# Is this a Cantor Set?

Suppose $\alpha:[0,1]\rightarrow\mathbb{R}^2$ is a continuous curve. Let $K\subset \alpha([0,1])$ be a cantor set, i.e. there exist a homeomorphism between K and the standard Cantor set in $\mathbb{R}$. Consider the set $$D=\{|x-y|:\ x,y\in K\}\subset\mathbb{R}$$

Is $D$ a Cantor set?

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If you look at the set of differences of the usual Cantor set, it contains an interval, so it's not a Cantor set itself. –  Gerry Myerson Jan 7 '13 at 14:04
Sorry, but i cant see why it contains an interval @GerryMyerson. Can you explain me please? –  Tomás Jan 7 '13 at 14:07
I think @Gerry is saying that the standard Cantor set $\times \{0\} \subset \mathbb{R}^2$ satisfies the assumptions. This $K$ is exactly the same as the standard Cantor set, and if its corresponding $D$ is not a Cantor set you will have a counterexample. –  Willie Wong Jan 7 '13 at 14:11
Tomas, see GEdgar's answer. –  Gerry Myerson Jan 7 '13 at 14:16
Interesting @GerryMyerson. Is this true in the general case? –  Tomás Jan 7 '13 at 14:26

The assumption that $K$ lies on a continuous curve is of no use, thanks to space-filling curves.

Claim. If $K\subset \mathbb R^2$ is a Cantor set with $\underline{\dim}_M K<1/2$, then the distance set $D=\{|x-y|:x,y\in K\}$ is also a Cantor set.

Here $\underline{\dim}_M K$ is a lower Minkowski dimension. By definition it is equal to $\liminf_{r\to 0}\frac{\log N(r)}{-\log r}$ where $N(r)$ is the minimal number of closed balls of radius $r$ required to cover $K$. There is also upper Minkowski dimension $\overline{\dim}_M K$ which uses $\limsup$ in place of $\liminf$. It is easy to see that Hausdorff $\le$ lower Minkowski $\le$ upper Minkowski. For general sets these inequalities may be strict. For self-similar sets like the standard Cantor set they coincide.

Lemma 1. Lipschitz maps do not increase any of the aforementioned dimensions: i.e., $\dim f(K)\le \dim K$ if $f$ is Lipschitz.

Proof: Easy exercise. $\Box$

Lemma 2. $\underline{\dim}_M (K\times K)\le 2\underline{\dim}_M K$.

Proof: If $K$ is covered by $N$ balls of radius $r$, then $K\times K$ is covered by $N^2$ balls of radius $r\sqrt{2}$: just take the products. $\Box$

Lemma 3. If $\underline{\dim}_M K<1$, then $K$ is totally disconnected.

Proof is a little easier in Euclidean spaces (using projection and Lemma 1), but here is a metric space version. Suppose $a\ne b$ are two points in the same connected component. Let $\delta=d(a,b)$. Given $r<\delta/3$, let $N=\lfloor \delta/(3r)\rfloor$. For each $k=0,\dots,N-1$ the set $K$ contains a point $x_k$ such that $d(x_k,a)=3rk$; otherwise $a$ and $b$ would be disconnected by $\{x:d(x,a)<3rk\}$ and $\{x:d(x,a)>3rk\}$. Since any ball of radius $r$ contains at most one of the points $x_k$, it follows that $N(r)\ge N$. Therefore, $\log N(r)\ge -\log r +\text{constant}$ and the claim follows. $\Box$

Proof of the Claim: Applying Lemmas 1-2-3 to the surjective Lipschitz map $f:K\times K\to D$ defined by $f(x,y)=|x-y|$, we see that $D$ is totally disconnected. Since $f$ is continuous, $D$ is compact.

