Prove inf(S)=-Sup(-S)?

1. Sep 20, 2012

SMA_01

Prove inf(S)=-Sup(-S)??

1. The problem statement, all variables and given/known data

Let S,T be subsets of ℝ, where neither T nor S are empty and both Sup(S) and Sup(T) exist.

Prove inf(S)=-sup(-S).

Starting with =>

I let x=inf(S). Then by definition, for all other lower bounds y of S, x≥y.

I'm stuck at this point...

Any help please?

Thanks

2. Sep 20, 2012

micromass

Staff Emeritus
Re: Prove inf(S)=-Sup(-S)??

You have to show two things:

• -x is an upper bound of -S
• If y is another upper bound of -S, then $-x\leq y$

So, in order to prove that -x is an upper bound of -S. Take an element -s from -S and prove $-s\leq -x$. Why is that true?

3. Sep 20, 2012

SMA_01

Re: Prove inf(S)=-Sup(-S)??

I'm not following along, why would -s≤-x? Unless you mean if I multiply both side by (-1), then it would be x≤s? Or am I totally off on a tangent?

4. Sep 20, 2012

jbunniii

Re: Prove inf(S)=-Sup(-S)??

If $-s \in -S$, then $s \in S$. And by definition, $x$ is a lower bound of $S$...

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