User:J D Bowen/Math710 HW1

1. Let $A, B \ $ be bounded, non-empty subsets of $\mathbb{R} \ $. Define $A+B = \left\{{ a+b : a\in A \land b \in B }\right\} \ $.

Suppose $q, r \ $ are upper bounds for $A, B \ $ respectively. Then for all elements $a\in A, b \in B \ $, we have $a<q, b<r \ $. We can add these two inequalities to get $a+b  \delta \ $.

Let $x\in A \ $ be such that $\text{sup}(A)-x<\delta/4 \ $; we know such a number exists because the alternative is that $\forall x \in A, \text{sup}(A)-x\geq \delta/4 >0 \ $, and so $\text{sup}(A)-\delta/5 \ $ would be a lower bound for the set which is less than $\text{sup}(A) \ $, which contradicts the definition of supremum.

Similarly, we can find $y\in B \ $ such that $\text{sup}(B)-y<\delta/4 \ $.

Then we have $\delta<\text{sup}(A)+\text{sup}(B)-\text{sup}(A+B)<x+y+\delta/2 -\text{sup}(A+B) \ $.

This leads to $\text{sup}(A+B)+\delta/2 \text{sup}(f)+\text{sup}(g) \ $. Then there is some value $x\in S \ $ such that $f(x)+g(x)>\text{sup}(f)+\text{sup}(g) \ $. From the definition of sup, we can be sure that for this value x, we have $f(x)\leq \text{sup}(f), g(x)\leq\text{sup}(g) \ $. Adding these two equations gives $f(x)+g(x)\leq \text{sup}(f)+\text{sup}(g) \ $. This contradicts our equation $f(x)+g(x)>\text{sup}(f)+\text{sup}(g) \ $, which we arrived at by assuming $\text{sup}(f+g)>\text{sup}(f)+\text{sup}(g) \ $; therefore, this assumption is false. It must be true that $\text{sup}(f+g)\leq\text{sup}(f)+\text{sup}(g) \ $.

(b)Let $f(x):[0,1]\to\mathbb{R} \ $ be defined

$ \begin{cases} 1,          & x \in [0,\tfrac{1}{4}] \\ \tfrac{2}{3}, & x \in (\tfrac{1}{4},\tfrac{3}{4}) \\ 0,           & x \in [\tfrac{3}{4},1], \end{cases} \ $

and $g(x)=f(-x) \ $. Then $\text{sup}(f+g)=4/3, \text{sup}(f)+\text{sup}(g)=2 \ $.

(c)Note that nowhere in our argument a did we use the assumption that f,g be non-negative. This condition was unnecessary.

3) Let $\left\{{x_n}\right\}, \left\{{y_n}\right\} \ $ denote sequences of real numbers, with finite limit superiors $L_x, L_y, L_{x+y} \ $, where the final value refers to the limit superior of $\left\{{x_n+y_n}\right\} \ $. This means that for any $\epsilon >0, \exists N \ $ such that for $n>N \ $, we have $|L_x-\text{sup}_{n>N} x_n| <\epsilon>|L_y-\text{sup}_{n>N} y_n| \ $, and $|L_{x+y}-\text{sup}_{n>N} (x_n+y_n)| <\epsilon \ $.

Suppose that $\lim \text{sup} (x_n+y_n)>\lim \text{sup}(x_n)+\lim \text{sup}(y_n) \ $. If we set $\epsilon < \tfrac{1}{3}(S_{x+y}-S_x-S_y) \ $, then we can find an $N \ $ such that

$\text{sup}_{n>N} (x_n+y_n)> \text{sup}_{n>N}(x_n)+ \text{sup}_{n>N}(y_n) \ $.

However, if we take our proof from 2a and set $f(n)=x_n, g(n)=y_n, S=\left\{{n\in\mathbb{N}:n>N}\right\} \ $, we have a demonstration that this is impossible. Therefore our assumption that $\lim \text{sup} (x_n+y_n)>\lim \text{sup}(x_n)+\lim \text{sup}(y_n) \ $ must be false. Hence, $\lim \text{sup} (x_n+y_n)\leq\lim \text{sup}(x_n)+\lim \text{sup}(y_n) \ $.

4) Let $p\in\mathbb{N}, \left\{{a_n}\right\} \ $ be a sequence such that $ 0\leq a_n< p, s_n=\sum_{i=1}^n a_i p^{-i} \ $.

