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