I've been asked to prove the following:
Prove that $C\times\{1,2,\dots,n\}$ is homeomorphic to $C$, where $C$ is the Cantor Set.
For convenience we can take the product metric to be $d_{\infty}((a,b),(c,d)) = \max\{d(a,b), d'(c,d)\}$.
I know that since $C\times\{1,2,\dots,n\}$ is compact, we need to only find a continuous bijection (the homeomorphism follows from the fact that $C$ is Hausdorff). Furthermore, it is clear that $C$ is the set of all reals in $[0,1]$ with a ternary expansion which only contains the numbers $0$ and $2$.
I think I've found a homeomorphism in the case that $n = 2^m$ for some natural $m$, where we write each number as its binary expansion, convert it to $0$s and $2$s and then attach it to the front of the ternary expansion. As an example, if $n=8$, then
$$f\bigg(\frac{1}{4},5\bigg) \implies f(0.0202\dots_3,101_2) = 0.2020202\dots$$
Since the binary expansion is unique, each number will have a unique inverse based on its first digits.
What is the homeomorphism if $n \neq 2^m$? Furthermore, can a homeomorphism also be drawn from the space $C\times\{\frac{1}{n}:n\in\mathbb N\}$ to $C$?
Best Answer
If $X$ is any zero-dimensional compact metric space without isolated points, $X \simeq C$ (a classical theorem due to Brouwer).
It follows that $C \simeq C^n$ for all $n$ and also $C \oplus C \simeq C$ (disjoint sum, any finite disjoint sum will do). Your fact is a direct consequence.