I need to prove that $\sup \{-x \mid x \in A\} = -\inf\{x \mid x \in A\}$ and am having trouble moving the $-x$ out of the $\sup$ to $\inf$. Another thing is that I don't quite know how to prove $b = \inf$ (see below). Any help would be appreciated! Here my attempt:

Let $a = \sup\{-x \mid x \in A\}$

Then by definition $a \geq -x$ for $x \in A \implies -a\leq x$ for $x \in A$

Thus $-a$ is a lower bound of $A$. Assume $b$ is a lower bound of $A$

Then $b \leq -a \implies a \geq -b$ and $a = \sup \{−x \mid x\in A\} = -\inf(A)$

  • $\begingroup$ When you reach "$-a$ is a lower bound of $A$": Why do we have $b\leq-a$? Also, you mean $-A$ in the last $\sup$ $\endgroup$ – AD. Oct 11 '13 at 6:00
  • $\begingroup$ I wanted to show that (1) it is a lower bound and brought in the b to show that (2) it is the lower bound. I thought this was the way to prove sup/inf? Also, the bottom should read sup {−x|x∈A}=−inf {x|x∈A}. Sorry, I'll fix that. $\endgroup$ – user2853043 Oct 11 '13 at 6:06
  • $\begingroup$ You already showed that $-a$ is a lower bound of $A$. You want to show $-a$ is the greatest lower bound of $A$. It suffices to prove that $b\leq -a$ for any given lower bound $b$ of $A$. Now, for such $b$ we have $b\leq x$ for all $x\in A$, and then $-x\leq-b$ for all $x\in A$ -- deduce $a\leq -b$ (because $a=\sup(-A)$). $\endgroup$ – AD. Oct 11 '13 at 6:56
  • $\begingroup$ Btw this is strongly related to math.stackexchange.com/questions/488022/infs-sup-s?rq=1 , but it is not identical since this is a question regarding a particular proof verification. $\endgroup$ – AD. Oct 11 '13 at 7:05
  • $\begingroup$ Thanks for your responses. I know this is a bit similar, but I didn't know what to do with sup{A} vs inf{A} instead of sup{A} vs inf{-A} at the time. Thinking back it might have been easier to just show sup{−x∣x∈A} = sup{x|x∈-A} I guess... $\endgroup$ – user2853043 Oct 11 '13 at 7:11

After you say that $-a\leq x$ for all $x\in A$, you can then conclude that $$ -a \leq b:= \inf A $$ and hence $$ a \geq -b $$ Similarly, $b\leq x$ for all $x\in A$, and hence $-b \geq -x$ for all $x\in A$, and so $$ -b \geq a $$

  • $\begingroup$ Thank you for your answer. Just to clarify (as I am still new to proofs), would I need to write both a ≥ - b and -b ≥ a to establish (1) that they are equal to each other and (2) that they are the inf? $\endgroup$ – user2853043 Oct 11 '13 at 5:55
  • $\begingroup$ You already defined $b$ to be the inf. So you only need to show that they are equal, and it suffices to show that $a\leq -b$ and $a\geq -b$ in order to do so. $\endgroup$ – Prahlad Vaidyanathan Oct 11 '13 at 6:03

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