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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?

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  • $\begingroup$ this proof is alright. $\endgroup$ – Abishanka Saha Sep 9 '13 at 12:56
  • $\begingroup$ Related and not a duplicate is this answer. $\endgroup$ – Git Gud Sep 9 '13 at 12:57
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The "proof" is not written in standard prose, capitalization, or punctuation. That is the first thing you should fix.

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 general, when a "for all ..." appears at the end of a sentence with other quantifiers, that final "for all ..." becomes the innermost quantifer.

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.

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  • $\begingroup$ 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. $\endgroup$ – Git Gud Sep 9 '13 at 13:14

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