User:J D Bowen/Math710 HW3

2.37) Let $x \ $ be any point in the set $[0,1]\backslash C \ $, where C is the Cantor set. Then in a ternary expansion $a_1, a_2, \dots \ $, there must exist at least one $a_i=1 \ $. Let $j(x) \ $ be the lowest such $i \ $.

$j=1 \implies x \in [1/3, 2/3)\backslash\left\{{1/3}\right\} = (1/3,2/3) \ $,

$j=2 \implies x \in [1/9,2/9)\cup[7/9,8/9) \backslash\left\{{1/9,7/9}\right\} \subset (1/9,2/9)\cup(3/9,4/9)\cup(5/9,6/)\cup(7/9,8/9) \ $,

and so on. Observe we must put $1/3, 1/9, 7/9, \ $ etc. back into the Cantor set, since they have alternative ternary expansions which do not have a 1 in the first (or second, or any) place.

In general, we see $j(x)=n \ $ implies $x\in\bigcup_{2 \ \text{does not divide} \ k} (k3^{1-n}+3^{-n},k3^{1-n}+2\cdot 3^{-n}) \ $.

Now form the sets $D_i = [0,1]\backslash \left\{{x:j(x)=i}\right\} \ $. Then, $D_i \ $ is the set of numbers in $[0,1] \ $ which have a 1 in the $i^{th}$ spot and not before it. We can form the sets

$E_i = \bigcup_{n=1}^i D_i \ $,

which will contain all numbers which have a 1 at or before the $i^{th}$ spot in their ternary expasions (both expasions, if they exist). Finally, note that this means that if we form

$C_i = [0,1]\backslash E_i \ $, we have

$i\leq j \implies C_j \subseteq C_i, \ $,

and

$\forall i, \ C\subset C_i$.

Since this process removes all points which are not in the Cantor set, we have

$C=\bigcap_{i=1}^\infty C_i \ $

2.38) By Ex. 37, 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 nth position indicating whether it is in the left or right interval from the nth 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) \ $.

2.43) 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 $[0,1] \ $ with ternary expansion $a_n(x) \ $. Let $N=\infty \ $ if none of the $a_i \ $ are 1, otherwise let N be the smallest value of n such that $a_n =1 \ $. Let $b_n(x) =(1/2)a_n(x) \ $ for $n<N, \ b_N = 1 \ $, and $j>N \implies b_j=0 \ $. 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 $[0,1] \ $. 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 $[0,1] \ $.

Independent of ternary expansion:

From Ex. 5 on homework 1, a number only has two ternary expansions if it has a terminating expansion

$x=\frac{y_1}{3^1}+\frac{y_2}{3^2}+\dots+\frac{y_n}{3^n} $, where $y_n\neq 0 $,

and the other expansion is

$x=\frac{y_1}{3^1}+\dots+\frac{y_n-1}{3^n}+\frac{2}{3^{n+1}}+\frac{2}{3^{n+2}}+\dots $

If we suppose that $\exists j < n : y_j=1 $, then the $b_j$ are the same for both expansions, since we stop before we reach the point where the expansions differ.

So, assume $\forall j<n: y_j\neq 1 $.

If we compute the b for the first (terminating) expansion, we get

$b_i = y_i/2 \ $ for $i<n$,

$b_n=1 $, since $y_n\in\left\{{1,2}\right\} \implies b_n=1 $,

$b_k =0 $ for $k>n $, since $y_k =0 \ $ for such k.

Now, if we compute the b for the second (infinite) expansion, we get

$b_i = y_i/2 $ for $i<n$,

$b_n=1 $ if $y_n=0$, and then $b_k=0 $ for $k>n$,

$b_n=0 \ $ if $y_n=1$, and then $b_k=1 $ for $k>n$.

Obviously, in the first case of the infinite expansion, the b sum is the same as in the finite expansion. But in this second case of the infinite expansion, we have

$\frac{0}{2^n}+\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\dots = \frac{1}{2^n} + 0+0+\dots \ $,

and so the b sum is again the same as in the finite expansion.

Hence $f(x) $ is independent of the particular expansion we use for x.

Constant on intervals:

An interval $(a,b) \subset [0,1]\backslash C \ $ must be removed at some point, so it is a subset of one of the intervals which are removed in a step of the construction of the Cantor set. Thus, we only need show $f \ $ is constant on the open intervals removed in the construction.

All intervals removed are of the form $(k3^{-n}, (k+1)3^{-n}) \ $

Suppose $x,y $ are in $(k3^{-n}, (k+1)3^{-n}) \ $, and we say that without loss of generality, $k3^{-n}<x\leq y <(k+1)3^{-n} $. We have

$x=\frac{a_1(x)}{3^1}+\frac{a_2(x)}{3^2}+\dots+\frac{a_n(x)}{3^{-n}}+\dots \ $.

Let $i\leq n $. Then $a_i(x)\neq a_i(y)\implies a_i(y)>a_i(x)$, but then

$y\geq x+3^{-i} \geq x+3^{-n}>(k+1)3^{-n} $.

