point, contradicting in either case one of our hypotheses). The two nonlinked bridges are denoted by L and R . Let A a 0, a 1 and B b 0, b 1 . Let ca 1-b 1 0 , and let da 1-b 0 0 (notice that d-c B ). Now (c,d) is a neighborhood of 0 , and (c,d) has been chosen so that the segments AC 1 and (BC 2) t are interweaved for all t(c,d) . To prove this, notice that if B A , then A and B t are in fact linked for all t(c,d) . On the other hand, if B A , then for t(a 0-b 0,d) , A and B t are linked, but for t(c,a 0-b 0 , we have AB t . However, when t(c,a 0-b 0 , C 1 and C 2 t are linked, so in order that C 1(C 2 t) , it must be that AC 1 and (BC 2) t are interweaved. So for all t(c,d) , we know that C 1(C 2 t) contains at least one overlapped point in A(B t) . When t d , the segments AC 1 and (BC 2) d are no longer interweaved, but C 1 and C 2 d are, so C 1(C 2 d) still contains at least one overlapped point. So when t d , it must be that at least one of the originally nonlinked intervals R and L d intersects with either A or B d . There are eight possible geometries'' of IU and (JV) d , depending on how either R intersects with B d , or L d intersects with A ; they are listed in Figure 2. For each configuration, we want to show that there is an neighborhood (a,b)(c,d) of 0 such that the density of in (a,b) has a lower bound that only depends on 1 and 2 . proof figure 1.6 A .85 U .4 R .8 1.5 .4 .4 .7 1.6 .9 .1 L d 1.2 1.3 B d Case 1a. 1.3 A .85 U .4 R .5 1.5 .4 .4 .1 .7 1.2 .9 .2 L d .9 1.2 B d Case 1b. 1.3 A .85 U .4 R .5 1.5 .4 .4 .6 .7 .7 .9 .7 L d .7 .9 B d Case 1c. 1.3 A 1 U .7 R .75 1.3 .5 1 .5 1.55 .8 .2 L d .9 1.1 B d Case 1d. 1.4 A .9 U .7 R .75 1.3 .5 1 .5 .5 1.05 .8 .5 L d .8 .9 B d Case 1e. 1.4 A .9 U .7 R .75 1.3 .5 1 1 .5 .55 .8 1 L d .55 .7 B d Case 1f. .8 A .95 U .7 R .3 1.1 .7 .8 .5 .9 .6 0 L d .7 .8 B d Case 1g. .8 A .95 U .7 R .3 1.1 .7 .8 .4 .5 .5 .6 .4 L d .55 .6 B d Case 1h. All the subcases of Case 1. figure proof Case 1a In this case, when t c , we get the following picture of IU and (JV) c . equation 2.3 A .75 U .4 R 1.7 1.2 .4 .4 1 .6 1.3 1 .1 L c 1.1 V c 1.1 B c equation And when t d , we get the following picture of IU and (JV) d . equation 2.3 A .75 U .4 R 1.7 1.2 .4 .4 1 1 .6 1.3 1 1 .1 L d 1.1 V d 1.1 B d equation For all t(c,d) , the segments in A and B t are interweaved. The intervals R and B t start out nonintersecting, then they are linked, then they become nonlinked but intersecting. By the Gap Lemma, the interweaved segments in A , B t , and the linked pair R , B t each guarantee us an overlapped point. However, the segments contained in the nonlinked but still intersecting pair A , B t need not be interweaved. So we restrict t to avoid this situation. Let ac , and let b(a 1 U R )-b 1 0 . When t b , we get the following picture of IU and (JV) b . equation 2.3 A .75 U .4 R 1.7 1.2 .4 .4 1 .8 .6 1.3 1 .8 .1 L b 1.1 V b 1.1 B b equation Now we can give a lower bound, for this case, on the density of those t in (a,b) for which C 1(C 2 t) contains at least two overlapped points. Notice that b-a ((a 1 U R )-b 1)-(a 1-b 1) U R . Then equation , (a,b) , b-a R U R 1 1 1 R U 1 1 1 1 1 1 1 . equation proof proof Case 1b This case is handled the same as Case 1a, since the interval L t was not used in that case, and everything else is the same. proof proof Case 1c Again, this case is the same as Case 1a. proof proof Case 1d In this case, let ac and bd , so b-a B . Then align , (a,b) , b-a B - U B 1-1 ( B V )( V U ) 1-1 ( B V )( A U ) (since A V ) 1-1 1 2 0. align proof proof Case 1e This is the most complicated case, and we handle it a bit differently. Let bd , a'a 0-b 0 0 , and a''a 1-b 1 0 . Notice that b-a' A , b-a'' B , and that AC 1 and (BC 2) t are interweaved for all t in either (a',b) or (a'',b) . The density of