User:J D Bowen/Math710 HW3

2.37) By Ex. 36, we have that the Cantor set $$C $$ is just the set [0,1] with the middle third removed, and the middle third of those intervals removed, etc. 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 $$C \sim S \sim S\backslash R \sim [0,1) \ $$.

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

Let $$C* \ $$ be the set of accumulation points of the Cantor set $$C \ $$. Then $$x\in C* \ $$ \implies $$\exists x_i \in C \ $$ such that $$x_i \to x \ $$.

Assume $$x\notin C \ $$. Then

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2.43) Let $$x\in \mathbb{Q}\cap[0,1] $$. 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 the set of fractions $$l/m, \ m,l\leq M \ $$ is finite, and that since $$x \ $$ is irrational and hence equal to none of them, there exists an $$\epsilon = \text{min} |x-l/m|>0 \ $$ such that no number of the form $$l/m, \ (l\in\mathbb{Z}, \ m\in\mathbb{N}, \ m\leq M) \ $$ is in $$(x-\epsilon,x+\epsilon) \ $$. 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.

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<N and b_N = 1.   Show

$$\sum_{n=1}^N \frac{b_n}{2^n} \ $$

is independent of the ternary expansion, if there are two, and that the function f(x) defined as the above expression is continuous, monotone on [01]. Show f is constant on each interval contained in the complement of the cantor set, and that f maps the cantor set onto the interval [01].

3.7) Let $$X \subset \mathbb{R} \ $$ be any set, and $$\Omega=\left\{{O_i}\right\}_{i\in I} \ $$ be any open cover of $$X \ $$. (Here $$I \ $$ is any index set, countable or uncountable, and $$O_i \ $$ is an open interval.)

Now consider the set $$X+r = \left\{{x+r:x\in X }\right\} \ $$. Let $$x \ $$ be any element of $$X \ $$. Then since $$\exists i\in I : x\in O_i \ $$, we can be sure that $$x+r\in O_i +r \ $$. Hence, the set $$\Omega+r = \left\{{O_i+r}\right\}_{i\in I} \ $$ forms an open cover for $$X+r \ $$.

Now suppose there is some collection $$\Omega+r = \left\{{O_i+r}\right\}_{i\in I} \ $$ of open sets which covers $$X+r \ $$. Then $$\exists i : x+r\in O_i+r \ $$, and so $$x+r-r \in O_i +r -r \ $$.

Hence we have $$X\subset \cup O_i \iff X+r\subset \cup(O_i+r) \ $$.

Observe that since each $$O_i=(a_i,b_i) \ $$, we have the length $$l(O_i) = b_i-a_i = b_i+r-a_i-r = (b_i+r)-(a_i+r) =l(O_i+r) \ $$.

Then, since $$m^*(X) = \text{inf} \sum l(O_i) \ $$ and since $$l(O_i+r)=l(O_i) \ $$, we have $$m^*(X+r)=\text{inf} \sum l(O_i+r) = \text{inf} \sum l(O_i) = m^*(X) \ $$.

3.8) Suppose $$m^*(A)=0, m^*(B)=x \ $$. \

Let $$\epsilon>0 \ $$ be any positive number. Then $$m^*(A)=0 \ $$ tells us we can find an open cover for $$A \ $$ such that the sum of the lengths of every interval in the cover is less than epsilon over two.

Let $$O^A=\left\{{O_i^A}\right\}_{i\in I} \ $$ be such an open cover for A; that is,

$$A\subset\bigcup_{i\in I} O^A_i, \ \sum_{i\in I} l(O^A_i)<\epsilon/2 \ $$.

Similarly, $$m^*(B)=x \ $$ tells us we can find an open cover for $$B \ $$ such that the sum of the lengths of every interval in the cover is less than $$x+\epsilon/2$$.

Let $$O^B=\left\{{O_j^B}\right\}_{j\in J} \ $$ be such an open cover for B; that is,

$$A\subset\bigcup_{j\in J} O^B_j, \ \sum_{j\in J} l(O^B_j)<\epsilon/2 \ $$.

Now we have $$y\in A\cup B \implies (y\in A \or y\in B) \implies (y\in\cup O^A_i \or y\in\cup O^B_j \implies A\cup B\subset ((\cup O^A_i)\cup(\cup O^B_j)) \ $$, so $$O^A\cup O^B \ $$ is an open cover for $$A\cup B \ $$.

But $$m^*(A\cup B) \leq \sum_{i\in I}O^A_i + \sum_{j\in J}O^B_j = x+0=x =m^*(B) \ $$.

