# Prove sup(AB) = supA*supB

1. Mar 25, 2013

### joecharland

Can anyone tell me if the below proof is correct? Also how can I format my TEX differently so that it all works properly on this site?

Lemma - if $A$ and $B$ are sets of positive real numbers, put $AB = \left\{ ab | a \in A, b \in B \right\}$.
Then $\sup AB = \sup A \sup B$.
\\ Clearly $\sup A \sup B$ is an upper bound for $AB$ since $a \leq \sup A$ and $b \leq \sup B$ implies that $ab \leq \sup A \sup B$. We must show that $\sup A \sup B$ is the LEAST upper bound for $AB$.
\\ \\ Given arbitrary $\delta > 0$ we want to find $\epsilon_1 > 0$ and $\epsilon_2 > 0$ such that
$$(\sup A - \epsilon_1)(\sup B - \epsilon_2) > \sup A \sup B - \delta$$
This is equivalent to choosing $\epsilon_1$ which satisfies the following:
$$\sup A - \epsilon_1 > \frac{\sup A \sup B - \delta}{\sup B - \epsilon_2} \mbox{ (if } \epsilon_2 < \sup B)$$
$$\iff - \epsilon_1 > \frac{\sup A \sup B - \delta}{\sup B - \epsilon_2} - \sup A$$
$$\iff \epsilon_1 < \sup A - \frac{\sup A \sup B - \delta}{\sup B - \epsilon_2}$$
If we choose an $\epsilon_1$ less than this quantity we are done. The question is for what values of $\epsilon_2$ can we choose such an $\epsilon_1$. Well if the quantity on the right side of the last inequality above is greater than zero, we know we CAN choose such an $\epsilon_1$. So we want:
$$\sup A - \frac{\sup A \sup B - \delta}{\sup B - \epsilon_2} > 0$$
$$\iff \frac{\sup A \sup B - \delta}{\sup B - \epsilon_2} < \sup A$$
$$\iff \sup A \sup B - \delta <\sup A (\sup B - \epsilon_2) \mbox{ (if } \epsilon_2 < \sup B)$$
$$\iff \sup A \sup B - \delta < \sup A \sup B - \epsilon_2 \sup A$$
$$\iff - \delta < - \epsilon_2 \sup A \iff \delta > \epsilon_2 \sup A \iff \epsilon_2 < \frac{\delta}{\sup A}$$
The above outline provides the basis for the following proof:
\\ Choose $\epsilon_2$ such that $0 < \epsilon_2 < \frac{\delta}{\sup A}$ and $\epsilon_2 < \sup B$. As shown above this means $\sup A - \frac{\sup A \sup B - \delta}{\sup B - \epsilon_2} > 0$. Hence we can choose $\epsilon_1$ such that $0 < \epsilon_1 < \sup A - \frac{\sup A * \sup B - \delta}{\sup B - \epsilon_2}$. Then $(\sup A - \epsilon_1)(\sup B - \epsilon_2) > \big[\sup A - \big( \sup A - \frac{\sup A * \sup B - \delta}{\sup B - \epsilon_2}\big)\big](\sup B - \epsilon_2) = \frac{\sup A * \sup B - \delta}{\sup B - \epsilon_2}(\sup B - \epsilon_2) = \sup A \sup B - \delta$. Now $\exists a \in A$ such that $\sup A - a < \epsilon_1$ and $\exists b \in B$ such that $\sup B - b < \epsilon_2$. This means $a > \sup A - \epsilon_1$ and $b > \sup B - \epsilon_2$. Hence $ab > (\sup A - \epsilon_1)(\sup B - \epsilon_2) > \sup A \sup B - \delta$
Now if $\alpha < \sup A \sup B$ then $\alpha < \sup A \sup B - \delta$ for some $\delta > 0$. But $\exists a \in A, b \in B$ such that $ab > \sup A \sup B - \delta$. Then $\alpha$ is not an upper bound for $AB$ and $\sup AB = \sup A \sup B$.
\\ \\ Now that we proved this lemma we return to proving that $b^{r+s}=b^rb^s$ for all real $r$ and $s$.
\\ We must show that $\sup B(r+s) = \sup B(r) sup B(s)$. If $x \in B(r+s)$ then $x = b^q$ for some rational $q < r + s$. There are rational numbers $c$ and $d$ such that $x = c + d$, $c < r$ and $d < s$. Then $b^c \in B(r)$ and $b^d \in B(s)$, and $x = b^q = b^{c+d}=b^cb^d$. This means $B(r+s) \subset B(r)*B(s)$. Clearly $B(r)*B(s) \subset B(r+s)$ so $B(r+s) = B(r)*B(s)$. Since these are sets of positive real numbers we can apply the above lemma to conclude that $\sup B(r+s) = \sup B(r) \sup B(s)$ which is what we wanted to prove.

2. Mar 25, 2013

### jbunniii

On this site, wrap your TeX with two # symbols to display it within the paragraph (math mode), for example, to typeset "for all $x \in X$", write
Code (Text):
for all $x \in X$
Use two \$ symbols to put it on its own line (display mode), for example
$$\int_{0}^{1} f(x) dx$$
is obtained using
Code (Text):
$$\int_{0}^{1} f(x) dx$$
I'm sure someone will be happy to check your proof after the typesetting is fixed, but it's very hard to read right now.