in (a',b) is bounded from below by equation , (a',b) , b-a' A - V A 1- V A , equation and density of in (a'',b) is bounded from below by equation , (a'',b) , b-a'' B - U B 1- U B . equation Since V can be arbitrarily close to A , or U can be arbitrarily close to B , we cannot say anything more about these last two estimates other than they are greater than zero. However, since 1 2 1 , we cannot have both V arbitrarily close to A , and U arbitrarily close to B ; as the lengths of A and V get close to each other, the lengths of U and V must be bounded away from each other, and vice versa. So there is a trade off between the density of in the intervals (a',b) and (a'',b) ; as one of the densities decreases, the other one must increase. We will analyze this trade off by introducing a rescaling of the Cantor set C 2 . To simplify the notation, make a couple of simple changes of variable so that d 0 and a 1 b 0 0 . Case 1e then looks like the following picture: equation 1.4 A .9 U .7 R .75 1.3 .5 1 .5 .5 1.05 .8 .7 L .7 V .65 0 .4 B equation where now A - A ,0 , B 0, B , (a',b) (- A ,0) , and (a'',b) (- B ,0) . We shall apply a linear rescaling'' transformation equation T(x) x with U B A V , equation to the Cantor set C 2 , and then compute the density of (C 1,C 2) in each of the intervals (a',b) and (a'',b) . (We do not need to consider A V and U B , since these are covered by Cases 1a or 1d, and Cases 1g or 1h.) A lower bound for the density of (C 1, C 2) in the interval (a',b) is given by equation , (a',b)(C 1, C 2) , b-a' A - V A 1- V A , equation and a lower bound for the density of (C 1, C 2) in the interval (a'',b) is given by equation , (a'',b)(C 1, C 2) , b-a'' B - U B 1-1 U B . equation What we want now is equation U B A V , 1- V A , 1-1 U B , . equation Since 1-( V A ) decreases and 1-( U B ) increases with , it suffices to solve for so that 1-( V A ) 1-( U B ) . This is solved by equation A U B V . equation If we plug this value of into our previous lower bounds, we get align ,(a',b) , b-a' , ,(a'',b) , b-a'' 1- V A A U B V 1-( A U B V ) -1 2 1-1 1 2 0. align This is our lower bound for the density of in one of the intervals (a',b) or (a'',b) , though we cannot say which one. proof proof Case 1f This case is the same as Case 1b, if we reverse the roles of C 1 and C 2 . proof proof Case 1g This case is the same as Case 1d, if we reverse the roles of C 1 and C 2 . proof proof Case 1h This case is the same as Case 1a, if we reverse the roles of C 1 and C 2 . proof proof Case 2 Suppose that the closure of U contains an endpoint of J , but the closure of V does not contain an endpoint of I . So we might have IU and JV positioned, relative to each other, as in the following picture. equation 1.5 .95 U .45 .75 1.5 .5 .5 .9 .4 1.3 .25 .8 V .7 equation However, in order that C 1 and C 2 not have two pairs of linked segments, V must contain an endpoint of U . Thus, we in fact have U and V positioned as in the following picture. equation .75 a 0 .75 A .7 a 1 .3 U .45 R 1 .75 1.5 .5 .5 1.8 .5 .3 0 b 0 1 B .75 b 1 .25 V .35 R 2 equation Notice that we have two linked bridges, which are denoted by A and B , and two nonlinked bridges, which are denoted by R 1 and R 2 . Let A a 0, a 1 and B b 0, b 1 . Let ca 0-b 1 0 , and let d ,a 0-b 0, a 1-b 1 , 0 . Notice that d-c B if B A , and d-c A if A B , and in either case d-c A . So (c,d) is a neighborhood of 0 , and (c,d) has been chosen so that the intervals A and B t are linked for all t(c,d) . So for all t(c,d) , we know that C 1(C 2 t) contains at least one overlapped point in A(B t) . When t c , A and B c are no longer linked, but C 1 and C 2 c are linked, so C 1(C 2 c) contains at least one overlapped point. So when t c , it must be that the interval R 2 c intersects with A . There are two possible