# Modified cantor set can be identified with binary sequences

I'm reading this paper on sets of uniqueness, and on page 15 the author constructs the following Cantor-like set:

The Cantor set in the interval $[0, 2π]$ is constructed by removing the middle $1/3$ open interval, then in each of the remaining two closed intervals remaining the middle $1/3$ open interval, etc. Now suppose we modify the construction by removing in the first stage the middle $1/2$ interval, at the second stage, the middle $1/3$ interval, at the third stage the middle $1/4$ interval, etc. Denote the resulting perfect set by $E_M$.

The author then claims that this set is in a canonical $1-1$ correspondence with the set of binary sequences, $2^\mathbb{N}$, and considers the coin-tossing measure on $w^\mathbb{N}$.

My question(s): How is he identifying $E_M$ with $2^\mathbb{N}$? I know the usual correspondence for the ternary Cantor set, so I imagine it must be a similar idea, but I'm not seeing it. Furthermore, what is the coin tossing measure? I tried looking for a definition online, but found nothing.

• Well, if you know the usual correspondence for the ternary Cantor set and $2^{\mathbb{N}}$, then you also know this correspondence. It is just a scaling of the "normal" correspondence. – Kyle Jul 10 '16 at 4:37
• A "coin-tossing" measure on $\{ 0, 1 \}^{\mathbb{N}}$ could be thought of as assigning a set $E$ the probability of the event: toss a fair coin infinitely many times in a row, and at the first step write $1$ if you got heads, $0$ if tails, and do so for each toss, and have the transcribed sequence be an element of $E$. – AJY Jul 10 '16 at 5:31

## 1 Answer

Label the original interval $[0,2\pi]$ with the empty sequence, $I_{\langle\rangle}$. At the first stage you remove the open middle half of the interval, splitting $I_{\langle\rangle}$ into a left subinterval $I_{\langle 0\rangle}$ and a right subinterval $I_{\langle 1\rangle}$, At the next stage you remove the open middle third of each of these intervals, splitting each into a left subinterval and a right subinterval; label these $I_{\langle 00\rangle},I_{\langle 01\rangle},I_{\langle 10\rangle}$, and $I_{\langle 11\rangle}$ from left to right. In general, if $\sigma$ is a finite sequence of $n$ zeroes and ones, you’ll remove the open middle $\frac1{n+2}$ of $I_\sigma$, splitting $I_\sigma$ into a left subinterval $I_{\sigma^\frown0}$ and a right subinterval $I_{\sigma^\frown1}$.

For each $x\in E_M$ there is a unique sequence $\sigma:\Bbb N\to\{0,1\}$ such that

$$\{x\}=\bigcap_{n\in\Bbb N}I_{\sigma\upharpoonright n}\;.\tag{1}$$

E.g.,

$$\left\{\frac{\pi}2\right\}=I_{\langle\rangle}\cap I_{\langle 0\rangle}\cap I_{\langle 01\rangle}\cap I_{\langle 011\rangle}\cap I_{\langle 0111\rangle}\cap\ldots\cap I_{\langle 011\ldots 1\rangle}\cap\ldots\;,$$

since $\frac{\pi}2$, the righthand endpoint of $I_{\langle 0\rangle}$, is in the righthand subinterval of every splitting after the first one. Here the sequence $\sigma$ is $\langle 0,1,1,1,1,\ldots\rangle$, i.e.,

$$\sigma:\Bbb N\to\{0,1\}:n\mapsto\begin{cases} 0,&\text{if }n=0\\ 1,&\text{otherwise}\;. \end{cases}$$

If $h:E_M\to 2^{\Bbb N}$ is the identification, $h(x)$ is simply the unique $\sigma\in 2^{\Bbb N}$ such that $(1)$ is true. Its inverse takes any $\sigma\in 2^{\Bbb N}$ to the unique $x\in E_M$ that is in the intersection of the intervals $I_{\sigma\upharpoonright n}$ for $n\in\Bbb N$.

• Wonderful, as always. Thank you very much! – Reveillark Jul 11 '16 at 1:04
• @Reveillark: You're welcome! – Brian M. Scott Jul 11 '16 at 1:47