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

Question

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 $\inf S = − \sup(−S)$.

Answer 1

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.

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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.

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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 $-S$.

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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. :)

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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.