# The Cantor set is homeomorphic to infinite product of $\{0,1\}$ with itself - cylinder basis - and it topology

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!)

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 $$b = (b_n)_n \in X$$, $$h^{-1}(b) = \sum_{n=1}^\infty\frac{2b_n}{3^n}.$$

• @nate: The answer to your first question is yes, provided that you look only at series in which all of the $a_n$’2 are $0$ or $2$. When you remove $(1/3,2/3)$, you get rid of all of the numbers whose only ternary expansions begin $0.1$. When you remove $(1/9,2/9)\cup(7/9,8/9)$ you get rid of those whose only ternary expansions begin $0.01$ or $0.21$. And so on. For your second question: no, $D_3 = \{0,1\}$. It’s just one of the factor spaces in the infinite product. A sequence of $0$’s and $1$’s is a member of that product. Oct 5, 2011 at 0:17
• I guess I am getting it, albeit slowly. My main confusion (for another posting when I get it worded well) is from the "cantor set" page on wikipedia where the author says to take the binary representation of $3/5_{10} \mapsto 0.10011001..._{2}$ and replace all the 1's by 2's. In base-3 (with 0,1,2), 3/5 is 0.12101210... I guess this will have to be another posting or look into it some more - this exchange 1's by 2's. Or is this a trick to just get rid of the middle thirds?
– nate
Oct 5, 2011 at 0:55
• @nate: It’s just a trick to get rid of the middle thirds. When you replace the $1$’s in a binary expansion by $2$’s and interpret the result in ternary, you will gave a different number. In fact, the two binary expansions of a dyadic rational will give you different numbers: $1/2_\text{ten}=0.10000\dots_\text{two}$ gives you $0.20000\dots_\text{three}=2/3_\text{ten}$, while $1/2_\text{ten}=0.01111\dots_\text{two}$ gives you $0.02222\dots_\text{three}=1/3_\text{ten}$: you’ve split $1/2_\text{ten}$ in two. Oct 5, 2011 at 1:11
• How can we prove that $h$ is continuous? I mean, what is $f^{-1} (B)$ with $B$ a basic open set of $\{0,1\} ^{\mathbb{N} }$? I don't see it. Nov 1, 2016 at 12:21
• @Whoknows: Let $\sigma=\langle b_0,\ldots,b_n\rangle$ be a finite sequence of zeroes and ones, and let $$B(\sigma)=\{x\in X:x_k=b_k\text{ for }k=0,\ldots,n\}\;;$$ the sets $B(\sigma)$ are a base for $X$. $h^{-1}[B(\sigma)]$ is the set of all points of $C$ whose ternary expansions begin $0.(2b_0)(2b_1)\ldots(2b_n)$, and it’s not hard to check that this is a clopen set in $C$. In fact it’s the set of points in $C$ that are in one of the closed intervals at stage $n+1$ of the usual construction. Nov 1, 2016 at 14:08

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.

• Thank you too for the effort. I need to read more about how to show $f$ is open. However, I am a little stuck on proving the bijectiveness. I got the injectivity by saying that any two different preimage elements result in different elements in the image set. But how about the surjectivity? That is still hard..... Any advice?
– nate
Oct 5, 2011 at 1:58
• @nate: If you know that the Cantor set is the same as $\{\sum_{n=1}^\infty a_n/3^n:\forall n\in\mathbb{N}(a_n\in\{0,2\})\}$, then the surjectivity is easy to see: pick $y=(a_n)_{n=1}^\infty\in\{0,1\}^\mathbb{N}$, then $x=\sum_{n=1}^\infty 2a_n/3^n$ is in the Cantor set and $f(x)=y$. Oct 5, 2011 at 9:33
• @nate: The proof of the last claim is a nice little interplay between compactness and closedness. First note that a bijection is open iff it is closed (follows easily from 'a subset is open iff the complement is closed'). Let $f:X\to Y$ be a continuous bijection between a compact and a Hausdorff space. Pick a closed subset $F$ of $X$. Since a closed subset of a compact space is compact and $f$ is continuous, $f(F)$ is compact. Since a compact subset of a Hausdorff space is closed, $f(F)$ is closed in $Y$. Hence $f$ is a closed (equivalently open) function. Oct 5, 2011 at 9:43

The Cantor set consists of numbers whose ternary expansion uses only $0$s and $2$s. So there's a "natural" bijection between the cantor set and $\{0,1\}^\omega$, or rather $\{0,2\}^\omega$. Everything else should just "work out".

Note that $\{0,1\}^\omega$ consists of all infinite sequences of $0$ and $1$.