# Confused by Baby Rudin chapter $3$ problem $19$

Associate to each sequence $$a={a_n}$$, in which $$a_n$$ is $$0$$ or $$2$$, the real number $$x(a)=\sum \frac{a_n}{3^n}$$. Prove all the $$x(a)$$ is precisely the Cantor set.

My attempt:

If I let $$a_n=2$$ then $$x(a)=1$$, $$a_n=0,\text{ then } x(a)=0$$. So, $$x(a)\in [0,1]$$. That problem looks right and I need a more precise proof. I notice when I take $$a_n=2$$, I do the sum from $$1$$ to $$n$$ it always the right point of interval

I know a technology that proof two set are equal if I prove $$A\subset B\,\text{ and }B \subset A$$ then $$A=B$$ but it seems like I can’t use that technology.

• A precise proof requires a precise definition of the Cantor set – Hagen von Eitzen Feb 28 '19 at 14:37
• @HagenvonEitzen i know how to construct the cantor set ,is the construction process a precise definition? – jackson Feb 28 '19 at 14:41

There's nothing to be said really. It's intuitively very clear because at each step you're throwing out the middle interval which has numerator $$1$$ in the corresponding digit in the base $$3$$ expansion. If you want to make it rigorous, it will only get ugly and will not give you any new insight. But here's an ugly proof for a simple beautiful fact:

Let's say that $$x \in C = \cap_{n=1}^{\infty}C_n$$. Then for each $$n\in\mathbb{N}$$, $$x \in C_n$$ where $$C_n$$ is the union of $$2^n$$ disjoint intervals of length $$\frac{1}{3^n}$$. This means that for each $$n\in\mathbb{N}$$ you can find a number $$0\leq y_n \leq 3^{n+1}-1$$ such that $$x \in [\frac{y_n}{3^{n+1}}, \frac{y_n+1}{3^{n+1}}]$$

It remains to show that $$y_n \equiv 0 \pmod{3^k} \,\,\text{ or }\,\, y_n \equiv 2 \pmod{3^k}\,\,\,\text{ for } k

You can prove this by induction. It is trivially true for the base case $$m=1$$. Assuming it's true for all $$m \leq n$$, consider step $$m=n+1$$ of the construction of Cantor's set. At step $$n$$, we have assumed that $$x \in [\frac{y_n}{3^{n+1}}, \frac{y_n+1}{3^{n+1}}]$$ where $$y_n = 3^{n-1}j_n + a_n$$ where $$a_n=0,2$$; so we know which subinterval in $$C_n$$ we're at. Let's divide this interval into three intervals of equal length. Our new $$y_{n+1}$$ will then be given by $$3y_{n}+a_{n+1}$$ where $$a_{n+1}=0,2$$. This proves that

$$y_n \equiv 0 \pmod{3^k} \,\,\text{ or }\,\, y_n \equiv 2 \pmod{3^k}\,\,\,\text{ for } k

Now we have found a sequence $$\{a_n\}_{n=1}^{\infty}$$ that satisfies the following relations

$$\frac{y_n}{3^{n+1}}\leq x < \frac{y_n+1}{3^{n+1}}$$ And $$y_n = 3y_{n-1} + a_n$$. This gives that $$y_n=3(3y_{n-2}+a_{n-1})+a_n=9y_{n-2}+3a_{n-1}+a_n$$. Continuing in this way, we have

$$y_n = 3^{n-1}y_1 + 3^{n-2}a_2 + \cdots + 3a_{n-1} + a_n$$

where $$y_1 = a_1$$. This shows that

$$\frac{\sum_{i=1}^n 3^{n-i}a_i}{3^{n+1}} \leq x < \frac{\sum_{i=1}^n 3^{n-i}a_i + 1}{3^{n+1}}$$

$$\sum_{i=1}^n \frac{a_i}{3^{i+1}} \leq x < \sum_{i=1}^n \frac{a_i}{3^{i+1}} + \frac{1}{3^{n+1}}$$

So, $$x \to 0.a_1a_2\cdots$$ in base $$3$$ where $$a_n = 0 \text{ or } 2$$ and $$\{a_n\}_{n=1}^{\infty}$$ is your desired sequence. Conversely, given a sequence $$\{a_n\}_{n=1}^{\infty}$$ which satisfies $$a_n=0$$ or $$a_n=2$$ for all $$n\in\mathbb{N}$$, the above inequalities and construction shows that $$x \in C$$. $$\fbox{Q.E.D.}$$

• Sorry I don’t understand what is $y_n\overset{3^k}\equiv 0\vee y_n \overset{3^k}\equiv 2$means – jackson Mar 1 '19 at 1:56
• @jackson Do you know about congruences? $a \equiv b \pmod{n}$ for example? – stressed out Mar 1 '19 at 1:57
• yes I know a little number theory.. – jackson Mar 1 '19 at 2:00
• @jackson OK. $a \overset{n}\equiv b$ is just another notation for $a \equiv b \pmod{n}$. Technically, it means that $a$ and $b$ have the same remainder when you divide them by $n$. Equivalently, you can say that $y_n\overset{3^k}\equiv 0\vee y_n \overset{3^k}\equiv 2$ is the same as $y_n = 3^kj_n+ 0$ or $y_n=3^kj_n + 2$. Is it clear now? – stressed out Mar 1 '19 at 2:04
• thanks a lot ,very clear – jackson Mar 1 '19 at 2:05