I would like to show that the Cantor middle-third set with subset topology inherited from $\mathbb R$ is homeomorphic to $\{ 0,1 \}^\infty$ under product topology. I am trying to construct a homeomorphism between the two spaces, but I am not clear what an open set (or closed set) of the second space generally look like. Could someone help me with that? Thanks.
[Math] Cantor set homeomorphic to $\{ 0,1 \}^\infty$
cantor setgeneral-topology
Related Solutions
I’m going to assume that Cantor set here refers to the standard middle-thirds Cantor set $C$ described here. It can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, i.e., real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$.
For each positive integer $n$ let $D_n = \{0,1\}$ with the discrete topology, and let $$X = \prod_{n=1}^\infty D_n$$ with the product topology. Elements of $X$ are infinite sequences of $0$’s and $1$’s, so $(0,0,0,1)$ and $(0,1,1,1,1,1,1)$ are not elements of $X$; if you pad these with an infinite string of $0$’s to get $(0,0,0,1,0,0,0,0,\dots)$ and $(0,1,1,1,1,1,1,0,0,0,0,\dots)$, however, you do get points of $X$. A more interesting point of $X$ is the sequence $(p_n)_n$, where $p_n = 1$ if $n$ is prime, and $p_n = 0$ if $n$ is not prime.
Your problem is to show that $C$, with the topology that it inherits from $\mathbb{R}$, is homeomorphic to $X$. To do that, you must find a bijection $h:C\to D$ such that both $h$ and $h^{-1}$ are continuous. The suggestion that you found is to let $$h\left(\sum_{n=1}^\infty\frac{a_n}{3^n}\right) = \left(\frac{a_1}2,\frac{a_2}2,\frac{a_3}2,\dots\right).$$ Note that $$\frac{a_n}2 = \begin{cases}0,&\text{if }a_n=0\\1,&\text{if }a_n=2,\end{cases}$$ so this really does define a point in $X$. This really is a bijection: if $b = (b_n)_n \in X$, $$h^{-1}(b) = \sum_{n=1}^\infty\frac{2b_n}{3^n}.$$
Yes, it is.
Note that the Cantor set is in fact a compact group, and the product measure $\mu$ is the Haar measure, that is, the invariant finite, regular Borel measure, which is by Haar's theorem unique up to a scaling factor.
That Lebesgue measure restricted to the fat Cantor set is invariant is an easy exercise (it follows easily from the fact that the Lebesgue measure is translation invariant).
It is also regular, because it is a restriction of a regular measure, so by Haar's theorem it is a scalar multiple of the product measure.
Best Answer
Your space $C$ is the collection of functions $f:\omega=\{0,1,2,\ldots\}\to \{0,1\}=X$, i.e. the collection of infinite binary sequences. I will try to illustrate how we can understand the space $C$ and its topology below, deriving some of its properties.
The basic open sets in this collection are (by definition) infinite products of open sets in $X$ where all but finitely many factors are $X$. Suppose the factors correspond to a sequence $k_1<k_2<\cdots<k_s$ of natural numbers. At these factors (assuming the set is not empty) there correspond either the open set $\{0\}$ or the open set $\{1\}$. So we can associate to each basic open set $O$ a function from a finite subset $F=\{k_1,\ldots,k_s\}$ of $\omega$ to $X$. This is a bijection. Note that if $f,g\in O$ and $F$ is associated to $O$; then $f\mid F=g\mid F$.
Now take a function $f:\omega\to X$. We can consider the open sets $O_i=\prod_{k=0}^i \{f(k)\}\times X\times \cdots$. Note that given any basic open set $O$ that contains $f$, there is $j$ such that $O_j\subseteq O$ (can you prove this?). This means that the collection $\mathscr B=\{O_i:i\in\omega\}$ is a neighborhood basis for $f$. This means our space is second countable! It is also separable, since the collection of functions that are eventually constant is countable and dense: given a basic neighborhood $O_i$ of $f$, the function that agrees with $f$ on $\{0,1,2,\ldots,i\}$ and is constant afterwards is in $O_i$. This proves this space is also perfect.
In particular $g\in \overline{\{f\}}$ iff $g$ agrees with $f$ on $\{0,1,\ldots,i\}$ with each $i$. This means $g=f$; so singletons are closed $\{f\}$. In fact your space is Hausdorff: if $g$ and $f$ do not agree at position $k$ (say one takes the value $0$ and the other $1$), then $\cdots\times X\times \{0\}\times X\times\cdots$ and $\cdots\times X\times \{1\}\times X\times\cdots$ are disjoint open sets that contain them.
Actually, your space is metrizable, and the neigborhood basis we obtained gives a decent idea of what this metric is: we define $d(f,g)=2^{-\nu(f,g)}$ where $\nu(f,g)$ is the first $k\in\omega$ such that $f(k)\neq g(k)$. We define $\nu(f,f)=\infty$, of course! Note then that $O_i=\{g\in C:d(f,g)<2^{-i}\}$. Check that $d(f,g)$ is indeed a metric.
Finally, your space is compact. Since we've shown it is metrizable, we can simply check it is sequentially compact. Now take a sequence of functions $(f_i)$ in your space. If we look at $f_i(0)$ for $i=1,2,3,\ldots$, we get a sequence of zeros and ones. Since this sequence is infinite, either the ones or the zeros repeat infinitely often. So we can take a subsequence indexed by $N_0$ that has all its first coordinates equal to say $x_0\in \{0,1\}$. Now look at the second coordinate of this sequence, and repeat the process to get $N_1\subseteq N_0$ and some $x_1$. We get a sequence of nested infinite subsets $N_0\supseteq N_1\supseteq N_2\supseteq \cdots$, and we may thus take a subsequence $f_{k_j}$ such that $k_i\in N_i-N_{i+1}$. You can check that this subsequence converges to the $f$ such that $f(i)=x_i$ (note this is just another diagonalization trick!).
Given we have exhibited your space as a metric space, I believe you can understand its topology quite nicely. Now, note that the Cantor set consists of real numbers whose ternary expansion contains only $0$s and $2$s. Does this give you an idea of what a possible homeomorphism $\eta:C'\to C$ may be?