Obviously, any open cover for $$A\cup B$$ must contain $$B \ $$, and so $$m^*(A\cup B)\geq m^*(B) \ $$.

Therefore, $$m^*(A\cup B)=m^*(B) \ $$.

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

E is measurable implies 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) Let $$C_n \ $$ be the $$n^{th}$$ step in the construction of the Cantor set $$C$$; that is, $$C_1 = [0,1/3]\cup[2/3,1], \ C_2=[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1] \ $$, etc.

Observe that $$C\subsetneq C_n \ $$ for any $$n \ $$, and we also have $$i0 \ $$. Then if each of the $$2^n \ $$ closed intervals $$[a_j,b_j]\in\left\{{ [a_1,b_1], \dots, [a_{2^n},b_{2^n}]}\right\} $$ is covered by an open interval $$(a_j-2^{-n-1}\epsilon, b_j+2^{-n-1}\epsilon) \ $$, we note that since $$b_j-a_j=\left({\tfrac{1}{3}}\right)^n \ $$, we can be sure each of these open intervals has length $$\left({\tfrac{1}{3}}\right)^n+2^{-n}\epsilon \ $$. Since $$\epsilon \ $$ was arbitrarily small, we have

$$m^*(C_n)\leq \left({\frac{2}{3}}\right)^n \ $$.

Now let $$\delta>0 \ $$. If we set $$N=\lceil \log_{2/3}(\delta) \rceil \ $$, then for $$n>N \ $$, we have

$$m^*(C_n)\leq \left({\frac{2}{3}}\right)^n \leq \left({\frac{2}{3}}\right)^{\lceil \log_{2/3}(\delta) \rceil} \leq \left({\frac{2}{3}}\right)^{ \log_{2/3}(\delta) } = \delta \ $$.

Since $$C\subset C_n \ $$, this implies $$m^*(C)\leq m^*(C_n)\leq \delta \ $$. Since we can find arbitrarily small delta greater than the measure of the Cantor set, the measure of the Cantor set must be 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

Measure: Define $$F(n) \ $$ as the nth step in the construction of that fat Cantor set. Then $$F=\bigcup_{j=1}^\infty F(n) \implies mF = \text{inf} (m(F(n))) \ $$.

Note that to form $$F(n) \ $$, we take $$F(n-1) \ $$ and remove from it $$2^{n-1} \ $$ intervals of length $$\alpha 3^{-n} \ $$. So $$mF(n) \ $$ is a strictly decreasing function of $$n \ $$, and hence $$\text{inf} (m(F(n))) = \lim_{n\to\infty}m(F(n)) \ $$.

Since $$m(F(0)) = m[0,1]=1 \ $$, we have

$$m(F(n)) = 1- \alpha/3^1 - 2(\alpha/3^2) - 2^2(\alpha/3^3)-\dots \ $$

$$=1- \sum_{j=1}^n \frac{2^{j-1}\alpha}{3^j} \ $$.

But $$\lim_{n\to\infty} \sum_{j=1}^n \frac{2^{j-1}\alpha}{3^j} =\sum_{j=1}^\infty \frac{2^{j-1}\alpha}{3^j} = \alpha \frac{1}{3}\sum_{j=0}^\infty \left({ \frac{2}{3}}\right) ^j $$

$$=\alpha \frac{1}{3} \frac{1}{1-\tfrac{2}{3}} = \alpha \frac{3}{3} = \alpha \ $$,

so

$$\lim_{n\to\infty} m(F(n)) = 1-\alpha \ $$

Dense: $$F_c=[0,1]\backslash F \ $$ is dense if $$F \ $$ contains no intervals. Again, define $$F(n) \ $$ as the nth step in the construction of that fat Cantor set. As we showed above, the total length of $$F(n) \ $$ is

$$1-\sum_{j=1}^n \frac{2^{j-1}\alpha}{3^j} \ $$

Since $$F(n) \ $$ is composed of $$2^n \ $$ intervals of equal length, the maximum length of any interval contained in $$F(n) \ $$ is

$$\text{maxLength}=\frac{\sum_{j=1}^n \frac{2^{j-1}\alpha}{3^j}}{2^n} \leq \frac{1}{2^n} \ $$

Suppose we had any interval $$(a,b) \ $$. Then $$b-a>0 \ $$, and so $$\exists N : b-a>2^{-N} \ $$. But then there can be no interval of length $$b-a \ $$ in $$F(N) \ $$, and so $$(a,b)\not \subset \bigcap_{n=1}^\infty F(n) = F \ $$. So $$F \ $$ contains no intervals, which implies $$F_c \ $$ is dense in $$[0,1] \ $$.