Let $C = \{\frac{a}{b}|a\in A, b\in B)$. Prove that $\sup C=\frac{\sup A}{\inf B}$

Question

Given $A,B \subseteq (1, 30).$

Let $C = \left\{\frac{a}{b}|a\in A, b\in B\right\}.$

Prove that $\sup C=\frac{\sup A}{\inf B}$

I know it seems trivial but I'm struggling with the formality.

Any help would be appreciated.

Answer 1

Let $a,b\in A,B$ respectively, then $b\geq\inf B$ and hence $\dfrac{1}{b}\leq\dfrac{1}{\inf B}$. And $a\leq\sup A$, so $\dfrac{a}{b}\leq\dfrac{\sup A}{\inf B}$, this shows $\sup C\leq\dfrac{\sup A}{\inf B}$.

Given $\epsilon>0$ and find some $a,b\in A,B$ respectively such that $\sup A-\epsilon<a$ and $\inf B+\epsilon>b$, then $\dfrac{\sup A-\epsilon}{\inf B+\epsilon}<\dfrac{a}{b}\leq\sup C$. Now let $\epsilon\downarrow 0$, then $\dfrac{\sup A}{\inf B}\leq\sup C$.

For the last part, another way: \begin{align*} \dfrac{\sup A-\epsilon}{\inf B+\epsilon}&<\sup C\\ b\sup A&<a\inf B+(a+b)\epsilon, \end{align*} and since $\epsilon>0$ is arbitrary, then \begin{align*} b\sup A&\leq a\inf B\\ \dfrac{\sup A}{\inf B}&\leq\dfrac{a}{b}, \end{align*} so $\dfrac{\sup A}{\inf B}\leq\sup C$.

Answer 2

Just do definitions.

If $c \in C$ then $c = \frac ab$ for some $a \in A$ and some $b \in B$.

Then $a \le \sup A$ and $b \ge \inf B$ so $\frac ab \le \frac {\sup A}{\inf B}$. (Note: This assumes $b > 0$ and $\inf B > 0$.)

So $ \frac {\sup A}{\inf B}$ is an upper bound of $C$.

If $0< x < \frac {\sup A}{\inf B}$ then $x*\inf B < {\sup A}$ so $x*\inf B$ is not an upper bound of $A$. so there exists and $\alpha \in A$ so that $x*\inf B < \alpha \le \sup A$. (Note: this assumes $\sup A > 0$.)

So $\inf B < \frac \alpha x$. So $\frac \alpha x$ is not a lower bound of $B$. So there exists a $\beta \in B$ so that $\beta < \frac \alpha x$.

So $x < \frac \alpha\beta \in C$. (Note: this relies upon $x > 0$.)

So $x $ is not an upper bound of $C$.

So $\frac {\sup A}{\inf B}$ is the least upper bound of $C$.

(Note: Don't need $A,B \subset (1,30)$ just need $A,B \subset \mathbb R^+$.)