If we let $\epsilon>0 \ $ be any positive real number, then define $q = -\log_p ( (1-p^{-1})\epsilon) \ $.

Observe that if $m,n\in\mathbb{N}, m>n>q \ $, we have

$s_m-s_n = \sum_{i=n}^m a_i p^{-i} < \sum_{i=n}^m p^{1-i} = p \sum_{i=n}^m  p^{-i} = p \left({ \sum_{i=0}^m p^{-i} - \sum_{i=0}^n p^{-i} }\right) \ $

$= p \left({ \frac{p^{-1-m}-p^{-1-n}}{p^{-1}-1} }\right) =\frac{p^{-n}-p^{-m}}{1-p^{-1}}<\frac{p^{-q}}{1-p^{-1}} \ $

$ =\frac{p^{\log_p( (1-p^{-1})\epsilon)}}{1-p^{-1}}= \frac{(1-p^{-1})\epsilon}{1-p^{-1}} = \epsilon \ $.

Hence, the sequence $s_n \ $ is Cauchy and therefore convergent.

5) a) Define $a_j = \lfloor \left({ x-\sum_{i=1}^{j-1} \frac{a_i}{p^i} }\right) p^j \rfloor \ $, where we accept the abuse of notation $\sum_{i=1}^0 a_ip^{-i} =0 \ $. This recursive definition allows for all $a_n \ $ to be computed.


 * Lemma: This will always be less than $p \ $.


 * Proof: Suppose to the contrary $\exists n \ $ such that $a_n\geq p \ $. Then


 * $a_n = \lfloor \left({ x-\sum_{i=1}^{n-1} \frac{a_i}{p^i} }\right) p^n \rfloor \geq p \ $


 * But then we can pull out the final term of the sum and divide by p to get


 * $\left({ x-\sum_{i=1}^{n-2} \frac{a_i}{p^i} }\right) p^{n-1} \geq 1+a_{n-1} \ $


 * This left-hand side is of course just


 * $a_{n-1}+\text{something in} \ [0,1) \geq 1+a_{n-1} \ $


 * which is impossible.

Define $s_n = \sum_{i=1}^n a_ip^{-i} \ $. Since both $a_i,p^{-i}>0 \ \forall i \ $, this series is increasing, and bounded above by $x \ $ by construction: at every point in the series, we add precisely as many $p^{-n-1} \ $ as will fit in $x-s_n \ $ without going over $x \ $:
 * Lemma: $s_n\leq x \forall n \ $
 * Proof: We have $s_1=a_1p^{-1}=\lfloor xp \rfloor p^{-1}\leq xpp^{-1}=x \ $. Suppose we have $s_j<x \ $ for some j.  By definition,  $s_{j+1}-s_j=a_{j+1}p^{-1-j} \ $.  But $a_{j+1}p^{-1-j} = \lfloor (x-s_j)p^{1+j}  \rfloor p^{-1-j} \leq (x-s_j) $.  So $s_{j+1}-s_j\leq x-s_j \implies s_{j+1}\leq x \ $.  Now suppose we have instead $s_j=x \ $.  Again we have $s_{j+1}-s_j=a_{j+1}p^{-1-j} \ $, but now $a_{j+1}p^{-1-j} = \lfloor (x-s_j)p^{1+j}  \rfloor p^{-1-j} = 0 \implies s_{j+1}=s_j=x \ $.  This completes the induction proof.

It remains to be shown this series converges to x. Observe that in the sum $s_{k-1}+a_kp^{-k}=s_k \ $, we have defined $a_k =\lfloor \left({ x-\sum_{i=1}^{k-1} \frac{a_i}{p^i} }\right) p^k \rfloor\ $ to count precisely how many $p^{-k} \ $ will fit in $x-s_{k-1} \ $. We could never have $x-s_k \geq p^{-k} \ $ because that would mean $a_k \ $ had undercounted by 1.

Therefore, $x-s_k < p^{-k}\ $.

Let $\epsilon >0 \ $. Then setting $z=-\log_p \epsilon \ $. Then if $N>z, \ x-s_N<p^{-N}<p^{\log_p} \epsilon = \epsilon \ $. Since $\left\{{s_k }\right\} \ $ is increasing, bounded above by $x \ $, and comes arbitrarily close to x, we have $\left\{{s_n}\right\} \to x \ $.

b) Let $\left\{{a_n}\right\} \ $ be the series defined in a, and let $b_n \ $ be some series of integers $0 \leq b_n < p$ such that $\left\{{t_n}\right\} \to x$, where $t_n = \sum_{i=1}^n b_ip^{-i}$.

