# What's wrong with my proof of $\inf S = -\sup(-S)$?

$$S\subset \mathbb{R}$$ 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?

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