Question:
I proved parts (a), (b) and (c) without any problem, however, I am struggling with d). After thinking for a while I concluded that $B(x)B(y) = \{ b^ab^s : a, s \in \mathbb{Q} \land a\le x \land s\le y\} = \{ b^{a+s} : a, s \in \mathbb{Q} \land a+s\le x+y\} = B(x+y)$ but I was unable to go any further.
Here is Rudin´s solution:
I understand the solution, however, I don´t see the motivation behind $m = \frac{1}{2}(\frac{c}{\sup B(x)}+\sup B(y))$, I can see that $\frac{c}{\sup B(x)} < \frac{1}{2}(\frac{c}{\sup B(x)}+\sup B(y)) < \sup B(y)$ and by using it you can get to the solution but I don´t understand how or why he came up with it. Furthermore, I found some notes that say that I could solve (d) using the inequality:
$\text{If } t,b > 1 \text{ and } n > \frac{b - 1}{t - 1} \text{, then } b^{\frac{1}{n}} < t.$
So I tried the following:
Fix $x,y \in \mathbb{R}$ and let $a, s \in \mathbb{Q}$, $a \le x$ and $s \le y$. Now, $b^{a+s} = b^ab^s \le b^xb^y$ so $\sup B(x) \sup B(y)$ is an upper bound of $\{ b^t : t \in \mathbb{Q} \land t \le x+y \} = B(x+y)$, next, consider any number $c$ such that $0<c<\sup B(x) \sup B(y)$ then for some $n$, $b^{\frac{1}{n}} < \frac{b^xb^y}{c} \implies c < \frac{b^xb^y}{b^{\frac{1}{n}}}$
but this takes me nowhere since what I am trying to prove is that if $0<c<\sup B(x) \sup B(y)$ then there exists some n such that $c < b^{x+y-\frac{1}{n}} < \sup B(x+y)$ to conclude that $\sup B(x) \sup B(y)$ is the least upper bound of $B(x+y)$.
I would appreciate any help, thanks in advance.
Edit:
I have been working on this question and I have finally found a way to prove it without using $m = \frac{1}{2}(\frac{c}{\sup B(x)}+\sup B(y))$ since I really didn´t understand how he came up with it. (Now I realize that it is just the middle point between $\frac{c}{\sup B(x)}$ and $\sup B(y)$).
Proof:
We know that $\sup B(x) \sup B(y)$ is an upper bound of $B(x+y)$ so know we want to show that any element $c < \sup B(x) \sup B(y)$ won´t be an upper bound of $B(x+y)$. $c < \sup B(x) \sup B(y) \implies \frac{c}{\sup B(y)} < \sup B(x)$ so for some $u \le x$, $b^u \in B(x)$ and $\frac{c}{\sup B(y)} < b^u \implies \frac{c}{b^u} < \sup B(y)$ so for some $v \le y$, $b^v \in B(y)$ and $\frac{c}{b^u} < b^v \implies c < b^vb^u = b^{v+u} \in B(x+y)$ so we conclude $\sup B(x+y) = \sup B(x) \sup B(y)$.
Still, I would appreciate any help finding a proof that uses the inequality $\text{If } t,b > 1 \text{ and } n > \frac{b - 1}{t - 1} \text{, then } b^{\frac{1}{n}} < t.$