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

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.