geometries'' of IU and (JV) c , depending on how R 2 c intersects with A ; see Figure 3. figure t 1.3 A .8 U .4 R 1 .75 1.2 .4 .5 .75 .6 .4 .15 B c .7 V c .6 R 2 c Case 2a. 1.3 A .8 U .4 R 1 .75 1.2 .4 .5 .75 1 .4 .2 B c .85 V c .7 R 2 c Case 2b. The two subcases for Case 2. figure proof proof Case 2a Let aa 1-(b 1 V R 2 ) , so c a 0 , and let bd . Notice that if A B , then b-a (a 1-b 1)-( a 1-(b 1 V R 2 )) V R 2 , and if B A , then align b-a (a 0-b 0)-(a 1-(b 1 V R 2 )) B V R 2 - A A V R 2 - A (since B A ) V R 2 . align In either case, a lower bound on the density of in (a,b) is given by equation , (a,b) , b-a R 2 V R 2 1 1 1 R 2 V 1 1 1 2 2 1 2 . equation proof proof Case 2b Notice that, by using both the fact that R 2 U 1 and the definition of thickness, we have equation A V R 2 V A U 1 2. 1 equation Now let ac , and bd , so b-a d-c A . Using inequality (1), a lower bound on the density of in (a,b) is given by equation , (a,b) , b-a A - V A 1- V A 1-1 1 2 . equation This concludes Case 2b, and also Case 2. Now that we have analyzed all the possible cases, let equation 1 1 1 1 , 2 2 1 2 , 3 1-1 1 2 , 4 1-1 1 2 , equation and let , 1, 2, 3, 4 , 0 . Then only depends on 1 and 2 . proof lem lem:7 Let C 1 , C 2 be linked Cantor sets, with thicknesses 1 , 2 , such that C 1C 2 contains a single overlapped point which is of the first or second kind. If 1 2 1 , then there is a constant ( 1, 2) 0 , which only depends on 1 and 2 , and neighborhoods (a n,b n) of 0 with n b n-a n 0 , such that for all n equation (a n,b n) b n-a n . equation lem proof In both Cases 1 and 2 of Lemma 6, after we removed the open intervals U and V from the closed intervals I and J , we were left with a pair of linked bridges which were denoted by A and B . The segments of C 1 and C 2 contained in A and B satisfy the hypotheses of Lemma 6. So we can apply Lemma 6 to these new Cantor sets, and get new linked bridges A 2 , B 2 , and another open neighborhood (a 2,b 2) of zero where the density of is bounded from below by . By induction, given linked Cantor sets C 1A n and C 2B n , we can apply Lemma 6 to get linked bridges A n 1 and B n 1 , and an open neighborhood (a n 1 ,b n 1 ) of zero where the density of is bounded from below by . Since 1 , 2 are lower bounds on the thicknesses of C 1A n , C 2B n , and depends only on 1 and 2 , the same value of works for all n . To show that n b n-a n 0 , it suffices to show that A n 0 and B n 0 as n , since (a n,b n)A n-B n (recall that A n and B n t are interweaved for all t(a n,b n) ). But A n n 1 is a sequence of bridges from C 1 that each contain the overlapped point x , so it must be that A n 0 , since C 1 is a Cantor set; similarly for the B n . proof Now we can give the proof of Lemma 5. proof Proof of Lemma lem:5 We need to show that ' has measure zero. Suppose that it has positive measure. By the Lebesgue density theorem, pp. 107--109 WZ , equation n '(a n,b n) b n-a n 1, equation for almost all t in ' , where ,(a n,b n) , n 1 is any sequence of intervals that shrink regularly to t . (The intervals (a n,b n) shrink regularly to t if (i) n b n-a n 0 , (ii) if D n is the smallest disk centered at t containing (a n,b n) , then there is a constant k independent of n such that D n k(b n-a n) .) Suppose that t 0' is a density point. By a simple change of variable, we can assume that t 0 0 . Let I , J denote the smallest closed interval containing C 1 , C 2 . proof theorem8 Without loss of generality, we can assume that I and J are linked. theorem8 proof To prove this claim, first notice that I and J cannot have a common endpoint; for if they did, the common endpoint would have to be either an overlapped point of the third kind, or a nonoverlapped point, which contradicts our assumption that 0' . Since IJ and I , J cannot