TITLE: Prob. 6 (d), Chap. 1, in Baby Rudin, 3rd ed: How to complete this proof?
QUESTION [16 upvotes]: Here is Prob. 6, Chap. 1, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Fix $b>1$.

Problem 6(a):

Let $m$, $n$, $p$, $q$ be integers such that $n>0$, $q>0$, and $r = m/n = p/q$. Then [I've managed to] prove that
$$ b^{m/n} = b^{p/q}.$$ So we can reasonably define $b^r$ as $$b^r \colon = \sqrt[n]{b^m}.$$

Problem 6(b)

[From this definition we can] prove that $$ b^{r+s} = b^r \cdot b^s, $$ where $r$ and $s$ are any rational numbers.

Problem 6(c):

Now for a rational number $r$, let the set $B(r)$ be defined as follows:
$$B(r) \colon= \{ \, b^t \colon \, t\in \mathbb{Q}, \, t \leq r \, \}. $$
Then it is clear that $$b^r = \sup B(r).$$ So for every real $x$, we can define $b^x$ as follows: $$b^x \colon= \sup B(x) = \sup \big\{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x \, \big\}. $$

Problem 6(d):

Using this definition, (we are required to) prove that, for every pair of real $x$ and $y$, the equation $$b^{x+y} = b^x \cdot b^y$$ holds.

My Attempt:

If $r$, $s \in \mathbb{Q}$ such that $r \leq x$ and $s \leq y$, then $r+s \in \mathbb{Q}$ also and $r+s \leq x+y$ so that we can write
$$ B(x) \cdot B(y)  = \{ \, b^r \cdot b^s \, \colon \, r \in \mathbb{Q}, s\in \mathbb{Q}, \, r \leq x, s\leq y \, \} \subseteq \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = B(x+y),$$ and hence
$$b^x \cdot b^y = \sup B(x) \cdot \sup B(y) = \sup \{ \, b^r \, \colon \, r \in \mathbb{Q}, \, r \leq x \, \} \cdot \sup \{ \, b^s \, \colon \, s \in \mathbb{Q}, \, s \leq y \, \} = sup \{ \, b^{r+s} \, \colon \, r \in \mathbb{Q}, \, s\in \mathbb{Q}, \,  r \leq x, \, s \leq y \, \} \leq \sup \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = \sup B(x+y) = b^{x+y}.$$
Here we have used the following definition: Given two non-empty sets $U$ and $V$, say of real numbers, we define the set $U \cdot V$ as follows:
$$U \cdot V \colon= \{ \, u \cdot v \, \colon \, u \in U, \, v \in V \, \}.$$

And we have also used the fact that if $W$ and $Z$ are two non-empty bounded above subsets of the set of positive real numbers such that $W \subseteq Z$, then we must have
$$ \sup W \leq \sup Z.$$
So far, we have shown that $$b^x \cdot b^y \leq b^{x+y}.$$
Now how to prove the reverse inequality using the machinery developed above?
That is, how to prove that $$b^{x+y} \leq b^x \cdot b^y?$$

REPLY [2 votes]: Right now, there are five answers: one that isn't actually an answer, two that use concepts like limits and continuity the Rudin book hasn't tackled by this point, and two that are correct but rely on redefining the set $B(x)$ from Rudin's definition $B(x) = \{b^t: t \in \mathbb{Q}, t \leq x\}$ to $\{b^t: t \in \mathbb{Q}, t < x\}$.
This redefinition is okay because $sup\ \{b^t: t \in \mathbb{Q}, t \leq x\} = sup\ \{b^t: t \in \mathbb{Q}, t < x\}$.  I thought it might be useful to show why this is true using the tools we have so far.
Let's define $B(x)$ as Rudin does, with $t\leq x$.  Let's define $C(x) = \{b^t: t \in \mathbb{Q}, t < x\}$.  I will show $sup\ B(x) = sup\ C(x)$.
Note that $C(x) \subset B(x)$. Thus, since $sup\ B(x) > y$ for all $y \in B(x)$, $sup\ B(x) > y$ for all $y \in C(x)$, and $sup\ B(x)$ is an upper bound of $C(x)$.
Pick $z < sup\ B(x)$.  We know $sup\ B(x) >0$ since $b^t>0$ for all $t\in \mathbb{Q}$.  Thus we can pick some positive real $w$ with $z<w<sup\ B(x)$.  Then $w$ is not an upper bound of $B(x)$ so there exists $p\in\mathbb{Q}$ with $p\leq x$ and $b^p>w$.
By the work in Problems 7a-c, we know there exists a positive integer $b$ such that $b^n < \frac{b^p}{w}$.  (Note that this is not a circular reference, since the work in 7a-c does not reference anything in Problem 6.)  Then $\frac{1}{b^n} > \frac{w}{b^p}$.  Then $b^p\frac{1}{b^n} > b^p\frac{w}{b^p}=w$.  Since $w>z$, we have $b^p\frac{1}{b^n} > z$.
Note $b^p\frac{1}{b^n} = b^{p-1/n}$, $p-\frac{1}{n}\in\mathbb{Q}$, and $p-\frac{1}{n}<p\leq x$.  Then $b^{p-1/n} \in C(x)$, and $z$  not an upper bound of $C(x)$.
Thus $sup\ B(x) = sup\ C(x)$, and it is equivalent to show that $sup\ C(x+y) = sup\ C(x)sup\ C(y)$ as two others have done above.