Bijection between Cantor set and binary sequences; Hunter

Question

I am reading John K. Hunter's lecture notes, specifically chapter 5, section 5.5., about the Cantor set. I have some questions about the proof of the theorem stating that the Cantor set has the same cardinality as the set of infinite binary sequences.

As is well known, the Cantor set is defined as the intersection of sets $F_n$ defined as follows

We define a nested sequence $\left(F_n\right)$ of sets $F_n \subset[0,1]$ as follows. First, we remove the middle-third from $[0,1]$ to get $F_1=[0,1] \backslash(1 / 3,2 / 3)$, or $$ F_1=I_0 \cup I_1, \quad I_0=\left[0, \frac{1}{3}\right], \quad I_1=\left[\frac{2}{3}, 1\right] . $$ Next, we remove middle-thirds from $I_0$ and $I_1$, which splits $I_0 \backslash(1 / 9,2 / 9)$ into $I_{00} \cup I_{01}$ and $I_1 \backslash(7 / 9,8 / 9)$ into $I_{10} \cup I_{11}$, to get $$ \begin{aligned} & F_2=I_{00} \cup I_{01} \cup I_{10} \cup I_{11}, \\ & I_{00}=\left[0, \frac{1}{9}\right], \quad I_{01}=\left[\frac{2}{9}, \frac{1}{3}\right], \quad I_{10}=\left[\frac{2}{3}, \frac{7}{9}\right], \quad I_{11}=\left[\frac{8}{9}, 1\right] . \end{aligned} $$ Then we remove middle-thirds from $I_{00}, I_{01}, I_{10}$, and $I_{11}$ to get $$ F_3=I_{000} \cup I_{001} \cup I_{010} \cup I_{011} \cup I_{100} \cup I_{101} \cup I_{110} \cup I_{111},$$ $$ I_{000}=\left[0, \frac{1}{27}\right], \quad I_{010}=\left[\frac{2}{27}, \frac{1}{9}\right], \quad I_{010}=\left[\frac{2}{9}, \frac{7}{27}\right], \quad I_{011}=\left[\frac{8}{27}, \frac{1}{3}\right], \tag{*} $$ $$ I_{100}=\left[\frac{2}{3}, \frac{19}{27}\right], \quad I_{101}=\left[\frac{20}{27}, \frac{7}{9}\right], I_{110}=\left[\frac{8}{9}, \frac{25}{27}\right], \quad I_{111}=\left[\frac{26}{27}, 1\right] .$$ Continuing in this way, we get at the $n$th stage a set of the form $$ F_n=\bigcup_{\mathbf{s} \in \Sigma_n} I_{\mathbf{s}} $$ where $\Sigma_n=\{\left(s_1, s_2, \ldots, s_n\right): s_n=0,1\}$ is the set of binary $n$-tuples.

Note that there is a typo in $(*)$. We have $I_{010}$ twice. The first occurrence of $I_{010}$ should be $I_{001}$.

The author goes on to prove the following theorem:

Theorem 5.66. The Cantor set has the same cardinality as $\Sigma$ [where $\Sigma=\{(s_1,s_2,\ldots,s_k,\ldots):s_k=0,1\}$ is the set of infinite binary sequences].

Prerequisites for the proof are that the diameter of a set $A$ ($\mathrm{diam} A$) is defined as $\mathrm{sup}\{|x-y|:x,y\in A\}$. If $\{A_n:n\in\mathbb{N}\}$ is a decreasing sequence of nonempty compact sets of real numbers (i.e. $A_1\supset A_2\supset \ldots$; this is referred to as a nested sequence) and $\mathrm{diam} A_n\to 0$ as $n\to\infty$, then $\bigcap_{n=1}^\infty A_n$ consists of a single point (this result is referred to as Theorem 5.43).

Proof. We use the same notation as above. Let $\mathbf{s}=\left(s_1, s_2, \ldots, s_k, \ldots\right) \in \Sigma$, and define $\mathbf{s}_n=\left(s_1, s_2, \ldots, s_n\right) \in \Sigma_n$. Then $\left(I_{\mathbf{s}_n}\right)$ is a nested sequence of intervals such that diam $I_{\mathbf{s}_n}=1 / 3^n \to 0$ as $n \to \infty$. Since each $I_{\mathbf{s}_n}$ is a compact interval, Theorem 5.43 implies that there is a unique point $$ x \in \bigcap_{n=1}^{\infty} I_{\mathbf{s}_n} \subset C . $$ Thus, $\mathrm{s} \mapsto x$ defines a function $f: \Sigma \rightarrow C$. Furthermore, this function is one-to-one: if two sequences differ in the $n$th place, say, then the corresponding points in $C$ belong to different intervals $I_{\mathbf{s}_n}$ at the $n$th stage of the construction, and therefore the points are different since the intervals are disjoint.

