How to derive closed formulas of Cantor set?

Question

The Cantor set $\mathcal{C}$ is defined as follows: $$\mathcal{C}:=\bigcap_{n=0}^{\infty}C_n$$ where $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.

From Wikiwand's page, The explicit formulas of Cantor sets are

$$\mathcal{C} = \bigcap_{n=0}^{\infty} \bigcup_{k=0}^{3^n-1} \left(\left[\frac{3k+0}{3^{n+1}}\,,\, \frac{3k+1}{3^{n+1}}\right] \cup \left[\frac{3k+2}{3^{n+1}}\,,\,\frac{3k+3}{3^{n+1}}\right] \right) \space (1)$$ and $$\mathcal{C} = [0,1] \setminus \bigcup\left\{\left(\frac {3k+1}{3^n}, \frac{3k+2}{3^n}\right) \,\middle\vert\, k,n\in \mathbb Z^+\right\} \space (2)$$

I have tried for several days to get $(1)$ and $(2)$ from the definition of Cantor set, but to no avail.

Could you please help me derive formulas $(1)$ and $(2)$? Thank you so much!

Answer 1

I have figured out the proof and posted it as an answer here. It will be great and beneficial if someone helps me verify my attempt. Thank you so much!

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Let $\mathbf{L}(x)=\dfrac{x}{3}$ and $\mathbf{R}(x)=\dfrac{x+2}{3}$. Let $I_0=[0,1]$ and $I_{n+1}=\mathbf{L}(I_n) \cup \mathbf{R}(I_n)$. Cantor set is defined as follows: $$\mathcal{C}:=\bigcap_{n=0}^{\infty} I_n$$

Theorem: $$\mathcal{C} = [0,1] - \bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^n-1} \left( \frac{3k+1}{3^{n+1}} ,\frac{3k+2}{3^{n+1}}\right)$$

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Let $T_n=I_0-I_n$ for all $n\in\Bbb N$.

Lemma 1: $$T_{n+1}=\mathbf{L}(T_n) \cup T_1 \cup \mathbf{R}(T_n)$$

Proof:

Notice that $\mathbf{R}(T_n)=\mathbf{R}(I_0-I_n)=\mathbf{R}(I_0)- \mathbf{R}(I_n) \subseteq \mathbf{R}(I_0) \subseteq I_0-\mathbf{L}(I_0)$. Thus $(I_0-\mathbf{L}(I_0)) \cap\mathbf{R}(T_n)=\mathbf{R}(T_n)$. Similarly, $\mathbf{L}(T_n) \subseteq \mathbf{L}(I_0) \subseteq I_0 -\mathbf{R}(I_0)$ and thus $(I_0-\mathbf{R}(I_0)) \cap\mathbf{L}(T_n)=\mathbf{L}(T_n)$. Moreover, $\mathbf{L}(T_n)\cap\mathbf{R}(T_n) \subseteq \mathbf{L}(I_0)\cap\mathbf{R}(I_0)=\emptyset$.

By the definition of $T_{n+1}$:

$\begin{align}T_{n+1} &=I_0-I_{n+1}\\ &=I_0-(\mathbf{L}(I_n) \cup \mathbf{R}(I_n))\\ &=(I_0-\mathbf{L}(I_n)) \cap (I_0 - \mathbf{R}(I_n))\\ &=[I_0-\mathbf{L}(I_0-T_n)] \cap [I_0 - \mathbf{R}(I_0-T_n)]\\ &=[I_0-(\mathbf{L}(I_0)-\mathbf{L}(T_n))] \cap [I_0 - (\mathbf{R}(I_0)-\mathbf{R}(T_n))]\\ &=[(I_0-\mathbf{L}(I_0)) \cup \mathbf{L}(T_n)] \cap [(I_0-\mathbf{R}(I_0)) \cup \mathbf{R}(T_n)]\\ &= [(I_0-\mathbf{L}(I_0)) \cap (I_0-\mathbf{R}(I_0))] \cup [(I_0-\mathbf{L}(I_0)) \cap\mathbf{R}(T_n)] \cup [\mathbf{L}(T_n)\cap(I_0-\mathbf{R}(I_0))] \cup [\mathbf{L}(T_n)\cap\mathbf{R}(T_n)]\\ &=[I_0 -\mathbf{(L}(I_0)\cup\mathbf{R}(I_0))]\cup \mathbf{R}(T_n) \cup \mathbf{L}(T_n)\\ &=(I_0-I_1)\cup \mathbf{R}(T_n) \cup \mathbf{L}(T_n)\\ &= T_1 \cup \mathbf{R}(T_n) \cup \mathbf{L}(T_n)\\ &=\mathbf{L}(T_n) \cup T_1 \cup \mathbf{R}(T_n)\end{align}$

Lemma 2: $$\bigcup_{m=0}^{n+1} \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right) = \left(\frac{1}{3},\frac{2}{3} \right) \cup \left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^{m+1}-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right]$$

Proof:

Let $t=m+1$. Then $RHS=\left(\frac{1}{3},\frac{2}{3} \right) \cup \left[\bigcup_{t=1}^{n+1} \bigcup_{k=0}^{3^t-1} \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right)\right]$.

