Ex3-2

=== Problem 3.2 ===
When b=3 in Problem 3.1, the expansion is called the triadic or ternary expansion of x.
(a) Show that the Cantor set \mathcal{C} consists of all x such that some triadic expansion for which every c_kis either 0 or 2.
=== Solution. ===
Define the sets D=\left[0,1\right]-\mathcal{C}=\bigcap^\infty_{k=1}\bigcup^{3^{k-1}-1}_{\ell=0}\left(\dfrac{3\ell+1}{3^k},\dfrac{3\ell+2}{3^k}\right)1,E=\left\{x\in\left[0,1\right] \ \biggr| \ x \text{ has a digit 1 in its ternary expansion} \right\} and S=\left\{\sum^\infty_{n=1}\dfrac{a_n}{3^n} \ \biggr| \ a_n\in\{0,2\}\right\}.
We suffice to show that x\in\mathcal{C} if and only if x\in S. Suppose that x\in S, then x=\sum^\infty_{n=1}\dfrac{a_n}{3^n} for some a_n\in\left\{0,2\right\}. 
Claim 1. The partial sum x_k=\sum^k_{n=1}\dfrac{a_n}{3^n} is the left endpoint of a closed subinterval of \mathcal{C}_k.
Proof. We prove it using induction. If k=1, then x_1 is either 0 or \dfrac{2}{3}, either of these is the left endpoint of \mathcal{C}_1=\left[0,\dfrac{1}{3}\right]\cup\left[\dfrac{2}{3},1\right]. Assume that x_{k-1} is a left endpoint of \mathcal{C}_{k-1}, then \left[x_{k-1},x_{k-1}+\dfrac{1}{3^{k-1}}\right]\subset \mathcal{C}_{k-1}. By the construction of the Cantor set \mathcal{C}, we trisect the interval \left[x_{k-1},x_{k-1}+\dfrac{1}{3^{k-1}}\right] and remove the middle closed interval to obtain the closed subinterval of \mathcal{C}_{k}:\left[x_{k-1},x_{k-1}+\dfrac{1}{3^k}\right]\cup\left[x_{k-1}+\dfrac{2}{3^k},x_{k-1}+\dfrac{1}{3^{k-1}}\right]\subset\mathcal{C}_kThis implies that x_{k - 1}+\dfrac{a_k}{3^k} is a left endpoint of the closed subinterval of \mathcal{C}_k for each a_k\in\left\{0,2\right\}. Since each x_n is now the left endpoint of \mathcal{C}_n, we have x_n\in\mathcal{C}_n. As the Cantor set \mathcal{C} is the arbitrary intersection of closed intervals \mathcal{C}_n, that is, C=\bigcap^\infty_{n=1}\mathcal{C}_n, so \mathcal{C} is closed and it contains all limit points of S, and finally we have x:=\lim\limits_{n\to\infty}x_n\in \mathcal{C} as desired. This shows S\subseteq\mathcal{C}.\square
On the other hands, suppose that x\not\in S for the sake of contrapositive, which implies that x\in E instead, then x appears to have the least number M such that a_M=1 for its ternary representation. Note that there cannot exists N\in\mathbb{N} such that x_n=0 or 2 for every n>N, otherwise x would have another ternary expansion without the digit 1, as 1.000\cdots_{10}=0.222\cdots_{3} holds by geometric series. Write x asx=\sum^{M-1}_{n=1} \dfrac{a_n}{3^n}+\dfrac{1}{3^M}+\sum^{\infty}_{n=M+1} \dfrac{a_n}{3^n}Now we rewrite \sum^{M-1}_{n=1} \dfrac{a_n}{3^n}=\dfrac{3}{3^{M}}\sum^{M-1}_{n=1}\dfrac{a_n}{3^{n-M+1}} and since 0<\sum^\infty_{n=M+1}\dfrac{a_n}{3^n}<\dfrac{1}{3^M} holds2 for a_n\in\left\{0,2\right\} and any M\in\mathbb{N}, it follows that0<\sum^\infty_{n=1}\dfrac{a_n}{3^n} - \sum^M_{n=1}\dfrac{a_n}{3^n}<\dfrac{1}{3^M}0<x-\sum^{M-1}_{n=1}\dfrac{a_n}{3^n}-\dfrac{1}{3^M}<\dfrac{1}{3^M}\dfrac{1}{3^M}\left(3\sum^{M-1}_{n=1}\dfrac{a_n}{3^{n-M+1}}+1\right)<x<\dfrac{1}{3^M}\left(3\sum^{M-1}_{n=1}\dfrac{a_n}{3^{n-M+1}}+2\right)
This shows that x\in\bigcap^\infty_{k=1}\bigcup^{3^{k-1}-1}_{\ell=0}\left(\dfrac{3\ell+1}{3^k},\dfrac{3\ell+2}{3^k}\right) for suitable \ell and k, hence x\not\in\mathcal{C}. This completes the proof.\blacksquare
----1 One can show the equality using induction and some topology facts. This is left as an exercise to the reader.
2 This can be proved by comparison test. Note that we do not allow a_n=0 or 2 for all n\in\mathbb{N}.
Category:Chapter 3