Proof for $-\sup(A) = \inf(-A)$

Let $A$ and $-A = \{ -x \mid x \in A \}$ be two bounded sets.

I have to prove that $-\sup(A) = \inf(-A)$, i did it in the following way and wish to know if it is sufficient:

$\exists x\in A$ such that $\sup(A) - \epsilon < x, \forall \epsilon > 0$, multiplying by -1 we get: $-\sup(A) + \epsilon > -x, \forall \epsilon > 0$, and this is the inequality for $\inf(-A)$ and since $\inf(-A)$ is unique, it follows that $-\sup(A) = \inf(-A)$

Is this wrong?

• this proof is alright. – Abishanka Saha Sep 9 '13 at 12:56
• Related and not a duplicate is this answer. – Git Gud Sep 9 '13 at 12:57

There is at least one error at the beginning. The sentence "there is an $x \in A$ such that $\sup(A) - \epsilon < x$ for all $\epsilon > 0$" means "there is an $x \in A$ such that for all $\epsilon > 0$, $\sup(A) - \epsilon < x$". This is not what you meant to say, and it is false for any set $A$ that has only one point.
In this case you want it to be the outermost quantifier, you want to say "for all $\epsilon > 0$, there is an $x \in A$ such that $\sup(A) - \epsilon < x$". In general, you can avoid this error by not writing "for all ..." at the end of a sentence until you have more practice with quantifiers. If you write the quantifiers at the beginning, it is easier to see how they are nested.
• In general, when a "for all ..." appears at the end of a sentence with other quantifiers, that final "for all ..." becomes the innermost quantifer. Golden rule which I had to figure out by myself after two years of studying. – Git Gud Sep 9 '13 at 13:14