Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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 $\inf(ST)= \sup (S) \sup (T)$.

share|improve this answer
    
Thanks. I'm a little confused by your notation,why did you write sup|s| and sup|t|? As in, are you referring to s as a set or an element of S? –  Alti Dec 5 '12 at 5:07
    
@Alti I edited it to make it clearer. –  Jebruho Dec 5 '12 at 5:09
    
"sup{|s|:s∈S} is attained when s=inf(S) and sup{|t|:t∈T} is attained when t=inf(T)" Can I automatically assume this or would I need to provide a proof? –  Alti Dec 5 '12 at 5:16
    
Well the absolute value is just the distance from the origin, and it is obvious that the "more negative" a number is the greater the absolute value so that the result becomes obvious. –  Jebruho Dec 5 '12 at 5:18
    
Okay I see what you mean, thank you for your help. –  Alti Dec 5 '12 at 5:27

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.

share|improve this answer
    
While these statements are true, they are obvious and do not address the issue of supremums or infimums at all, which is really the key, more difficult part of the problem. –  Jebruho Dec 5 '12 at 4:54
    
While I agree with the first part of your statement, I do not tend to agree with the second. Yes, they are obvious, but it may not be clear that they are the properties in play. I believe that the extension from the qualities named to supremums and infimums is obvious as well, and that someone who is in a place to be approaching these sort of problems would, pedagogically, be better suited by being pointed in a useful direction, rather than being given the full elaboration of the answer. –  deftfyodor Dec 5 '12 at 5:23
    
Thank you for your help. –  Alti Dec 5 '12 at 5:25
    
@deftfyodor Yes, but your first statement isn't all that relevant to the proof and the second Alti referenced himself: "Since multiplying two negative numbers is positive." –  Jebruho Dec 5 '12 at 5:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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