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

Question

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 $$b^{x+y} \leq b^x \cdot b^y?$$

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

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, $b^x\cdot b^y \ge b^{x+y}$.

Answer 5

If you have shown that $b^xb^y \leq b^{x+y}$ then it is enough to to show that $b^xb^y \lt b^{x+y}$ is false. To show this, define: $$b^x = sup\{b^t | t \in \Bbb Q \land t \lt x\}$$ which is different from Rudin's but is equivalent.

Suppose $b^xb^y \lt b^{x+y}$. Then there must exist $r \in \Bbb Q$ with $r \lt x+y$ such that $b^{t_1}b^{t_2} \lt b^r$ for all $t_1, t_2 \in \Bbb Q$ with $t_1 \lt x$ and $t_2 \lt y$.

Note: Since $b \gt 1$, if $u, v \in \Bbb Q$ and $u \lt v$ then $b^u < b^v$.

So if we find $s, p \in \Bbb Q$ with $s\lt x$ and $p\lt y$ such that $r\lt s+p$, we would prove that no such $r$ as described above exists, because $r\lt s+p \implies b^r \lt b^{s+p} = b^sb^p$.

So let $\varepsilon = x+y-r$. By the Archimedean property, there exists a natural number $n$ such that $n\varepsilon \gt 1$ so $\varepsilon \gt 1/n$. From $r = x+y-\varepsilon$ it follows that $r + \frac 1n \lt x+y$ so $$r\lt x+y -\frac 1n$$ Now choose $s$ such that $x- \frac 1{2n} \lt s \lt x$ and $p$ such that $y- \frac 1{2n} \lt p \lt y$. So: $$p+s \gt x - \frac 1{2n} + y - \frac 1{2n}=y+x- \frac 1n \gt r$$ Just like we wanted, so we can conclude that $b^xb^y \lt b^{x+y}$ is false. And therefore $b^xb^y = b^{x+y}$.