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Let $-S$ be the set $\left\{ -s:s \in S \right\}$ where $-S$ is the set that contains negatives of the members of $S$. We want to prove that $\inf(S) = -\sup(-S)$

Here is how I proved it Let $s_0= \sup(-S)$. That is for all $-s_1\in -S$ then $-s_1 \leq s_0$. Multiplying both sides by $-1$ we get $-s_0 \leq s_1$ for all $s_1 \in S$. So $\inf(S)=-s_0=-\sup(-S)$ It looks short and sweet. Not sure if its right though.

user60887
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    Looks like Brian M. Scott posted a few seconds ahead of me because I spent some time editing the question. Where I found $sup(-S)$ and $inf(-S)$ I wrote $\sup(-S)$ and $\inf(-S)$, coded as \sup(-S) and \inf(-S), with a backslash. This (1) prevents italicization, (2) provides proper spacing in experssions like $a\sup b$, which would otherwise look like $a sup b$, and (3) in "displayed" contexts, affects positions of subscripts in things like $\displaystyle\sup_{x\in A}f(x)$. – Michael Hardy May 27 '13 at 05:05
  • OK thanks. Looks like ill need to go back and figure it out. I just thought of it right now. – user60887 May 27 '13 at 05:06
  • Could I just add in there that $\inf(S)=-s_0$ since $-s_0 \leq s_1$ for all $s_1 \in S$. – user60887 May 27 '13 at 05:14

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It’s fine as far as it goes, but it’s incomplete. You’ve shown that $-\sup(-S)$ is a lower bound for $S$, but not that it’s the greatest lower bound. If you assume that it’s not the greatest lower bound, you can use the same basic idea to get a contradiction; I’ll leave it to you to try that on your own, but I’ll be happy to say more if necessary.

Brian M. Scott
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This argument is incomplete. If $s_0=\sup(-S)$, the certainly for all $s_1\in S$ we have $-s_1\le s_0$. But that says only that $s_0$ is an upper bound of $-S$, not that is is the least upper bound. You have to account for "leastness".

Michael Hardy
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