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.

  • $\begingroup$ Very nice indeed. $\endgroup$ – Algebra geek Jan 27 '19 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.