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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.

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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.

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  • $\begingroup$ Very nice indeed. $\endgroup$ – Wesley Strik Jan 27 at 8:45

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