User:J D Bowen/Math710 HW3

2.37) Notice that any point $$x\in C \ $$ must be in one of the two intervals from the first step; then, it must be either the left or the right interval of THAT interval; then, it must be either the left or the right interval of THAT interval, and so on. Hence, we can divide the Cantor set into a collection of non-empty sets which can be identified by an infinite string of 0s and 1s, the bit in the $$n^{th}$$ position indicating whether it is in the left or right interval from the $$n^{th}$$ step.

Since the Cantor set is totally disconnected, each of these sets is a singleton, and hence, every point in the Cantor set can be represented in this way. Hence the Cantor set $$C \ $$ can be put in one-to-one correspondence with the set of all sequences $$\left\{{a_i}\right\}_{i=1}^\infty, \ a_i\in\left\{{0,1}\right\} \ $$. Call this set of sequences $$S \ $$.

Let $$X=\left\{{x_1, \dots, x_i, \dots}\right\}$$ be any countable subset of $$S$$ (realize each $$x_i$$ is a sequence of 1s and 0s). If form the new set $$Y=\left\{{x_2, x_4, x_6, \dots }\right\} \ $$, we can easily see that X and Y are in one-to-one correspondence, since both are countable. Hence we can take $$S$$, remove a countable set of points $$E$$, and the two sets $$S, \ S\backslash E $$ will be in one-to-one correspondence.

Now return to problem 4 from the first homework, where we demonstrated that any number $$x \in [0,1) \ $$ can be represented by a sum $$\Sigma a_ip^{-i} \ $$. If we take $$p=2$$, then we have an infinite sequence of 0s and 1s representing any point in [0,1).

Let $$R\subset S \ $$ be the set of all sequences ending in an infinite string of 1s. In 4b from homework 1, we demonstrated that all sequences in $$R $$ represent the some number $$x\in[0,1)$$ that is already represented by a sequence in $$S$$ which terminates in an infinite string of 0s. We also showed that this is the only situation where this happens.  Hence, $$S\backslash R$$ can be put in one-to-one correspondence with [0,1).

Next, note that since every sequence in $$R$$ is all 1s after some number $$N$$, and since there are only $$2^N$$ possible sequences with all 1s after the $$N^{th}$$ term, the set $$R$$ is countable.

Then $$C \sim S \sim S\backslash R \sim [0,1) \ $$.

2.38) Show the set of acc. points is the cantor set itself.

workspace
2.43) f(x) is x if x is irrational, p sin 1/q if x=p/q in lowest terms.

where is f continuous

Let $$x\in \mathbb{Q} $$. Since $$\mathbb{R}\backslash\mathbb{Q} \ $$ is dense in $$\mathbb{R} \ $$, we can find a sequence $$\left\{{a_n}\right\} \ $$ in $$\mathbb{R}\backslash\mathbb{Q} \ $$ such that $$a_n \to x \ $$. Since $$\forall a_n, \ f(a_n)=a_n, \ f(a_n)\to x \ $$. Since for any point $$x\in \mathbb{Q}, \ f(x)=p\sin(q^{-1}) \ $$, an irrational number, $$f(x)\neq x \ $$. Hence, $$f \ $$ fails to be continuous at any rational number.

Now suppose $$x\in\mathbb{Q}\backslash\mathbb{R} \ $$. Then $$f(x)=x \ $$.

Consider any Cauchy sequence $$\left\{{b_n}\right\} \ $$ which goes to $$x \ $$. Since it is Cauchy, it has precisely one limit point. If this sequence contains an infinite number of irrational numbers, then the series $$\left\{{f(b_n)}\right\} \ $$ contains a subsequence which goes to $$x \ $$. Therefore, the limit of the whole series goes to $$x \ $$.

Now suppose $$\left\{{b_n}\right\} \ $$ is any Cauchy sequence converging to $$x \ $$, and that this sequence has only a finite number of irrational terms. Let the furthest out irrational term be $$b_N \ $$, and define a new series $$\left\{{a_n}\right\} \ $$ as $$a_j = b_{N+j} \ $$. Obviously, $$a_n\to x \ $$ as well, and every term is rational, with $$a_n=p_n/q_n \ $$ being the lowest terms expression. Then

$$f(a_n)=p_n\sin(q_n^{-1}) \ $$.

Pick some integer $$M\in\mathbb{N} \ $$, and consider that since $$x \ $$ is irrational, there exists an $$\epsilon \ $$ such that no number of the form $$l/m, \ (l\in\mathbb{Z}, \ m\in\mathbb{N}, \ m\leq M) \ $$ is closer than $$\epsilon \ $$ to $$x \ $$. Given this $$\epsilon>0 \ $$, the fact that $$a_n\to x \ $$ implies $$\exists N : \forall n>N, \ |x-a_n|<\epsilon \ $$.

Hence $$\forall n>N, \ q_n>M \ $$.

Therefore, $$\lim_{n\to\infty} q_n^{-1}=0 \ $$, and so we have $$\lim_{n\to\infty} \sin(q_n^{-1})=0 \ $$. Then for any $$\epsilon>0, \ \exists N: \forall n<N, \ q_n^{-1}-\sin(q_n^{-1})<\epsilon \ $$. So as $$n\to\infty \ $$, we have $$f(a_n)=p_n\sin(q_n^{-1})\to p_n q_n^{-1} \ $$. Therefore, $$\lim_{n\to\infty} f(a_n)= \lim_{n\to\infty} a_n = x \ $$.

So, any Cauchy sequence \left\{{x_n}\right\} converging to $$x \ $$ also satisfies $$f(x_n)\to f(x) \ $$, and hence $$f \ $$ is continuous at any irrational point.

end workspace
2.48) Let x be in [01] with ternary expansion a_n. Let N=infinity if none of tha ai are 1, otherwise let N be the smallest value of n such that a_n =1.  let b_n =(1/2)a_n for n m*(A cup B)=m*B

3.13a) Show that for m*E < infinity,

E is measurable impluies given e>0 there is an open set O containing E with m*(O\E)0, there is a finite union U of open intervals st m*(U delta E) < e

3.14a) Show that the cantor ternary set has measure zero

3.14b) Let F be a subset of [01] constructed like the cantor set except that each of the intervals removed at the nth step has length a3^{-n} with 0<a<1. Show F is closed, F~ is dense in [01], and mF=1-a