We wish to show that $a_n=b_n \forall n \ $, unless $x=qp^{-k} \ $ for some $k\in\mathbb{N} \ $. Assume to the contrary that there are terms which do not agree and let $b_m \ $ be the first term of the b sequence which does not agree with the a sequence.

Then $b_m > a_m \lor b_m < a_m$. Let us consider the first case. Since we know that $a_m $ counts precisely how many $p^{-m} \ $ can be added to $s_{m-1} \ $ without exceeding $x \ $, we can be sure that if $b_m>a_m \ $, then $s_{m-1}+b_mp^{-m}=t_{m-1}+b_mp^{-m}=t_m >x \ $, and since $\left\{{t_n}\right\} \ $ is always increasing, it can never converge to $x \ $.

Now consider the second case, $b_m<a_m \ $. First, we will need a lemma:


 * Lemma: $( \exists N : \forall n\geq N, \ a_n=0) \iff (x=qp^{1-N} ) \ $
 * Proof:
 * ($\Rightarrow$)
 * Suppose $\exists N : \forall n\geq N, \ a_n=0 \ $. Then $x=\sum_{n=1}^\infty a_np^{-n} = \sum_{n=1}^{N-1} a_np^{-n} \ $.  But $a_np^{-n} = (a_np^{N-1-n} p^{1-N} \ $. Since $x \ $ is a sum of these terms of $p^{N-1} \ $, we must have $x=qp^{N-1} \ $ for some $q\in\mathbb{N} \ $.
 * ($\Leftarrow$)
 * Suppose $x=qp^{1-N}$. Observe that since $p^{1-N}|s_{N-1} \ $, we must have $s_{N-1}=x \ $.  Since $\left\{{s_n}\right\} \to x \ $ and is strictly increasing, we must have all successive terms equal to zero.

Now suppose that $x=qp^{-k} \ $ for some k. We wish to show that there are only two series which converge to x; the series $\left\{{a_n }\right\} \ $ as defined above, and another series we describe now.

Consider the series $\left\{{a_n }\right\} \ $ when $x=qp^{-k} \ $. Now, if we define

$b_n= \begin{cases} a_n,          & nk, \end{cases} \ $

Then we see that $\sum_{j=1}^\infty b_jp^{-j} = \left({ \sum_{j=1}^{k-1} b_jp^{-j} }\right) + (a_k-1)p^{-k} + \sum_{j=k+1}^\infty (p-1)p^{-j} \ $

$= \left({ \sum_{j=1}^{k-1} a_jp^{-j} }\right) + a_kp^{-k}-p^{-k} + \sum_{j=k+1}^\infty (p-1)p^{-j} = \left({ \sum_{j=1}^{k} a_jp^{-j} }\right) - p^{-k} + \sum_{j=k+1}^\infty (p^{1-j}-p^{-j}) \ $

$=x-p^{-k} + \sum_{j=k+1}^\infty (p^{1-j}-p^{-j})=x-p^{-k} +p^{-k}\sum_{j=0} \left({ p^{-j}}\right) - p^{-k-1}\sum_{j=0} \left({ p^{-j} }\right) \ $

$=x-p^{-k} +p^{-k} \left({ \frac{1}{1-p^{-1}} }\right) - p^{-k-1}\left({ \frac{1}{1-p^{-1}} }\right) = x-p^{-k}+\frac{p^{-k}-p^{-k-1}}{1-p^{-1}} = x-p^{-k}+p^{-k}\frac{1-p^{-1}}{1-p^{-1}}=x $

So, this series converges to $x \ $ as well.

Let us suppose, finally, that $x\neq qp^{-k} \ $ for any k. We have already shown that if the first differing term of another series $b_n \ $ is greater than the corresponding term $a_n \ $, the sum series cannot converge to x. Now we examine the case $b_m < a_m \ $ at the first differing term. As we saw above, if the first term to differ is only one less, ie, $b_m = a_m-1 \ $, then it is necessary for every other term afterwards to be increased from $0 \ $ to $p-1 \ $ in order to make up for this deficit. The remaining terms of course, cannot be increased more than this, or they would violate the condition that all terms be less than $p \ $. Since in the case $x\neq qp^{-k} \ $, there are no infinite strings of zeroes, we cannot decrease any one term and increase the succeeding terms by $p-1 \ $.