So, all points $x,y \ $ in such an interval have $a_i(x)=a_i(y) \ $ for $i\leq n \ $.

Additionally, for such an interval to be removed in the nth step, we must have $a_n(x) =a_n(y)=1 \ $. Hence $b_i(x)=b_i(y) \ $ for all relevant $i \ $, and so $f(x)=f(y) \ $.

Monotone:

Suppose we have $x,y\in[0,1], \ x<y \ $. Then the first $j \ $ such that $a_j(x)\neq a_j(y) \ $ must be such that $a_j(x)<a_j(y) \ $. Therefore,

$f(x)=\sum_{k=1}^{\text{max}(N(x),N(y)} \frac{b_k(x)}{2^k} \leq \sum_{k=1}^{\text{max}(N(x),N(y)} \frac{b_k(y)}{2^k} = f(y) \ $,

with equality only if $j>\text{max}(N(x),N(y) \ $.

Surjective:

Let $y\in[0,1] \ $. Then there is a binary expansion for $y \ $

$y=\sum_{n=1}^\infty e_n 2^{-n} \ $,

where $e_n \in \left\{{0,1}\right\} \ $.

Define $h_n=2e_n \ $. Then the point $x=\sum_{n=1} h_n 3^{-n} \ $ is in the Cantor set by definition, and satisfies $f(x)=y \ $. Hence $f(C)=[0,1] \ $.

Continuous:

Since $f \ $ is monotone, the only possible discontinuities are removable and jump discontinuities. But if there was a such discontinuity, the monotonicity of $f \ $ means the points not mapped to in the discontinuity could never be returned to; and since $f \ $ is surjective, this is impossible. Hence $f \ $ cannot have a discontinuity.

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 \lor y\in B) \implies (y\in\cup O^A_i \lor 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) Suppose $m^* E<\infty $. Suppose further that $E \ $ is measurable, and let $\epsilon>0 \ $. Since $E \ $ is measurable, and $m^*\neq \infty \ $, there is an open cover $O'=\cup_{i\in I} O_i \ $ such that

$\left({ \sum_{i\in I} l(O_i) }\right) - m(E) < \epsilon \ $.

Define the open set $O=\cup O_i \ $. Then we have $E\subset O = \cup O_i \implies m(E)\leq m(O) \leq \sum m(O_i) <m(E)+\epsilon \ $, and so $O \ $ is an open set containing $E \ $ satisfying $m(O\backslash E) <\epsilon \ $.

Let $m^*E<\infty \ $ and suppose there is an open set $O \ $ containing $E \ $ such that $M^*(O\backslash E)<\epsilon/2 \ $. This implies that $m^*O<\infty \ $.

Since any open set is a countable collection of open intervals, let $O=\cup_{n\in\mathbb{N}} O_n \ $. Since $m^*O<\infty \ $, we must have $\sum m^* O_n <\infty \ $.

This implies

$\exists N: \sum_{n=N}^\infty mO_n <\epsilon/2 \ $,

or in other words, if we define $U=\bigcup_{n=1}^N \ $, we will have

$ m^*(O\backslash U)<\epsilon/2 \ $.

But since $E,U\subset O \ $, we certainly have $m^*(U\backslash E)\leq m^*(O\backslash E) \ $ and $m^*(E\backslash U)\leq m^*(O\backslash U) \ $. Hence,

$\epsilon/2+\epsilon/2>m^*(O\backslash E)+m^*(O\backslash U) \geq m^*(U\backslash E)+m^*(E \backslash U) \ $.

Since $U\backslash E, E \backslash U \ $ are disjoint, this is just $m^*(U\Delta E)<\epsilon \ $.

Consider the other direction; suppose there is a set $E \ $ such that $m^* E < \infty \ $ and for all epsilon, there is a union of open intervals $U \ $ such that $m^*(U\Delta E)<\epsilon \ $.

We have $m^*(U\Delta E)=m^*(U\backslash E \cup E \backslash U) \ $, but since $U\backslash E, E \backslash U \ $ are disjoint, this is just

$m^*(U\backslash E )+m^*( E \backslash U) < \epsilon \ $

$\implies m^*(E\backslash U)<\epsilon \ $

$\implies \exists \left\{{O_i }\right\}:O_i \ \text{is open}, \ E\backslash U \subset \cup O_i, \ \Sigma mO_i<\epsilon \ $

$\implies O=U\cup \cup O_i \ \text{is open}, \ E\subset O, \ m^*(O\backslash E)=m^*O-m^*E<\epsilon+m^*E-m^*E =\epsilon \ $.

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 $i<j \implies C_j\subsetneq C_i \ $.

Note that $C_n \ $ is composed of $2^n \ $ closedintervals, each of length $\left({\tfrac{1}{3}}\right)^n \ $.

Let $\epsilon>0 \ $. 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)

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{1-\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] \ $.

Closed: Since at each step we remove a finite collection of open intervals, each "step" is a closed set. Therefore, the fat Cantor set, as a countable intersection of closed sets, is closed.