Prove: $-M = \inf(-E)$

Question

I have to prove this exercise for my math study:

Let $E \subset \mathbb{R}$, $\mathbb{R} \ni M = \sup(E)$ and $-E := ${ $-x$ | $x \in E$ }.

Prove: $-M = \inf(-E)$

I think I've completed the proof by contradiction:

Assume $-M \neq \inf(-E)$
$\Rightarrow \exists$ $-x \in -E$ for which $-x < -M$ (and $x \in E$)
$\Rightarrow$ $\exists$ $x \in E$ for which $x > M$
This is a contradiction because $M = \sup(E)$
So $-M = \inf(-E)$

Is this proof correct? It looks a little bit simple to me. If it's not correct, could you tell me how to proof this exercise? Thanks in advance!

Answer 1

Take $x \in E$
$\Rightarrow x \leqslant M$
$\Rightarrow -x \geqslant -M$
$\Rightarrow -x \in -E$ if $x \in E$, so $-M$ is a lower bound of $-E$

Now assume $-M' > -M$ and $-M'$ is a lower bound of $-E$
$\Rightarrow -M' > -M$
$\Rightarrow M' < M$
$\Rightarrow -x \geqslant -M'$
$\Rightarrow x \leqslant M'$
So, $M'$ is an upper bound that is lower than $M$. This is a contradiction because $M = sup(E)$.
$\Rightarrow -M'$ is not an upper bound of $-E$
$\Rightarrow -M = inf(-E)$ $\Box$