Conversely, if $x \in C$, then $x \in F_n$ for every $n \in \mathbb{N}$ and there is a unique $\mathbf{s}_n \in \Sigma_n$ such that $x \in I_{\mathbf{s}_n}$. The intervals $\left(I_{\mathbf{s}_n}\right)$ are nested, so there is a unique sequence $\mathbf{s}=\left(s_1, s_2, \ldots, s_k, \ldots\right) \in \Sigma$, such that $\mathbf{s}_n=\left(s_1, s_2, \ldots, s_n\right)$. It follows that $f: \Sigma \rightarrow C$ is onto, which proves the result.

1. What exactly does the author mean that $\left(I_{\mathbf{s}_n}\right)$ is a nested sequence of intervals? Not every $\mathbf{s}_3\in\Sigma_3$ makes $I_{\mathbf{s}_3}$ a subset of $I_{\mathbf{s}_2}$ for a given $\mathbf{s}_2\in\Sigma_2$, e.g. $I_{111}\not\subseteq I_{01}$.
2. How come $\bigcap_{n=1}^\infty I_{\mathbf{s}_n}\subset C$?
3. I do not understand the part below. Why does this follow from the intervals being nested?

The intervals $\left(I_{\mathbf{s}_n}\right)$ are nested, so there is a unique sequence $\mathbf{s}=\left(s_1, s_2, \ldots, s_k, \ldots\right) \in \Sigma$, such that $\mathbf{s}_n=\left(s_1, s_2, \ldots, s_n\right)$.

Answer 1

Your objection in item 1 is not relevant, because the author does not assert that "every $\mathbf{s}_3 \in \Sigma_3$ makes $I_{\mathbf{s}_3}$ a subset of $I_{\mathbb{s}_2}$".

Instead, the author starts with a particular $x \in C$, and from that $x$ they then constructs an $\mathbf{s}_n$ for each $n$, and using those particular $\mathbf{s}_n$ that are constructed from $x$ one gets nested intervals $I_{\mathbf{s}_1} \supset I_{\mathbf{s}_2} \supset I_{\mathbf{s}_3} \supset \cdots$.

---

Regarding 2, I assume you have jumped backwards to the proof of the forward implication of the proof. By definition $C = \bigcap_{n=1}^\infty F_n$. Also, by construction $I_{\mathbf{s}_n} \subset F_n$ for all $n$. Therefore, $\bigcap_{n=1}^\infty I_{\mathbf{s}_n} \subset \bigcap_{n=1}^\infty F_n = C$.

---

Regarding 3, because the intervals are nested, i.e. because $I_{\mathbf{s}_1} \supset I_{\mathbf{s}_2} \supset I_{\mathbf{s}_3} \supset I_{\mathbf{s}_4} \cdots$, it follows that $\mathbf{s}_1$ is the first term of $\mathbf{s}_2$, and that $\mathbf{s}_2$ is the first two terms of $\mathbf{s}_3$, and that $\mathbf{s}_3$ is the first three terms of $\mathbf{s}_4$, and so on and so on .......

Therefore, there exists an infinite sequence $\mathbf{s}$ whose first term is $\mathbf{s}_1$ and whose first two terms are $\mathbf{s}_2$ and whose first three terms are $\mathbf{s}_3$ and whose first four terms are $\mathbf{s}_4$ and so on and so on ......

Answer 2

The "usual" construction of the Cantor set starts with an interval, then removes the middle third of that interval, leaving a "left interval" and "right interval". Then the same process is applied to the two subintervals created in that first step, then to the four subintervals created in the second step, and so on.

The idea is that a word $s$ gives us the "address" of a point in the Cantor set by specifying a sequence of intervals which converge to that point. The interpretation is that if the $n$-th letter in the word is a $0$, then "choose the left interval" in the $n$-th stage of the construction of the Cantor set. Similarly, $1$ means "choose the right interval". So, for example, take $\vec{s} = 001\ldots)$.

The first several intervals $I_{s_n}$ are then \begin{align} I_0 &= [0, 1/3], && \text{(the left subinterval in the first step)} \\ I_{00} &= [0, 1/9], && \text{(the left subinterval of $I_0$)} \\ I_{001} &= [2/27, 1/9]. && \text{(the right subinterval of $I_{00}$)} \end{align} Note that $I_{s_n 0}$ and $I_{s_n 1}$ are always subintervals of $I_{s_n}$, hence the sequence of intervals generated in this manner is nested.