Find for given set $A$, $|A|$ set's supremum & bounds.

Question

Need help in vetting my answers for the questions in CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller.

1. Let $A$ be a nonempty subset of the real numbers. Define the set $|A|$ to be $|A|:= \{|x| : x \in A\}$.
   (a) If $A$ is bounded above, is $|A|$ necessarily bounded above? If not, give an example. If so, by what? Explain.
   (b) If the set $A$ is bounded, is the set $|A|$ bounded? If not, give an example. If so, by what? Explain.
   (c) If the maximum of set $A$ exists, does $|A|$ have maximum? If so, what would it be? If not, give an example.
   (d) If the minimum of set $A$ exists, does $|A|$ have minimum? If so, what would it be? If not, give an example.
   (e) If the supremum and infimum of $A$ exist, does $|A|$ have a supremum? If not, give an example. If so, what would it be? Does $|A|$ have a infimum? If not, give an example. If so, what would it be?

(a) No, it may happen that the negative value of $|A|$ is unbounded. Say, $A=(-\infty,10)$. Here, $|A|$ has upper bound $=\infty$.

(b) Yes, if set $A$ is bounded, then there are both ends bounded, & hence supremum & infimum exist. Hence, $|A|$ will be bounded by the largest absolute magnitude of the bound for the set $|A|$.

*Edit : The edit is still informal one.
It is made a separate post (for fear of not attracting responses now, here) at : Supremum, infimum of $|A|$.

(c) Yes, but it may happen that the negative value of $x$ in $|A|$ is unbounded. In that case, the case (a) example applies. Else, it is possible that the negative values (of $x$) in $|A|$ take a higher value in magnitude than the positive value taken by $x$. Say, $A=[-20,10)$, then $|A|$ has maximum $20$.

(d) Yes, but it may happen that any positive value of $x$ in $|A|$ takes a smaller value in magnitude than a negative value taken by $x$. Say, $A=[-20,10)$, then $|A|= [0,20]$ has minimum $0$. Even if the positive set of values taken by $x$ in $A$ have un-bounded values, there would still exist minimum. In this regard, the answer differs from (c).

(e) Supremum & infimum are bounded values. So, if both exist, then the set $A$ is bounded. If bounded, then on taking the set $|A|$ all values are in bounds. So, $|A|$ has supremum, although it is possible that the infimum of $A$ is supremum of $|A|$. Say, $A= [-3,2]$. Here, supremum of $|A|=3$.
Same logic applies for infimum of $|A|$, i.e. it would exist, but it is possible to have supremum of $A$ as infimum of $|A|$. Say, $A= [-3,2]$. Here, infimum of $|A|=0$, as $|A|= [0,3]$.

*Edit : The edit is still informal one.
(e) Supremum & infimum are merely bounds, & it never means that a set possessing them has either of these in it as a member. E.g. $(-3,3)$ has infimum = $-3$, supremum $=3$ with none of them in the set.

Answer 1

For (a), your answer is technically correct, but the proof needs work. How do you know that $|A|$ is not bounded above in that case?

For (b), your answer is again correct, but the proof is not. You claim that $|A|$ "is bounded by the highest value in the set $|A|$. How do you know that that highest value even exists? For example, if $A=(0,1)$, then "the highest value of $|A|$ does not exist!

For (c), your answer is not correct. $A$ can have a maximum while $|A|$ can still be unbounded.

For (d), if $A=[-20, 10)$, then $|A|=[0,20]$, and the minimum of $|A|$ is not $10$.

For (e), your answer is correct, but your proof is not a proof at all. You claim:

Supremum & infimum are bounded values. So, if both exist, $|A|$ has supremum

how do you know this?