# $A, B$ nonempty, bounded subsets of $(0, \infty)$, find $\inf(C)$ where $C=\{ \frac{b}{a} \mid b \in B , a \in A\}$ and where $\sup(A)=\inf(B)$

I want to find the infimum of $$C$$ where $$C=\{ \frac{b}{a} \mid b \in B , a \in A\}$$ and where $$\sup(A)=\inf(B)$$. We also know that $$A$$ and $$B$$ are both nonempty subsets of $$(0,\infty)$$ and therefore bounded.

Clearly the sets $$B$$ and $$A$$ have some "ordering". We know that on the real number line, $$A$$ is to the left of $$B$$, we now consider the ratios of numbers in these sets, this will have a lower bound, we expect that when the smallest number in $$B$$ is divided by the highest number in $$A$$ this will give a really small number. We thus expect the greatest lower bound to be $$\frac{\inf(B)}{\sup(A)}=\frac{\sup(A)}{\sup(A)}=1$$. So we expect the infimum of $$C$$ to be $$1$$. Indeed we have that for all $$a\in A$$ and for all $$b\in B$$

$$\frac{b}{a} \geq \frac{\inf(B)}{a}\geq \frac{\inf(B)}{\sup(A)}=1$$ So $$1$$ is a lower bound.

The hardest part is of course to prove it is the greatest lower bound. I know two approaches:

$$1)$$ show that no greater lower bound can exist, so suppose we have found one, and then forcing a contradiction.

$$2)$$ Showing there exists a number $$c\in C$$ such that $$c < \inf(c) + \epsilon$$ for all $$\epsilon >0$$

I don't think the second approach is very useful, but I'm not sure how to get a contradiction in the first version of the infimum either. I need to somehow use that I know that they are related.

For the expression $$\frac{\inf B}{\sup A}$$ to make sense, you should first verify that $$\sup A\neq0$$. This is of course not hard because $$A\subset (0,\infty)$$.

Now suppose a greater lower bound exists, say $$1+\varepsilon$$ where $$\varepsilon>0$$. Then for all $$a\in A$$ and $$b\in B$$ $$\frac{b}{a}\geq 1+\varepsilon\qquad\text{ and so }\qquad b\geq (1+\varepsilon)a.$$ In particular it follows that $$b\geq(1+\varepsilon)\sup(A)$$ for all $$b\in B$$, and hence that $$\inf B\geq(1+\varepsilon)\sup A>\sup A,$$ contradicting the fact that $$\inf B=\sup A$$. So no lower bound greater than $$1$$ exists.

• Very nice indeed. – Wesley Strik Jan 27 at 8:45