# Set of distance between elements of Cantor set is equal to [0,1]

Let K the Cantor set. Show that {$$|x-y|;x\in K,y\in K$$}$$=[0,1]$$

Hint: Note that the set $$D=$${$$d ;d=|x-y|$$}, with $$x,y \in K$$ is compact and it contains all the proper fractions with the denominators are power of $$3$$.

• You can express the elements of the cantor set as trinary expansions of a number without a $1$ in them. – Dean Young Oct 10 '19 at 13:46

You can also just manipulate terms to see that:

$$\{|x-y|;x\in K,y\in K\}=|\sum_{i=1}^\infty \frac{x_i - y_i}{3^i}|=|\sum_{i=1}^\infty \frac{z_i}{3^i}|$$

where $$x_i,y_i \in \{0,2\}$$ from the ternary expansion, in such a way that $$z_i \in \{-2,0,2\}$$. Now lets just manipulate a bit more: $$|\sum_{i=1}^\infty \frac{z_i}{3^i}|=|\sum_{i=1}^\infty \frac{z_i +2}{3^i}-\frac{ 2}{3^i}|=2\cdot|\sum_{i=1}^\infty \frac{(z_i +2)/2}{3^i}-\frac{1}{3^i}|$$

Now define $$w_i=(z_i +2)/2$$. Notice that $$w_i \in \{0,1,2\}$$, and notice $$\sum_{i=1}^\infty \frac{1}{3^i}=1/2$$, hence

$$2\cdot|\sum_{i=1}^\infty \frac{(z_i +2)/2}{3^i}-\frac{1}{3^i}|=|2\cdot\sum_{i=1}^\infty \frac{w_i}{3^i}-1|$$

But, since $$w_i \in \{0,1,2\}$$, the sum term is simply the ternary expansion for any number in [0,1].

Finally, we can see that {$$|x-y|;x\in K,y\in K$$} is giving us a function in the form of $$f:[0,1]\to [0,1]$$, $$x\mapsto |2x-1|$$

Go ahead and plot its graph. You will have exactly the image you want in the y-axis.