# Bounded sets, infimum, supremum proof

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.

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For all $s,t$, $st=|s||t|$. Because $S$ and $T$ are bounded, $\sup\{|s|:s\in S\}$ is attained when $s=\inf S$ and $\sup \{|t|:t \in T\}$ is attained when $t= \inf T$. Clearly $\sup ST= (\inf S)(\inf T)$, both of which exist (we are in $\mathbb{R}$). Then, since $st>0$ for all $s \in S, t \in T$, it follows that for all $a,b \in ST, d(a,b)\leq (\inf S) (\inf T)$ so that $ST$ is bounded. A similar argument shows that $\inf(ST)= \sup (S) \sup (T)$.
If $|a|<|c|$, then $|ab|<|bc|$ for any $b$. Also, if $a<0,b<0$ then $ab>0$. Those two properties come together pretty nicely to give you the answer.