have a common endpoint, it must be that either they are linked, in which case we are done, or one of I or J is contained in the interior of the other. Suppose that J is contained in the interior of I , so I and J are positioned, relative to each other, as in the following picture. equation 1.25 .4 I 3 1 1.5 1.5 J equation Let U be the longest gap of C 1 that intersects with J . So IU and J might be positioned, relative to each other, as in the following picture. equation 1.25 .85 U 2 .25 .75 1 1.5 1.5 J equation But in order that C 1 and C 2 not have two linked segments, and hence two overlapped points in C 1C 2 , it must be that U contains an endpoint of J , i.e., IU and J are in fact positioned, relative to each other, as in the following picture. equation 1.25 A 1.1 U 2.25 .25 .5 1 1.5 1.5 J equation The interval to the left of U , which is denoted by A , is linked with J . The segment C 1A has thickness at least 1 , and (C 1A)C 2 contains a single overlapped point, which is still of the first or second kind. So, without loss of generality, we can replace C 1 with C 1A , and also I with A , and then I and J are linked. So C 1 and C 2 are linked Cantor sets such that 0' , and their thicknesses satisfy 1 2 1 . By Lemma 7, we have neighborhoods (a n, b n) of 0 with n b n-a n 0 , such that for all n equation (a n,b n) b n-a n , equation for some constant 0 which is independent of n . Since 0(a n,b n) for all n , the intervals (a n,b n) shrink regularly to 0 (let k 2 in the definition of shrink regularly). Since 0 is a density point of ' , we can choose an n so that equation '(a n,b n) b n-a n 1-. equation Since and ' are disjoint, these last two inequalities contradict each other, so it must be that ' has measure zero. proof For some intuition on what ' can look like see K2 , where the structure of ' is examined in detail using symbolic dynamics for the special case where C 1 C 2 is a middle- Cantor set with 1 3 . We end this paper with a couple of conjectures. Since the proofs of both the Gap Lemma and Theorem 1 are essentially renormalization arguments, and since renormalization often leads to critical phenomena, we can conjecture that the condition 1 2 1 on thicknesses is some kind of critical boundary for difference sets of Cantor sets. Since 1 2 1 implies both that C 1-C 2 is a union of intervals and that C 1(C 2 t) contains a Cantor set for almost all tC 1-C 2 , we can conjecture the following phenomena for the condition 1 2 1 . proof Conjecture 1 For any positive real numbers 1 and 2 with 1 2 1 , there exist Cantor sets C 1 , C 2 with thicknesses 1 , 2 such that C 1-C 2 does not contain any intervals (and hence it is a Cantor set). proof proof Conjecture 2 For any positive real numbers 1 and 2 with 1 2 1 , there exist Cantor sets C 1 , C 2 with thicknesses 1 , 2 such that C 1(C 2 t) does not contain a Cantor set for almost all real numbers t . proof Notice that neither of these conjectures implies the other. For any (0,1) , let C denote the middle- Cantor set in the interval 0,1 . Since a middle- Cantor set will minimize Hausdorff dimension among all Cantor sets of a given thickness ( pp. 77--78 PT2 and p. 23 K1 ) it would seem reasonable to expect them to be good candidates for solving the above conjectures. So we can make the following more specific conjectures. proof Conjecture 1' For any real numbers 1, 2(0,1) with 1 3 1 2 2 1 , there exists a real number 0 such that C 1 - C 2 does not contain any intervals. proof proof Conjecture 2' For any real numbers 1, 2(0,1) with 1 3 1 2 2 1 , there exists a real number 0 such that C 1 (C 2 t) does not contain a Cantor set for almost all real numbers t . proof These conjectures are related to Problem 7 from p. 151 PT2 . These conjectures are very easy to prove when 1 2 ; see K3 . amsalpha