The binary map is surjective.

Question

The following is the question (and its solution) that I was reading in the following link The binary map is bijective. but I did not fully understand the proof of surjectivity there.

The Question:

Consider the probability space $X = \{0,1\}^{\mathbb N}$ of sequences $x = (a_1, a_2, a_3, \dots)$ where each $a_i$ is either $0$ or $1.$ Take $\sigma$ to be the smallest $\sigma$-algebra that contains all cylinder sets $$C(n,A) = \{\textbf{x}| x_n \in A, A \subset \{0,1\}\}$$ that is, those sequences whose $n^{th}$ term belongs to some subset $A$ of $\{0,1\}.$ Let $\mu$ be the measure so that, given a finite collection $A_1, \dots , A_n \subset \{0,1\},$ we have $$\mu (\{\textbf{x} \,|\, x_i \in A_i, i = 1, \dots , n \}) = \Pi_{i=1}^{n}\frac{|A_i|}{2}.$$In the first step, it was proved that the set $X_{\infty}$ has measure zero.

Consider the map $\phi: X\setminus X_{\infty} \to [0,1)$ defined by $$\phi(a_1, a_2, \dots , a_n, \dots) = \sum_{i=1}^{\infty}a_i2^{-i}$$ so $\phi$ sends each sequence $(a_1, a_2,\dots)$ to the point $x$ in $[0,1)$ with that binary decimal expansion. Check that $\phi$ is a bijection. We may assume that $\phi$ is measurable with measurable inverse.

My questions are (The original OP question):

First: To check that it is 1-1: Assume that $\phi(\textbf{x}) = \phi(\textbf{y}) $ then $$\sum_n a_n 2^{-n} = \sum_n b_n 2^{-n}$$ but then how can I conclude that $a_i = b_i$ for every $1\leq i \leq n.$

For being surjective, I think this is immediate. Right?

$X_{\infty}$ definition.

the set $X_{\infty} = \{ (a_1, a_2, a_3, \dots )\,|\,\exists N s.t. \forall n \geq N, a_n = 1 \}$

The solution to the surjectivity part:

For surjectivity, you need to give a procedure/algorithm for how to find a sequence $(a_1,a_2,\ldots)$ in $X\setminus X_\infty$ such that $\phi(a_1,a_2,\ldots) = x$. I would suggest thinking about exactly how one does this in decimal. For instance how do you find the decimal representation of a number like $\sqrt{2}$. Well, you build it decimal place by decimal place. You first note that $1<\sqrt{2}<2$, so the first digit is 1. Then you note that $0.4 < \sqrt{2} - 1 < 0.5$, so the second digit is 4. Then $0.01 < \sqrt{2} - 1.4 < 0.02$, and so on. How might this process look in binary instead of base $10$, and can you formalise it?

You need to then justify why this sequence of finite approximations converges to the correct number. That is, show that $(x-\sum_{i=1}^n a_n 2^{-n}) \to 0$ as $n \to \infty$, and therefore conclude that $x=\sum_{i=1}^\infty a_n 2^{-n}$. You also need to explain why the sequence you generate is not in $X_\infty$.

My questions are:

I do not know how to come up with the algorithm in case of the binary, any clarification will be greatly appreciated!

Answer 1

Let $x\in [0,1)$ be any real number. Then $x$ can be represented as an infinite binary number $0.a_1a_2a_3\dots$, like the decimal case. For instance, in the decimal case we have $1/3=0.333\dots$, $1/2=0.500\dots$, $\sqrt{2}-1=0.4142\dots$, $\pi-3=0.14159\dots$. In the binary case we have $1/3=0.010101\dots$, $1/2=0.1000\dots$, $\sqrt{2}-1=0.011010\dots$, $\pi-3=0.001001\dots$. If ${\bf x}=(a_1,a_2,\dots)\in \{0,1\}^{\mathbb N}\setminus X_\infty$ then $\phi({\bf x})$ is naturally related with the binary representation $x=0.a_1a_2a_3\dots$ of $x$, namely, $$\phi({\bf x})=\phi(a_1,a_2,\dots)=0.a_1a_2a_3\dots=x.$$ To study the injectivity and surjectivity of the map $\phi$ we need to study when $x$ can have different binary representations. This is also similar to the decimal case. For instance, $0.5=0.5000\dots=0.4999\dots$. The number $x$ has a unique binary representation, but the case when $x>0$ is binary-rational, that is there exists a natural number $N$ such that $x2^N$ is integer. In the latter case $x$ has exactly two binary representations, the first, where all but finitely many digits are zeroes, and the second, where all but finitely many digits are ones. For instance, $1/2=0.1000\dots=0.01111\dots$. In this case we choose the first representation. Now we define the map $\psi$ from $[0,1)$ to $X$ by putting $\psi(x)=(a_1,a_2,\dots)$, where the binary representation $0.a_1a_2a_3\dots$ of $x$ is chosen as above. It is easy to see that this choice ensures that the range $\psi([0,1))$ of $[0,1)$ is exactly $\{0,1\}^{\mathbb N}\setminus X_\infty$ and the construction of the map $\psi$ ensures that both compositions $\phi\circ\psi$ and $\psi\circ\phi$ are the identity maps (of $\{0,1\}^{\mathbb N}\setminus X_\infty$ and $[0,1)$, respectively). Hence both maps $\psi$ and $\phi$ are bijections.