I am given set $A$ which lays in $\mathbb{R}$, and is bounded above and below:
$$ |A|=\{|a|:a\in A\}. $$
Now I have to prove that $$\sup|A|=\max\{-\inf A,\,\sup A\}.$$ However, I have to do that using the following steps:
1. Use $|a| = \max\{−a, a\in$ to show that $|a| ≤ \max\{− \inf A,\,\sup A\}$ for all $a \in A.$
2. Prove that if $|a| ≤ b$ for all $a ∈ A$, then $\max\{− \inf A,\,\sup A\} \le b.$
I've been struggling with this for a while now. How do I get from the set $A$ to its elements, or well, small $a?$ Isn't it obvious that the absolute value of element $a$ should be larger than $-\inf A ?$ Since it's positive only?
Can someone help me maybe? Thanks!
Note that, for $\forall a\in A$, $$ \inf A\le a\le \sup A\tag{1}$$ and hence $$ -\sup A\le -a\le -\inf A.\tag{2}$$ From (1) and (2), one has $$ |a|\le \max\{-\inf A,\sup A\} $$ and hence $$ \sup |A|\le \max\{-\inf A,\sup A\}. \tag{*}$$ On the other hand, for $a\in A$, $$ |a|\le \sup|A|. $$ Since $a\le|a|$ and $-a\le|a|$, one has $$ a\le \sup|A|, -a\le\sup|A| $$ which implies $$ \sup A\le\sup|A| \tag{3}$$ and $$ \sup(-A)\le\sup|A|. \tag{4} $$ Noting $\sup(-A)=-\inf A$, one has, from (4) $$ -\inf A\le\sup|A|. \tag{5} $$ $$ a\le\sup |A|. $$ Now (3) and (5) imply $$ \max\{\sup A, -\inf A\}\le\sup|A|. \tag{**}$$ From (*) and (**), one has $$ \sup |A|= \max\{-\inf A,\sup A\}. $$