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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 $3$.

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

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