User:J D Bowen/Math710 HW1

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

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 $$a+b <q+r \ $$, and so $$q+r \ $$ is an upper bound for $$A+B \ $$. Since $$\text{sup}(A)+\text{sup}(B) \ $$ is a sum of an upper bound for $$A \ $$ and an upper bound for $$B \ $$, it follows that it is an upper bound for $$A+B \ $$.

All that remains to be shown is that is it the least upper bound. Suppose to the contrary that $$\text{sup}(A+B) < \text{sup}(A)+\text{sup}(B) \ $$. By definition, this means we can find a positive number $$\delta \ $$ such that $$\text{sup}(A)+\text{sup}(B)-\text{sup}(A+B) > \delta \ $$.

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 $$0\leq b_na_m \or b_ma_m \ $$, 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 $$p-1 \ $$.