Let $S$ and $T$ be bounded sets of negative real numbers. Define $ST=\{st|s \in S, t\in T\}$. Show that $ST$ is bounded and that $\inf(ST)=\sup(S)\sup(T)$.
I tried to get an idea how to prove this by considering a small, finite set of negative numbers. So, for instance $S=\{-5,-4,-3,-2,-1\}$ and $T=\{-10,-9,-8,-7,-6\}$. The $\sup(S)=-1$, and $\inf(S)=-5$, $\sup(T)=-6$, and $\inf(T)=-10$. Since multiplying two negative numbers is positive, it's obvious that the $\inf(ST)=(-1)(-6)=6$. I tried doing this with infinite sets, where I assumed the set was in order from least to greatest and then I would take the absolute value and show that the order changed, but when I wrote it out, it was wrong because the sets are not necessarily countable. How would I prove this?
Thank you.