Moreover, $\bigcup_{t=0}^0 \bigcup_{k=0}^{3^t-1} \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right) = \bigcup_{k=0}^{3^0-1} \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right) = \bigcup_{k=0}^0 \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right)=$ $\left(\frac{1}{3},\frac{2}{3} \right)$.

It follows that $RHS = \left [ \bigcup_{t=0}^0 \bigcup_{k=0}^{3^t-1} \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right) \right] \cup \left[\bigcup_{t=1}^{n+1} \bigcup_{k=0}^{3^t-1} \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right)\right]=$ $\bigcup_{t=0}^{n+1} \bigcup_{k=0}^{3^t-1} \left(\frac{3k+1}{3^{t+1}}, \frac{3k+2}{3^{t+1}}\right) = \bigcup_{m=0}^{n+1} \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right) =LHS$.

Lemma 3: $$T_{n+1}=\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right)$$

Proof:

We prove the assertion by induction on $n$. It is clear that it trivially holds for $n=0$. Let it hold for a positive integer $n$.

Notice that $\frac{1}{3} < \frac{3k+1}{3^{m+2}} < \frac{3k+2}{3^{m+2}} < \frac{2}{3}$ for all $3^m \le k \le 2.3^m-1$. It follows that $\bigcup_{m=0}^n \bigcup_{k=3^{m}}^{2.3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right) \subseteq \left(\frac{1}{3},\frac{2}{3} \right)$ and thus $\left(\frac{1}{3},\frac{2}{3} \right)= \left [ \left(\frac{1}{3},\frac{2}{3} \right) \cup \left( \bigcup_{m=0}^n \bigcup_{k=3^{m}}^{2.3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right) \right)\right ]$.

By Lemma 1,

$T_{(n+1)+1}=T_{n+2}=\mathbf{L}(T_{n+1}) \cup T_1 \cup \mathbf{R}(T_{n+1})$

$=\mathbf{L}\left(\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right)\right) \cup \left(\frac{1}{3},\frac{2}{3} \right) \cup \mathbf{R}\left(\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right)\right)$

$=\left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right] \cup \left(\frac{1}{3},\frac{2}{3} \right) \cup \left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1+2.3^{m+1}}{3^{m+2}}, \frac{3k+2+2.3^{m+1}}{3^{m+2}}\right)\right]$

$=\left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right] \cup \left(\frac{1}{3},\frac{2}{3} \right) \cup \left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3(k+2.3^m)+1}{3^{m+2}}, \frac{3(k+2.3^m)+2}{3^{m+2}}\right)\right]$

$=\left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right] \cup \left(\frac{1}{3},\frac{2}{3} \right) \cup \left [\bigcup_{m=0}^n \bigcup_{k=2.3^{m}}^{3^{m+1}-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right]$

$=\left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right] \cup \left [ \left(\frac{1}{3},\frac{2}{3} \right) \cup \left( \bigcup_{m=0}^n \bigcup_{k=3^{m}}^{2.3^m-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right) \right)\right ] \cup \left[ \bigcup_{m=0}^n \bigcup_{k=2.3^{m}}^{3^{m+1}-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right]$

$= \left(\frac{1}{3},\frac{2}{3} \right) \cup \left[\bigcup_{m=0}^n \bigcup_{k=0}^{3^{m+1}-1} \left(\frac{3k+1}{3^{m+2}}, \frac{3k+2}{3^{m+2}}\right)\right]$ $=\bigcup_{m=0}^{n+1} \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right)$ by Lemma 2

We proceed to prove our main theorem.

$\mathcal{C}:=\bigcap_{n=0}^{\infty} I_n=\bigcap_{n=0}^{\infty} (I_0-T_n)=I_0-\bigcup_{n=0}^{\infty} T_n=I_0-\bigcup_{n=1}^{\infty}T_n$ [since $T_0=\emptyset$]

$=I_0-\bigcup_{n=0}^{\infty}T_{n+1}=I_0-\bigcup_{n=0}^{\infty}\bigcup_{m=0}^n \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right)$

$=I_0-\bigcup_{m=0}^{\infty} \bigcup_{k=0}^{3^m-1} \left(\frac{3k+1}{3^{m+1}}, \frac{3k+2}{3^{m+1}}\right)$

$=I_0-\bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^n-1} \left(\frac{3k+1}{3^{n+1}}, \frac{3k+2}{3^{n+1}}\right)$.

This completes the proof.

Answer 2

The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = \bigcap_n C_n$ by definition.

Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.