Prob. 6 (d), Chap. 1, in Baby Rudin, 3rd ed: How to complete this proof?

By definition $b^{x+y} = \sup B(x+y)$, where $B(x+y)$ is the set of all numbers $b^t$ with $t$ rational and $t<x+y$. Then any rational number $t < x+y$ can be written as $r+s$ where $r,s$ rational and $r<x,s<y$ (Think about why this is true). Hence we can write $B(x+y)$ as the set of all numbers $b^rb^s$ with $r<x,s<y$, this means that $B(x+y)$ is the set of all products $uv$ where $u\in B(x)$ and $v\in B(y)$.

Since any such product is less than $\sup B(x) \sup B(y)$, we can write $M= \sup B(x) \sup B(y)$ is an upper bound for $B(x+y)$. On the other bound, suppose $0<c<\sup B(x) \sup B(y)$, then $c/(\sup B(x)) < \sup B(y)$. Let $m=\frac{1}{2}\left( \frac{c}{\sup B(x)}+\sup B(y)\right)$. Then $c/(\sup B(x)) < m < \sup B(y)$, and $\exists u \in B(x), v \in B(y)$ s.t. $c/m < u, m<v$ Hence we have $c=(c/m) \times m < uv \in B(x+y)$, and so $c$ is not an upper bound of $B(x+y)$. It follows that $\sup B(x) \sup B(y)$ is the least upper bound of $B(x+y)$. Then we have this equality $b^{x+y}=b^xb^y$ as required.

I am stuck in the same question. Maybe we can think together. :)

I've divided into two cases: $x+y\notin \Bbb{Q}$ and $x+y\in\Bbb{Q}$.

Case 1:$x+y\notin \Bbb{Q}$

Well, as you've done, we have that $B(x)\cdot B(y)\subset B(x+y)$. Ok.
So, let $b^t\in B(x+y)$. Then $t\in \Bbb{Q}$ and $t<x+y$, since $x+y\notin \Bbb{Q}$. So, we have $t-y<x$ and, therefore, $\exists r\in(t-y,x)\cap\Bbb{Q}$. Set $s=t-r\in\Bbb{Q}$. Thus, we have $$t=r+s<x+y, \text{ with }r<x \text{ and } s<y.$$ And so, $b^t=b^{r+s}=b^rb^s$, by item (b) of this exercise 6. But $b^r\in B(x)$, and $b^s\in B(y)$. So $b^t=b^rb^s\in B(x)\cdot B(y)$.

We've proved then that $B(x+y)\subset B(x)\cdot B(y)$ and, hence, $B(x+y)=B(x) \cdot B(y)$.
Then $$\sup B(x+y)=\sup B(x) \cdot B(y)$$ $$=\sup B(x) \cdot \sup B(y),$$ since $B(x),B(y) \subset [0, \infty )$.

But, what about the case $x+y\in \Bbb{Q}$? I haven't done great advances on it.. :( Can you guess anything?

Since $b^{x+y}=\sup\{b^t:t\in \mathbb Q,\ t\le x+y\}$, we know that $\forall \epsilon>0$, $\exists r\in\mathbb Q$, $r\le x+y$, such that $b^r>b^{x+y}-\epsilon$.

Claim: $\exists r'\in\mathbb Q$, $r'\lt x+y$, such that $b^{r'}>b^{x+y}-2\epsilon$.

If $x+y \notin \mathbb Q$, $r'=r$ satisfies the condition. Let's suppose $r=x+y\in\mathbb Q$, and consider $r'=r-\frac 1 n<r$. Since $b^r-b^{r'}=b^r(1-b^{-\frac 1 n})$, we only need to show that $\lim_{n\to\infty}b^\frac 1 n = 1$, and then choosing $n$ large enough, $r'$ is what we need to find. This is an easy exercise in calculus.

Now we can prove part(d) of the problem. Write $r'$ as $p+q$, where $p\le x$ and $q\le y$ are both rational. This is possible because $r'<x+y$ and rational numbers are dense. Hence, $b^x\cdot b^y\ge b^p\cdot b^q = b^{r'}>b^{x+y}-2\epsilon$. Since $\epsilon$ is arbitrary, $b^x\cdot b^y \ge b^{x+y}$.