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 $3$.

  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.