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!)

  • $\begingroup$ For some bibliography: Topology by James Dugundji, chapter IV section 4 has a detailed proof along with an application to Peano Curves. $\endgroup$ Commented Sep 9, 2022 at 5:09

3 Answers 3


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

  • $\begingroup$ @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. $\endgroup$ Commented Oct 5, 2011 at 0:17
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    $\begingroup$ @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. $\endgroup$ Commented Oct 5, 2011 at 1:11
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    $\begingroup$ 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. $\endgroup$
    – Who knows
    Commented Nov 1, 2016 at 12:21
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    $\begingroup$ @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. $\endgroup$ Commented Nov 1, 2016 at 14:08
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    $\begingroup$ Is it only I who wonders what's so interesting about $\sum3^{-p_n}$? $\endgroup$ Commented Jul 25, 2018 at 17:26

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.

  • $\begingroup$ 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? $\endgroup$
    – nate
    Commented Oct 5, 2011 at 1:58
  • $\begingroup$ @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 $f(x)=y$. $\endgroup$
    – LostInMath
    Commented Oct 5, 2011 at 9:33
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    $\begingroup$ @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. $\endgroup$
    – LostInMath
    Commented 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 $1$.


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