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

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

## Homework Statement

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...

Thanks

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?

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?

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