The binary map is injective.

Question

I was reading the following question here The binary map is bijective.:

But I did not get the idea of the proof of injectivity in details there in details.

Here is the question and the solution of injectivity given 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 part, it was proved that $X_{\infty}$ has measure zero. Now, 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(of the OP who asked the questions there):

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 show injectivity, suppose $a_i \neq b_i$ for some $i \geq 1$ and let $i$ be the least index at which they differ (so we assume that $a_j=b_j$ for all $1\leq j <i$. That means one of $a_i$ or $b_i$ is $1$ and the other is $0$. WLOG suppose that $a_i=1$ and $b_i=0$. So we have $$\phi(a) = \sum_{j=1}^{i-1} a_j2^{-j} + 2^i +\sum_{n=i+1}^\infty a_n 2^{-n}$$ and $$\phi(b) = \sum_{j=1}^{i-1} a_j2^{-j} +\sum_{n=i+1}^\infty b_n 2^{-n}.$$ From this, can you find a lower bound for $\phi(a)$ and an upper bound for $\phi(b)$? What must each of the entries $b_n$ for $n \geq {i+1}$ be in order for $\phi(b)$ to achieve this upper bound?

Now, my own questions are:

1- I think the lower bound for $\phi(a)$ will be achieved when $a_{i+1} = 0$ and in this case the sum of this geometric series part will be zero. Am I correct?

2- And the upper bound of $\phi(b)$ will be achieved when $b_{n+1}$ is $1$ and in this case the sum of the geometric series will $1/2^i$. Am I correct?

3- lastly, why we are thinking of the lower bound of $\phi(a)$and not the upper bound? And same question for $\phi(b)$,why we are searching for the upper bound of it?

4- What exactly the idea of the contradiction I will get in proving injectivity?

Answer 1

Much simpler: First, let us write quite suggestively $$ \phi(a_1, a_2, \dots ) = \sum_{i=1}^{\infty}a_i2^{-i}\color{red}{=0.a_1a_2\cdots}, $$ which is the usual form of writing the number in base $2$. For a sequence in $X_\infty$ there are two ways of writing $\sum_{i=1}^{\infty}a_i2^{-i}$: if $N$ is the smallest integer in the definition of $X_\infty$, then $a_{N-1}=0$ and so $$ \sum_{i=1}^{\infty}a_i2^{-i}=0.a_1\cdots a_{N-2}0111\cdots=0.a_1\cdots a_{N-2}1000\cdots. $$ Therefore, for $(a_1, a_2, \dots )$ and $(b_1, b_2, \dots )$ in $ X\setminus X_\infty$, if $$ \phi(a_1, a_2, \dots )=\phi(b_1, b_2, \dots ),$$ which is the same as $$ 0.a_1a_2\cdots=0.b_1b_2\cdots, $$ $a_i=b_i$ for all $i$ (since we excluded $X_\infty$). This shows that $(a_1, a_2, \dots )$ and $(b_1, b_2, \dots )$ are equal.