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Need help in vetting my answers for the questions here 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.

jiten
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1 Answers1

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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?

5xum
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  • Thanks for pointing out the error in (d), it was similarly repeated in last line. But, edited the post to correct errors, as it is more for reference for me. I hope the part (e) proof is now fine. If not, then, please point it, will try again. Also, attempted to correctly state the proof in part (b). – jiten Apr 25 '19 at 11:54
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    @jiten Your proof for (b) is still just as incorrect as it was. For example, if I was a TA correcting an exam, your answer for (b) would get $0$ points. Same for (c). For (d), your answer is still incorrect. Sure, the answer "yes" is correct, but the justification is very very very lacking. It would get, if I were in a good mood, a quarter of a point. Examples are good when they are counterexamples. When proving some statement that says something is true for all sets $A$, showing that it is true for some $A$ is more or less worthless. – 5xum Apr 25 '19 at 12:01
  • I feel lost. Might be a part's (say, part (b)) answer / hint can tell me how to proceed. I am sorry, but clueless. But, got clear that examples do not work here, formal proof is needed. But, for me formal proof is till this only, unless get initiated by some proof. Part (b) proof seems most abstract of all, & hence any clue for that is the best. – jiten Apr 25 '19 at 12:09
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    @jiten Your first problem is that you asked 5 questions in this post. This site works best if you have one question per post. That way, you can discuss that one question in detail. For $b$, you want to prove that the answer to the question is yes, which means you must start with an arbitrary set $A$, and prove that $|A|$ is bounded. That is, you must look at the definition of "bounded", and prove that $|A|$ satisfies the definition. – 5xum Apr 25 '19 at 12:34
  • Please tell me where to look, as I have delved and found that it links with epsilon; as here : https://math.stackexchange.com/a/2461842/424260. I submitted so many answers without knowing the depth involved for the proofs. Had I known the depth, it would have been impossible to submit so many answers in a go by me. If possible, kindly take only part (b) & provide its answer. I hope to learn a ton from that. I request the answer to be detailed enough to promote further reading. – jiten Apr 25 '19 at 13:07
  • @jiten The comments are not the place to hold prolonged discussions. I explained to you that you should only ask one answer per post. – 5xum Apr 25 '19 at 13:14
  • What can be done if mistake is made once. I can in no way frame part (b) (as a new question) without any background knowledge. Only you can help. Else, even direction of approach is still not clear. Just 'bounded sets' can lead to another way easily, at least for me. So, some help is definitely needed to remain close to the goal, and know what all to look at. – jiten Apr 25 '19 at 13:18
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    @jiten Why couldn't you frame (b) as a new question? – 5xum Apr 25 '19 at 13:20
  • I am ready with draft, but worried as it relies too much on your answer to build correct proof. Please accept it else will modify it: https://i.stack.imgur.com/DvVqZ.png. I will delete this comment in any case. I cannot pursue this new post without your approval. – jiten Apr 25 '19 at 14:05
  • Have put for part (b) at : https://math.stackexchange.com/q/3217092/424260 – jiten May 07 '19 at 12:23