Let $ A $, $ B $ two subsets of the real line; let, for at least one $ a>0 $, $ a\in A $. I've tried to prove that $ \inf\left(B/A_{>0}\right)=\inf{B}/\sup{A} $, were $ A_{>0} $ is meant to be $ \mathbb{R}_{>0}\cap A $. The thesis sounds "geometrically" evident to me.
My poor attempt of proof: It's clear that $ \inf{B}/\sup{A} $ should be a lower bound for $ B/A_{>0} $. The first thing that came to my mind to show that, given $ x>\inf{B}/\sup{A} $, there exists $ b'/a'\in B/A_{>0} $ such that $ b'/a'<x $, was to consider $ \epsilon>0 $ then let $ x=\inf{B}\epsilon $ and $ y=\sup{A}(1/\epsilon) $. There exists consequently $ b'<\inf{B}\epsilon $ and $ a'>\sup{A}(1/\epsilon) $ and $$ \frac{b'}{a'}<\frac{\inf{B}}{\sup{A}}\cdot\epsilon^2 $$ actually proving the statement.
This approach is obviously wrong, since $ x $ and $ y $ need not to be respectively greater than $ \inf{B} $ and lower than $ \sup{A} $.
Is this statement at all true? If so, may I have a hint or a proof?
This is not true. If $B=[-1,0]$ and $A=(0,1]$, $\inf B/\sup A= -1$ even though $\inf (B/A)=-\infty$. If we assume $B\subset (0,\infty)$, then $\inf B/\sup A\le b/\sup A\le b/a$ for $a\in A_{>0}$ so $\inf B/\sup A\le \inf (B/A_{>0})$. Since $a\inf (B/A_{>0})\le b$ for all $a\in A_{>0}$, $a\inf (B/A_{>0})\le \inf B$. If $\inf (B/A_{>0})=0$ then $\inf B=0$ or $\sup A=\infty$ and we are done (assuming $b/\infty=0$). Otherwise, $\sup A\le \inf B/\inf(B/A_{>0})$, which gives the required reverse inequality $\inf(B/A_{>0})=\inf B/\sup A$.