How to show two properties about the Cantor Set

Question

Define $C_0=[0,1]$ and for $n\in\mathbb{N}$, define $$C_n=C_{n-1}\setminus\bigg(\bigcup_{k=0}^{3^{n-1}-1}\bigg(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\bigg)\bigg) $$ Then the Cantor set is defined as $$C=\bigcap_{n\in\mathbb{N}}C_n$$

Things I need to show:

(1) Each $C_n$ is the disjoint union of $2^n$ closed sub-intervals of length $3^{-n}$ and that the endpoints of each $C_n$ are in $C$.

(2) For any distinct $x,y\in C$, there exists non-empty, disjoint, open sets (open in in $C$) $A,B\subset C$ such that $A\cup B=C$ with $x\in A$ and $y\in B$.

For (1), using induction seemed to me to be the best way to go about proving (1), since I am unsure how to go about proving it for an arbitrary $n\in\mathbb{N}$. However, using the inductive hypothesis with this definition turned out to be unwieldy and awkward, and did not yield any results with the methods I tried.

Proof outline for (2):

Let $x,y\in C$ and WLOG, let $x<y$. Choose $N\in\mathbb{N}$ such that $$3^{N-1}(y-x)>2$$ Now $$3^{N-1}x<3^{N-1}x+2<3^{N-1}y $$ so there is an integer $k\in\mathbb{N}$ such that $$3^{N-1}x<k<k+1<y3^{N-1}$$ Then $$x<\frac{3k}{3^{N}}<\frac{3(k+1)}{3^{N}}<y $$ and finally, we have $$x<\frac{3k}{3^{N}}<\frac{1+3k}{3^N}<\frac{2+3k}{3^N}<\frac{3(k+1)}{3^N}<y $$

The idea here was to find one of the "deleted" intervals which separated $x$ and $y$. If we set $A=(-1,z)$ and $B=(z,2)$ for some $z$ between $\frac{1+3k}{3^N}$ and $\frac{2+3k}{3^N}$, then $A\cap C$ and $B\cap C$ will be open in $C$, they will be disjoint, and their union will equal $C$.

Questions:

How should I proceed in proving (1)? Is this proof for (2) sufficient?

Answer 1

Here we look at the induction proof of the claim

Let \begin{align*} C_n&=C_{n-1}\setminus\bigcup_{k=0}^{3^{n-1}-1}\bigg(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\bigg)\qquad\qquad n\geq 1\tag{1}\\ C_0&=[0,1]\\ \\ C&=\bigcap_{n\in\mathbb{N}}C_n \end{align*} then each $C_n$ is the disjoint union of $2^n$ closed sub-intervals of length $3^{-n}$ and that the endpoints of each $C_n$ are in $C$.

Base step: $n=0$

$C_0=[0,1]$ is the disjoint union of $2^0=1$ closed sub-intervals of length $3^{0}=1$. We also see the endpoints $0$ and $1$ are in $C$, since the open intervals which are subtracted in the definition of $C_n$ do not contain these points.

Induction hypothesis: $n=N$

We assume the claim is valid for $n=N$. In order to apply the hypothesis we look somewhat closer at $C_n$ and derive a further representation. We see (more or less clearly) that in each step the middle third of intervals is subtracted. This indicates to use a representation of the reals at base $b=3$. We start with small $n=0,1,2$ and get \begin{align*} \color{blue}{C_0}&=[0,1] \color{blue}{= [0_3,0.\dot{2}_3]}\tag{2.0}\\ \color{blue}{C_1}&=[0,1]\setminus\left(\frac{1}{3},\frac{2}{3}\right) =\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]\color{blue}{=[0_3,0.0\dot{2}_3]\cup[0.2_3,0.\dot{2}_3]}\tag{2.1}\\ \color{blue}{C_2}&=C_1\setminus \left(\left(\frac{1}{9},\frac{2}{9}\right)\cup\left(\frac{4}{9},\frac{5}{9}\right) \cup\left(\frac{7}{9},\frac{8}{9}\right)\right)\\ \\ &\,\,\color{blue}{=\left[0_3,0.00\dot{2}_3\right]\cup\left[0.02_3,0.0\dot{2}_3\right] \cup\left[0.2_3,0.20\dot{2}_3\right]\cup\left[0.22_3,0.\dot{2}_3\right]}\tag{2.2} \end{align*} We observe in (2.0) - (2.2) the numbers in $C_n$ can be written in ternary representation (indicated with an index $3$) without the digit $1$ at the first $n$ positions after the comma. We can generally write \begin{align*} C_n=\left\{z\in[0,1]\bigg|z=\sum_{j=1}^\infty a_j\cdot 3^{-j}, a_k\in\{0,2\}, 1\leq k\leq n\right\}\tag{3} \end{align*}

We are now well prepared for the induction step.

Induction step: $n=N+1$

We obtain \begin{align*} C_{N+1}&=C_N\setminus\bigcup_{k=0}^{3^{N}-1}\left(\frac{1+3k}{3^{N+1}},\frac{2+3k}{3^{N+1}}\right)\\ &=C_N\setminus\bigcup_{k=0}^{3^{N}-1}\left(\frac{k}{3^N}+\frac{1}{3^{N+1}},\frac{k}{3^N}+\frac{2}{3^{N+1}}\right)\tag{4.1}\\ &=\left\{z\in[0,1]\bigg|z=\sum_{j=1}^\infty a_j\cdot 3^{-j}, a_k\in\{0,2\}, 1\leq k\leq N+1\right\}\tag{4.2} \end{align*} and the claim follows.

Comment:

• In (4.1) we observe the intervals $\left(\frac{k}{3^N}+\frac{1}{3^{N+1}},\frac{k}{3^N}+\frac{2}{3^{N+1}}\right)$ which are subtracted from $C_N$ have length $\frac{1}{3^{N+1}}$. Furthermore these are intervals which are precisely the numbers which can be written with a $1$ at position $N+1$ in ternary representation.

  According to the induction hypothesis $C_N$ is the union of $2^N$ disjoint intervals. Since we subtract from each of these intervals the middle part, the number of disjoint intervals is doubled, giving $2^{N+1}$ intervals.

• In (4.2) we use the result from (4.1) together with the induction hypothesis to obtain this representation. We observe the endpoints of the intervals in $C_{N+1}$ can be written without using the digit $1$ and are this way justified as points in $C$.