# Prob. 6, Chapter 1 in Baby Rudin [duplicate]

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Here's problem $6$ in Chapter $1$ in the book Principles of Mathematical Analysis by Walter Rudin, $3$rd edition:

Fix a real number $b$, such that $b > 1$.

$(a)$ If $m, n, p, q$ are integers, $n > 0$, $q > 0$, and $r = m/n = p/q$, then (I've managed to show that)

$$\left( b^m \right)^{1/n} = \left( b^p \right)^{1/q}.$$

Hence it makes sense to define

$$b^r \colon= \left(b^m\right)^{1/n}.$$

$(b)$ (I've also managed to show that) for any rational numbers $r, s$, we have $$b^{r+s} = b^r b^s.$$

$(c)$ For any real number $x$, define the set $B(x)$ as follows:

$$B(x) \colon= \left\{ \ b^t \ \colon \ t \ \mbox{ is rational}, \ t \leq x \ \right\}.$$

We can then prove that

$$b^r = \sup B(r)$$

when $r$ is a rational number.

Hence it makes sense to define

$$b^x \colon= \sup B(x) = \sup \left\{ \ b^t \ \colon \ t \ \mbox{ is rational}, \ t \leq x \ \right\}$$ for every real number $x$.

$(d)$ How to prove (USING THE MACHINERY DEVELOPED HERE) that $$b^{x+y} = b^x b^y$$ for all real numbers $x$ and $y$?

I've already posted this question. Here's the link. However, none of the answers there seems to work for me.

My Work

Let $X, Y$ be any two non-empty subsets of $\mathbb{R}$. Then we have the following facts:

$(1)$ If $X \subset Y$ and if $Y$ is bounded above in $\mathbb{R}$, then so is $X$ and we have the inequality $$\sup X \leq \sup Y.$$

$(2)$ If $X, Y$ are non-empty subsets of the set of non-negative real numbers and if both $X$ and $Y$ are bounded above, then so is the set

$$\{ \ xy \ \colon \ x \in X, \ y \in Y \ \};$$ moreover, we have the identity

$$\sup \{ \ xy \ \colon \ x \in X, \ y \in Y \ \} = \sup X \cdot \sup Y.$$

Thus,

\begin{align} b^x b^y &= \sup B(x) \sup B(y) \\ &= \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, r \leq x \ \} \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 \ \}. \end{align}

But if $r \in \mathbb{Q}$, $s \in \mathbb{Q}$, $r \leq x$, and $s \leq y$, then the sum $r+s \in \mathbb{Q}$ and $r + s \leq x + y$ also. So

$$\{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r \leq x, \ s \leq y \ \} \subseteq \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \}.$$ Therefore,

\begin{align} b^x b^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}. \end{align}

Now we have to show that

$$b^x b^y \geq b^{x+y}.$$

Case I. When $x+y$ is irrational:

If $t$ is a rational number such that $t \leq x+y$, then since $x+y$ is irrational, we can also conclude that $t < x+y$.

So we can find a rational number $r$ such that

$$t-y < r < x.$$

Then

$$t-r < y < x+y -r.$$

Let's take $s \colon = t-r$. Then $s \in \mathbb{Q}$. And $s < y$.

Thus, we have written $t$ as $t = r + s$, where $r, s \in \mathbb{Q}$, with $r < x$, $s < y$.

So, when $x + y \in \mathbb{R} - \mathbb{Q}$, then we have

\begin{align} b^{x+y} &= \sup B(x+y) = \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \} \\ &= \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t < x+y \} \\ &= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r< x, \ s < y \} \\ &= \sup \{ \ b^r b^s \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r< x, \ s < y \} \\ &= \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, \ r < x \ \} \sup \{ \ b^s \ \colon \ s \in \mathbb{Q}, \ s < y \} \\ &\leq \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, \ r \leq x \ \} \sup \{ \ b^s \ \colon \ s \in \mathbb{Q}, \ s \leq y \} \\ &= \sup B(x) \sup B(y) \\ &= b^x b^y. \end{align}

Case II. What if $x + y \in \mathbb{Q}$?

In this case, we can conclude that both $x$ and $y$ are irrational.

How do we proceed with showing the remaining reverse inequality in this particular case?

## marked as duplicate by Alex M., Claude Leibovici, N. F. Taussig, SchrodingersCat, Kamil JaroszFeb 1 '16 at 15:43

This is another instance of proving stuff with one hand tied behind your back, limiting knowledge of the reals to what is in Rudin ch. 1. For this reason I write out a complete proof, even though most of it is given by the OP. We assume $b>1$ real.

Lemma 0. If $p>0$ is rational, $b^p>1$.

Lemma 1. If $s<x+y$, $s$ rational, then $s=p+q$ with $p$ and $q$ rational, $p<x$, $q<y$.

Proof. Pick $s'$ rational with $s<s'<x+y$. Pick $p$ rational with $x-(s'-s)<p<x$. Let $q=s-p$. We then have $q=s-p <s-x+(s'-s)<s-x+x+y-s=y$ as desired.

Lemma 2 (Rudin ch. 1 ex. 1.7(c)). If $t>1$ and $n>(b-1)/(t-1)$ then $b^{1/n}<t$.

Lemma 3. For any $x$ we have $b^x=\sup \{b^r:r<x, r\in\mathbb{Q}\}$.

Proof. If $x$ is irrational this follows from the definition and the fact that $r\ne x$ for rational $r$, so the set of rational $r\le x$ is really the set of rational $r<x$. If $x$ is rational, by Lemma 0 $b^x/b^r=b^{x-r}>1$ so $b^x$ is an upper bound of $\{b^r: r<x, r\in\mathbb{Q}\}$. Suppose there were a smaller upper bound $u$. Choose $n$ so that $b^{1/n}<b^x/u$, using Lemma 2. We then have $b^{x-1/n}>u$, with $x-1/n$ rational and $<x$, contradiction.

Lemma 4. If $X$ and $Y$ are sets of reals, $\sup X\cdot\sup Y = \sup \{xy:x\in X, y\in Y\}$. (Stated to be proven by the OP.)

Now we're ready for the main result, $b^{x+y}=b^x b^y$.

$$b^{x+y}=\sup\{b^r: r\in\mathbb{Q}, r<x+y\}=\sup\{b^{p+q}:p,q\in\mathbb{Q}, p<x, q<y\} =\\ \sup\{b^p:p\in\mathbb{Q}, p<x\}\cdot\sup\{b^q:q\in\mathbb{Q}, q<y\}=b^xb^y$$

The first and last equalities follow from Lemma 3. The second follows from Lemma 1. The third follows from Lemma 4. Fingers crossed I haven't made a mistake.

• Lemma 4 is wrong. $X$ and $Y$ need to be sets of positive reals, with $\sup X \ge 1$ and $sup Y \ge 1$ Luckily the sets we're dealing with satisfy those properties! – Quantaliinuxite Jul 4 '17 at 7:47
• Oops. Very sorry about that. – ForgotALot Jul 4 '17 at 16:26