TITLE: The Cantor set is homeomorphic to infinite product of $\{0,1\}$ with itself - cylinder basis - and it topology
QUESTION [21 upvotes]: I know the Cantor set probably comes up in homework questions all the time but I didn't find many clues - that I understood at least.
I am, for a homework problem, supposed to show that the Cantor set is homeomorphic to the infinite product (I am assuming countably infinite?) of $\{0,1\}$ with itself.
So members of this two-point space(?) are things like $(0,0,0,1)$ and $(0,1,1,1,1,1,1)$, etc.
Firstly, I think that a homeomorphism (the 'topological isomorhism') is a mapping between two topologies (for the Cantor sets which topology is this? discrete?) that have continuous, bijective functions.
So I am pretty lost and don't even know what more to say! :( I have seen something like this in reading some texts, something about $$f: \sum_{i=1}^{+\infty}\,\frac{a_i}{3^i} \mapsto \sum_{i=1}^{+\infty}\,\frac{a_i}{2^{i+1}} ,$$ for $a_i = 0,2$. But in some ways this seems to be a 'complement' of what I need.... Apparently I am to use ternary numbers represented using only $0$'s and $1$'s in; for example, $0.a_1\,a_2\,\ldots = 0.01011101$?
Thanks much for any help starting out!

Here is the verbatim homework question:

The standard measure on the Cantor set is given by the Cantor $\phi$ function which is constant on missing thirds and dyadic on ternary rationals.
Show the Cantor set is homeomorphic to the infinite product of $\{0,1\}$ with itself.
How should we topologize this product?
(Hint: this product is the same as the set of all infinite binary sequences)
Fix a binary $n$-tuple $(a_1,\ldots, a_n)$ (for e.g., $(0,1,1,0,0,0)$ if $n = 6$).
Show that the Cantor measure of points ($b_k$) with $b_k=a_k$ for $k \leq n$ and $b_k \in \{0,1\}$ arbitrary for $k>n$, is exactly $1/2^n$. These are called cylinders. (They are the open sets, but also closed!)

REPLY [11 votes]: Note that the $1/3$-Cantor set in $[0,1]$ can be represented as the set of real numbers of the form $\sum_{n=1}^\infty a_n/3^n$ where $a_n\in\{0,2\}$ for each $n\in\mathbb{N}$. A homeomorphism you are looking for is the function $f$ which maps the point $\sum_{n=1}^\infty a_n/3^n$ in the Cantor set to the sequence $(a_n/2)_{n=1}^\infty$ in the product $\{0,1\}^\mathbb{N}$. The product $\{0,1\}^\mathbb{N}$ consists of countably infinite sequences of $0$'s and $1$'s. Note that no finite tuple such as $(0,0,0,1)$ is in $\{0,1\}^\mathbb{N}$. The product is topologized so that each factor $\{0,1\}$ is given the discrete topology and then the product is given the product topology.
You want to prove that $f$ is a continuous and open bijection. The bijectiveness is very easy to show. For the continuity you may want to use the fact that the product topology of $\{0,1\}^\mathbb{N}$ is generated by the sets of the form $U(N,a)=\{(a_n)_{n=1}^\infty\in\{0,1\}^\mathbb{N}:a_N=a\}$ where $N\in\mathbb{N}$ and $a\in\{0,1\}$, and hence it suffices to show that the preimages of these sets $U(N,a)$ are open in the Cantor set. Finally to show that $f$ is open you can use the following general fact: a continuous bijection from a compact space to a Hausdorff space is open.