# Proof of Inf S= -sup(-S)

• September 30th 2012, 05:28 PM
JSB1917
Proof of Inf S= -sup(-S)
I'm trying to prove that inf S = -sup(-S) where -S={-s : s is in S} and S is a nonempty subset of the Reals.

I was able to do the bounded above and below case, and the bounded below case and not above, but I having trouble with the not bounded below and bounded above and not bounded at all cases.

Help Appreciated.
• September 30th 2012, 06:58 PM
hollywood
Re: Proof of Inf S= -sup(-S)
If S is not bounded below, then -S is not bounded above. What do we say inf S and sup(-S) are in this case?

- Hollywood
• September 30th 2012, 07:05 PM
JSB1917
Re: Proof of Inf S= -sup(-S)
Quote:

Originally Posted by hollywood
If S is not bounded below, then -S is not bounded above. What do we say inf S and sup(-S) are in this case?

- Hollywood

Yes, I understand that if S is not below then inf S=-infinity and if -S is not above, then sup -S = infinity. But is it really just a matter of saying then since inf S = -infinity and sup -S= infinity then -sup -S is -infinity also?

Or am I trying to make it harder than it has to be?
• September 30th 2012, 09:04 PM
hollywood
Re: Proof of Inf S= -sup(-S)
Yes, that's all it is.

To be 100% technically correct, the problem statement would have to be "Inf S = -Sup(-S) when they are finite and Inf S = $-\infty$ if and only if Sup(-S)= $+\infty$" since the expression $-(+\infty)$ is undefined. But that's a technicality - no one thinks $-(+\infty)$ is anything other than $-\infty$.

The final case is easy now, right?

- Hollywood