# Drawing a Cantor-based fractal set

I am trying to draw the following set $$E=\{(x,y)\in\Bbb R^2:x+y,x-y\in F\}$$ where $F$ is the middle third Cantor set: I think this is what the first iteration looks like: but I am really not sure (I'm sorry for the pathetic quality by the way, I was forced to use Paint - I will happily create a nicer image if it is the correct one).

Could someone tell me if I am correct and if not, what it should llook like please? Thank you.

• what is middle third in cantor set? Aug 20 '17 at 16:32
• @SantoshLinkha is the set pictured above, the ones with the lines. Each iteration you remove a middle third of each line segment of the previous iteration. Aug 20 '17 at 16:36
• I'm not sure on your notation but yes I think so. $E_{0}=[0,1]$, $E_{1}=[0,\frac{1}{3}]\cup[\frac{2}{3},1]$,.... Aug 20 '17 at 16:55

The region bounded by the lines $x+y=0$, $x+y=1$, $x-y=0$ and $x-y=1$ is a diamond (rotated square) with opposite corners at $(0,0)$ and $(1,0)$. The zeroth iteration of $E$ is simply that filled-in diamond.
$x+y$ and $x-y$ are perpendicular, so further iterations of $E$ are the intersection of two Cantor sets extended into stripes running across the diamond. Indeed, the first iteration of $E$ should look like this: $E$ itself is known as Cantor dust, and looks like the below image (rotated of course): • That's very nice, thank you. I still don't quite understand why, for example, $(0,1)$ isn't in the set since $x+y=0+1=1\in F$. Could you explain please? Aug 20 '17 at 16:35
• @JSharpee $x-y=-1$ then, but the Cantor set (at least the canonical one) is only defined over $[0,1]$. Aug 20 '17 at 16:36