Given a set $S\in \mathbb{R}$, let us write $−S$ for the set $\{−x \mid x\in S\}$. Prove that if $S$ is bounded below then $−S$ is bounded above.
This is not a hard problem, but I am puzzled by the follow:
I assumed $x$ is an arbitrary element in $S$, and let $L$ be an lower bound for $S$. Thus, $x\geq L.$ so $-x\leq -L.$ Now the problem is how do you show $-x$ covers all the elements of $-S$. I mean, $x$ is an element of $S$, but now how do you show $-x$ also represent every element of $-S$?
You already proved it:
$$x\ge L\;\;\text{forall}\;\;x\in S\iff -x\le -L\;\;,\;\text{forall}\,-x\in-S$$
You take an arbitrary element $y\in -S$ and you prove that $y\leq -L$.
Now, since $y\in -S$ you know that there exists such an $x\in S$ that $y=-x$. You know this because the very definition of $S$ is:
All such values that can be written as $-x$ for some value of $x\in S$
And since $x\geq L$, you know that $-x \leq -L$, thus, $y\leq -L$ is true for an arbitrary $y\in-S$.
