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There is a question in my textbook that goes as follows:

Let $A \subset \mathbb{R}$ be a non empty-set that is bounded from below. Show that $-A$ = {$-x | x \in A $} is a non-empty set and bounded from above.

This is my short proof. (In my textbook they worked with the infimum and supremum, which I didn't do since I thought it was unnecessary. I think my approach is actually better, because it's up to another question from the textbook to ask me to prove that $\sup(-A) = - \inf (A)$.)

It is obvious that $-A$ is not empty if $A$ is not empty. Because of the fact that $A$ is bounded below, there exists an $x \in \mathbb{R}$ such that: $a \geq x, \forall a \in A$. If we multiply this expression with $-1$, we get $-x \geq -a$, so we see that $-x$ is then an upper bound for $-A$. So $-A$ is bounded from above. $\blacksquare$

hearot
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user34
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    I think it's OK. At this level I am usually very pedantic, so I'd personally prove that $-A$ is non-empty and not just say it's obvious. (Of course it is obvious ;-) ) – ancient mathematician Aug 16 '23 at 08:53
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    Your proof is indeed right. There is no need to show that $\sup(-A) = -\inf(A)$ (even though it suffices to prove that to get to the same conclusion). In fact, if $A$ is not empty, $-A$ cannot be empty as well, trivially (but I appreciate the comment from @ancient mathematician). So letting $m$ be a lower bound for $A$, $m \leq a$ $,\forall a \in A$; then $-m \geq -a$ $,\forall {-a} \in {-A}$. – hearot Aug 16 '23 at 08:55
  • @ancientmathematician I understand what you're saying and usually I don't like these kind of sentences like "this is trivial" or "obviously" but the assistants of this course do it sometimes, so I guess I copied the behaviour. But I don't like it because you assume something about the prior knowledge of your reader, I think. Or it could be that you try to omit the proof for the wrong reasons. – user34 Aug 16 '23 at 11:59
  • I reckon it's worth being hyper-careful when you first meet sups/infs but after a month or two, once you're sure that you're not kidding yourself, then you get to claim that the obvious is indeed obvious! – ancient mathematician Aug 16 '23 at 15:00

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