# Prove that $m$ is a lower bound for $S$ if and only if $−m$ is an upper bound for $−S$.

Given : $$−S = \{−s : s \in S\}.$$

Prove that $$m$$ is a lower bound for $$S$$ if and only if $$−m$$ is an upper bound for $$−S$$. And prove that if $$S$$ is bounded below then its greatest lower bound satisfies $$\inf S = − \sup(−S)$$.

• This is just an application of $x \le y \iff -x \ge -y$. Where did you get stuck? – Theo Bendit Oct 10 '18 at 4:09

Since these proofs are pretty similar to each other, I'll complete one and leave the rest incomplete, with the first proof serving as a hint.

Prove that $$m$$ is a lower bound for $$S$$ if and only if $$−m$$ is an upper bound for $$−S$$.

To expound upon what was noted by Theo Bendit in the comments, we essentially use the definition of upper/lower bounds and the fact $$x \leq y \Leftrightarrow -x \geq -y$$ to complete the proof.

Proof ($$\Rightarrow$$):

Suppose $$m$$ is a lower bound for $$S$$. Then, for all $$s \in S$$, by definition of lower bound, $$m \leq s$$. Then, as a result, $$-m \geq -s$$ for all $$s \in S$$.

Recall the set $$-S$$ is defined by $$-S = \{ -s | s \in S\}$$.

Since $$-m$$ meets that condition - that it is greater than or equal to all elements of $$-S$$ - then $$-m$$ is by definition an upper bound for the set $$-S$$.

Proof ($$\Leftarrow$$):

The converse -- that $$-m$$ being an upper bound for $$-S$$ implies $$m$$ is a lower bound for $$S$$ -- follows very similar logic to the previous, and thus is an exercise left to the reader. :)

And prove that if $$S$$ is bounded below then its greatest lower bound satisfies $$\inf S = − \sup(−S)$$

This one mirrors the previous in that it basically exploits $$x \leq y \Leftrightarrow -x \geq -y$$ and the definitions of supremum and infimum. So I'll again leave this proof incomplete, just with the hints that:

• $$\inf S$$ is the element which is the greatest of the lower bounds.
• $$\sup S$$ is the element which is the least of the upper bounds.
• To show that an element is either, you not only want to show it is an upper/lower bound by the definition, but also that it is the least/greatest such bound respectively. How you choose to do this is up to you: supposing not, and then finding a contradiction, is how I'd go about it.

Thus, you want to show $$\inf S$$ is not only an upper bound of $$-S$$ (hint: use the previous proof) and also show it is the least upper bound.