The fact that $D$ has no isolated points is somewhat subtle. Indeed, this is false if we use $\ell_1$ or $\ell_\infty$ distance instead of Euclidean. It is also false for the Euclidean distance in dimensions higher than $2$. The proof for $\mathbb R^2$ with Euclidean distance relies on the fact that any two distinct circles have finite intersection. Indeed, if $d\in D$ is isolated then there is a point $x_0\in K$ such that the set $K_d:=K\cap \{x:|x-x_0|=d\}$ is infinite. But there are other points of $K$ arbitrarily close to $x_0$, and their distances to elements of $K_d$ are close to, but not always equal to, $d$. $\Box$

Remark. For the upper Minkowki dimension Lemma 2 can be generalized to $\overline{\dim}_M (A\times B)\le \overline{\dim} A+\overline{\dim} B$. For the lower Minkowki dimension this does not work: it's a good exercise to see why. Both of these are actual exercises in Mattila's book Geometry of sets and measures in Euclidean spaces.

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1 - Do you know anything about the case $\underline{dim}_M K\in(\frac{1}{2},1)$? –  Tomás Jan 11 '13 at 16:07
@Tomás The middle-third set has dimension $\log_3 2\approx 0.63$ (in any of the senses mentioned above). Thus, in the range $\dim\ge \log_3 2$ the distance set is not necessarily a Cantor set. I don't know what happens in the range $[1/2, \log_3 2)$. –  user53153 Jan 13 '13 at 3:43

If $C$ is the usual middle-thirds Cantor set, then $\{|x-y|\,\mid x,y\in C\}$ is the entire closed interval $[0,1]$.

It is easier to see this in the form $\{x+y\mid x,y\in C\}=[0,2]$ (which implies the original because the Cantor set is symmetric about $1/2$.

The Cantor set consist of all numbers that have a base-3 representation using the digits 0 and 2 only. In claim that every number on $[0,2]$ is the sum of two such numbers. The main problem is to make the sum contain 1s at the right places, but we can do it this way:

  0.22222000000222022220002002222...
+ 0.00002000000002000020002000000...
---------------------------------------
1.00001000001001100010011010000....


The second of the addends here consists of only 0s except with 2s at every second of the positions where we want 1 in the result. If we want to change any of the 0 in the sum to 2s, we can simply insert additional 2s in the corresponding positions in the second addend.

On the other hand, for example, a middle-8/10s Cantor set (homeomorphic to the middle-third one) does not have this property.

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Henning: off hand, do you know what the difference set for the middle 8/10 Cantor set is? For the general question the OP asked, since $K$ is perfect and compact, we have the $D$ is also perfect and compact using that the distance function is continuous on $(\mathbb{R}^2)^2$. So whether $D$ is a Cantor space really only depends on disconnectedness. I wonder if there's anything we can say about neighborhoods of zero in the general case. –  Willie Wong Jan 7 '13 at 15:12
@WillieWong: It's a "generalized Cantor set" in the sense that the decimal expansions of its members can be generated by a finite automaton. My intuition suggests that except for isolated points, such a set is either empty, contains an interval, or is homeomorphic to the Cantor set, but I don't have the beginnings of a proof of that. –  Henning Makholm Jan 7 '13 at 15:24
@Willie The difference set of C is a Lipschitz image of the square of C. If C has dimension less than 1/2 then the difference set has dimension less than 1, hence is totally disconnected. In particular, this applies to the middle 8/10 set, which has dimension about 0.3. (One has to choose a notion of dimension that is subadditive under products, such as upper Minkowski). –  user53153 Jan 10 '13 at 17:19
@Willie The continuity of distance function does not by itself imply that $D$ is perfect. For example, consider the $\ell_\infty$ distance and $K$ being the union of two standard Cantor sets lying on horizontal lines $y=\pm 1$. Here $D=[0,1]\cup\{2\}$. For Euclidean distance we get that $D$ is perfect using a bit of geometry, as in my answer. –  user53153 Jan 10 '13 at 21:42
@Pavel: of course. Some injectivity property is needed. Thanks. –  Willie Wong Jan 11 '13 at 9:03

Proofs of the difference property for the standard Cantor set... http://www.cut-the-knot.org/do_you_know/cantor3.shtml

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