Show the Cantor set $C$ is equal to its product $C\times C$.

Question

I've been reading up on the Cantor set, and it is simple to show a bijection $C \to [0,1]$.

I was thinking that it would be easy to show that there exists some space filling curve by showing that $C = C \times C$ (equality in the sense of homeomorphism). Because then I would have $[0,1] = [0,1] \times [0,1]$.

But, how would I go about showing that the cantor set is homeomoprhic to its product? I'm guessing there's a standard such mapping somewhere out there but I can't seem to find it.

Thanks!

Answer 1

A member of the Cantor set can be written uniquely as $$x = \sum_{i=1}^\infty d_j 3^{-j},\ \text{where all} \ d_j \in \{0,2\}$$ Map this to the pair $$(y,z) = \left(\sum_{j=1}^\infty d_{2j-1} 3^{-j}, \sum_{j=1}^\infty d_{2j} 3^{-j}\right)$$

Answer 2

It seems you can do this recursively. The Cantor set is constructed by repeatedly removing the middle third, starting with the unit interval. Now if you halt that process after the second removal, what you have are four small intervals.

Similarily, $C\times C$ may be constructed from the unit square by repeatedly removing one horizontal strip and one vertical strip, each accounting for one third of the width of the square from which they have been removed. If you stop this process after the first removal, you have four small squares.

You could assign one square to each interval, say like this: $$ \begin{array}[cccc] 11&2&3&4 \end{array}\\ \Downarrow\\ \begin{array}[cc] 22&3\\1&4 \end{array} $$ Then for each of those intervals and squares, repeat the process to have each of sixteen intervals assigned to each of sixteen squares. It should look like this: $$ \begin{array}[cccc] 11&2&3&4& &5&6&7&8&&9&10&11&12&&13&14&15&16 \end{array}\\ \Downarrow\\ \begin{array}[ccccc] 66&7&&10&11\\ 5&8&&9&12\\ \\ 2&3&&14&15\\ 1&4&&13&16 \end{array} $$Note that the first four intervals (which were all part of the first interval in the first iteration) all go to the four squares in the bottom left-hand corner (which were all part of the square numbered $1$ in the first iteration), and similarily for the other quadruples of intervals. We see that if we keep this up, this process naturally uses the self-similarity of the Cantor set to define how the assigment should be one level down.

If you keep this going ad infinitum, you get a bijection from $C$ to $C\times C$, and I believe it is a homeomorphism.

Answer 3

An indirect way: by a theorem of Brouwer says that every compact metrisable space that is zero-dimensional (or equivalently has no connected subsets except singletons) and has no isolated points (no $x$ with $\{x\}$ open)is homeomorphic to the Cantor set. Now $C \times C$ satisfies these conditions, just as $C$ does. The same goes for $C^3$ or even $C^\mathbb{N}$, or $\{0,1\}^\mathbb{N}$, or $\{0,1,2\}^\mathbb{N}$ etc.

Answer 4

Just to be a bit more concrete than the other answers: Recall that a number $x$ between $0$ and $1$ is in the Cantor set if and only if it can be expanded in base $3$ with all digits equal to either $0$ or $2$.

A member of $C\times C$ is a pair $(x,y)$ of members of the Cantor set. $$ \begin{array}{cccccccccccccc} x & = & 0 & . & 2 & & 0 & & 0 & & 2 & & 2 & & 2 & & 0 & & 2 & & \ldots & \in C \\ & & & & \downarrow & \nearrow &\downarrow & \nearrow &\downarrow & \nearrow &\downarrow & \nearrow &\downarrow & \nearrow &\downarrow & \nearrow &\downarrow & \nearrow &\downarrow & \nearrow & \\ y & = & 0 & . & 0 & & 0 & & 2 & & 2 & & 2 & & 0 & & 0 & & 2 & & \ldots & \in C \end{array} $$ Follow the arrows to get a new sequence of $0$s and $2$s, thus a new member of the Cantor set.