Let $(C,|\cdot|)$ be a Cantor set metric space and $P=\prod_{k=1}^\infty \left\{0,\frac1{2^k}\right\}$ with metric from $(\ell_1, \|\cdot\|)$. Prove that there exists a continuous bijection $f:C\to P$.

Here is how I would like to proceed:

  • Show that $P$ is compact.

  • Construct a bijective function $g:P\to C$.

  • Show that $g$ is continuous.

  • Deduce that $g^{-1} := f$ exists.


1. $P$ is compact.

Let $(x_m^l)^l\subset P$ be a sequence of sequences, where $m$ denotes the $m$-th element of a sequence, and $l$ denotes the $l$-th sequence in the sequence of sequences. Then $(x_m^l)^l$ contains a subsequence which converges to the constant sequence $(0,0,\dots,0,\dots)$, since we can choose a sufficiently large $l$ in every $x_m^l$ such that $k$ in $\frac1{2^k}$ is sufficiently large.

2. Construct a bijective function $g:P\to C$.

This is where I'm lost. Would appreciate some hint.

  • $\begingroup$ Do you know that the Cantor set is homeomorphic to $\{ 0, 1 \}^{\mathbb{N}}$ ? $\endgroup$
    – Adayah
    Oct 27, 2017 at 7:57
  • $\begingroup$ @Adayah Unfortunately, I don't know much about differential geometry. But, anyway, would appreciate some details. $\endgroup$
    – sequence
    Oct 27, 2017 at 8:04
  • 1
    $\begingroup$ Your strategy is right. If you are willing to use Tychonoffs Theorem ( product of compact spaces is always compact) then you can omit Step 1. For Step 2 you have to know that any point of $C$ can be written uniquely as $\sum \frac {a_n}2^{-n}$ where each $a_n$ is 0 or 1. Map this point to $\{a_n}(2^{-n}\}$. To prove that f is continuous let $x,y$ be in $C$ and consider the first unequal term in their representations. Some simple calculation is needed to complete Step 2. $\endgroup$ Oct 27, 2017 at 8:23
  • $\begingroup$ @KaviRamaMurthy I think you should write a full answer, since your outline is simple and straightforward. $\endgroup$
    – sequence
    Oct 28, 2017 at 6:02

1 Answer 1


We prove a slight generalisation.

Let $M$ be a non-empty compact metric space. Define a "piece" of $M$ to be any compact non-empty subset of $M.$

Lemma 1: Given $\varepsilon >0, \, \exists \, k \in \mathbb{N}$ and pieces $M_i $ (where $i=1, \ldots, 2^k$ ) with $\text{diam} M_i \leq \varepsilon$ such that $M = \displaystyle\bigcup_{i=1}^{2^k}M_i$

Let $a =a_1a_2\ldots a_n$ be a word of length $n$ formed by the letters $a_i$ where $a_i \in \{0,2\}.$ We call $a$ an address string and denote its length by $|a| =n.$ Let $a|k$ denote the truncation of the $a$ to its first $k$ letters.

Define a dyadic filtration of compact metric space $M$ to be a collection $F = \{M_a\}$ of pieces of $M$ such that:

a) $a$ varies over all finite words of length $n$ with letters in $\{0,2\}.$

b) For each $n \in \mathbb{N}, M = \displaystyle\bigcup_{|a| =n} M_a.$

c) If $a$ is expressed as a "compound word" $a=bd$ then $M_a \subset M_b.$

d) $\displaystyle \lim_{n \to \infty} \max\{\text{diam} \,{M_a: |a|=n}\} =0$

Lemma 2: Every non-empty compact metric space $M$ has a dyadic filtration.

Theorem:If $M$ is a non-empty, compact metric space and $C$ is the standard middle-thirds Cantor set, then there exists a continuous surjection from $C$ onto $M.$

Proof of theorem: If $M$ is a compact metric space, then by lemma $2$ we have a dyadic filtration $\{M_a\}$ of $M.$ Since $C$ is homeomorphic to $\{0,2\}^{\mathbb{N}}$ we can represent each point $p \in C$ as $p= p(\omega)$ where $\omega$ is an infinite address string consisting of $0$ and $2.$

Define $f: C \to M$ by $f(p) = q(\omega)$ where $p =p(\omega)$ and $q(\omega) = \displaystyle\bigcap_{n \in \mathbb{N}}M_{\omega|n}.$ To see that $f$ is a surjection, fix $k \in \mathbb{N}$ and note that by definition of a dyadic filtration, $q \in M = \displaystyle\bigcup_{|a|=k}M_a.$ So there exists some address string $l_1$ of length $k$ such that $q \in M_{l_1}.$ Corresponding to $k+1$, there exists an address string $l_2$ of length $k+1$ such that $q\in M_{l_2}.$ We must necessarily have $l_2 = l_1 x$ for some $x \in \{0, 2\}$ and thus $M_{l_2} \subset M_{l_1}.$ Proceeding like this we obtain a nested sequence of compact sets $M_{l_i}$ whose diameter is tending to $0$ (by condition d) of being a dyadic filtration) and thus $\displaystyle \bigcap_{i=1}^{\infty}M_{l_i}$ consists of exactly one point $q.$ Thus the infinite address string $\omega$ defined by $\omega|n =l_n$ is such that $q = q(\omega)$ and $p = p(\omega)$ is such that $f(p) = q(\omega) =q$ and $f$ is surjective.

To see that $f$ is continuous, let $\varepsilon >0$ be given. Choose $k \in \mathbb{N}$ such that $\max\{\text{diam} \,{M_a: |a|=k}\} < \varepsilon$ and choose $\delta >0$ such that $\delta < \frac{1}{3^k}.$ This implies that if $C^{n}$ denotes the $n^{th}$ level of $C$ then the intervals $C_a$ in $C^k$ lie further apart then $\delta.$ Then if $p_1 , p_2 \in C$ such that $|p_1 -p_2|< \delta,$ it follows that there exists a string $a,$ with $|a|=k$ such that $p_1, p_2 \in C_a$. Then if $\omega_1$ and $\omega_2$ are the infinite addresses of $p_1$ and $p_2$ respectively we have $\omega_1|k = \omega_2|k =a.$ Therefore $f(p_1) (= q(\omega_1))$ and $f(p_2) (= q(\omega_2))$ both belong to the same piece $M_a$ and thus $d(f(p_1), f(p_2) < \max\{\text{diam} \,{M_a: |a|=k}\} < \varepsilon$ which proves continuity.

The statement of the theorem can be strengthened to ensure the map is injective too, if $M$ is totally disconnected as well (then for a fixed $n$ all the pieces $M_a$ with $|a| =n$ are disjoint), which is the case for $P$ defined in your question. Then we have a continuous bijection from $C$ to $P$ and since both spaces are compact, we get a homeomorphism.

Note: The proof of lemma 1 and lemma 2 can be found in Pugh's Real Mathematical Analysis Chapter 2, pg 99-101.


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