$S\subset \mathbb{R}$ is non-empty. We define $-S:=\{-s:s\in S\}$.

First, I prove $\inf S \geq -\sup(-S)$.

$-s\leq \sup(-S)$ for every $-s\in -S$ because supremum is an upper bound.

$\implies s \geq -\sup(-S)$ for every $s\in S$. So we have a lower bound for $S$.

$\implies \inf S \geq -\sup(-S)$ because infimum is the greatest among lower bounds.

Now, I prove $\inf S \le -\sup(-S)$ and it suffices to conclude the equality.

$s\ge \inf S$ for every $s\in S$ because infimum is an lower bound.

$\implies -s \le -\inf S$ for every $-s\in -S$. So we have an upper bound for $-S$.

$\implies \sup(-S)\le -\inf S$ because supremum is the smallest among upper bounds.

$\implies \inf S \le -\sup(-S)$.

Is this proof acceptable? If it was a quiz or a test, how much score would I get?

And how do I improve it?

  • $\begingroup$ Where you say ‘So we have an upper bound for $S$’ you mean ‘… an upper bound for $-S$’, but otherwise it looks fine. $\endgroup$ Mar 30 '20 at 4:25
  • $\begingroup$ @BrianM.Scott It was typo, thanks. I edited it. $\endgroup$
    – user642721
    Mar 30 '20 at 4:26
  • $\begingroup$ "$\implies\sup(-S)\leq-\inf S$ because supremum is the greatest among upper bounds" should be "the smallest among upper bounds". $\endgroup$
    – trisct
    Mar 30 '20 at 4:26
  • $\begingroup$ @trisct Thanks. I edited it. $\endgroup$
    – user642721
    Mar 30 '20 at 4:27
  • $\begingroup$ It's not wrong, looks good. $\endgroup$
    – Zduff
    Mar 30 '20 at 4:33

Yes, this proof is acceptable. you will get decent score. Be aware that $\Bbb R$ is complete so for a bounded set supremum and infimum always exist by completeness property. If $S$ is unbounded then you have to consider extended real number system.


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