Supremum, infimum of $|A|$.

Question

Let $A$ be a nonempty subset of the real numbers. Define the set $|A|$ to be $|A|:= \{|x| : x \in A\}$.
If the set $A$ is bounded, is the set $|A|$ bounded? If not, give an example. If so, by what? Explain.

An open bound is also bounded one, & both bounds can be open; still the set is bounded. Also, for one point sets, supremum (let, $s$) = infimum (let, $i$).
Yes, if a set is bounded, its modulus set too is. The possible cases for both supremum & infimum are:

(i) supremum $\le 0$:
As supremum $\ge $ infimum, infimum $\le 0$. $|A|$ is not including origin then. Say, $A = [-2, -1)$, hence does not contain $0$. The modulus should have all values in range |A|=(1,2]$

(ii) supremum $\ge 0$:
As supremum $\ge $ infimum, options are :

1. infimum $\le 0$. Say, $A = [-2, 1)$, so $0$ is in the set. Hence, $|A|=[0,2]$, but element $1$ is not in the set $|A|$. So, confused as the set $|A| = [0,1)\cup (1,2]$. So, in fact have two sets in $|A|$ if an open bound is now lying inside the new bounds. So, there should be infimum, supremum for two sets separately in $|A|$.

2. infimum $\ge 0$. Here, same as for $A$.

Answer 1

(i) supremum $\le 0$:
As supremum $\ge $ infimum, infimum $\le 0$. $|A|$ is not including origin then. Say, $A = [-2, -1)$, hence does not contain $0$. The modulus should have all values in range $|A|=(1,2]$

This is not a proof. While the statement you are trying to prove is correct, what you did here is only a demonstration of the claim on one particular set $A$. And, yes, sure, for $A=[-2,-1)$ it is true that $|A|$ is a bounded set. But how do you know that this is true for every bounded set $A$? How do you know there doesn't exist some very strange exotic set $A$ which is bounded, but $|A|$ is not bounded?

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To actually prove the claim, you should:

1. Take an arbitrary set $A$.
2. Assume that $A$ is bounded.
3. Prove that $|A|$ is therefore also bounded

In steps $1$ and $2$, you are not allowed to decide what $A$ will look like. All you are allowed to assume is that $A$ is a bounded set.

The actual steps of the proofs can differ, but the easiest way to prove it would be something like this. Note, I will not be doing all the work here. I am just giving you guidelines.

1. Assume $A$ is bounded.
2. Write down, in detail and exactly, what the definition of bounded means. (that is, $A$ is bounded if there exists some such thing that some other thing is true for all things in somethingsomething).
3. Write down, in detail and exactly, what the sentence "$A$ is bounded" means
4. Write down, in detail and exactly, what the sentence "$|A|$ is bounded" means
5. Prove that if sentence 3 is true, then sentence 4 is also true.

remember, in sentence $4$, you will get a sentence with this structure:

$|A|$ is bounded if, for every $x$ in $|A|$, somethingsomething.

This can be proven by taking an arbitrary $x\in|A|$ and doing something with it.

Answer 2

I'm not sure why you dive into supremum and infimum. In any case, your text about these is wrong. For example, if $\sup=\inf=0$, $A$ still can include the origin.

Anyway, for the original question: $A$ being bounded means there is $M>0$ such that $|x|\leq M$ for all $x\in A$. But the absolute value of $|x|$ is still $|x|$, so this implies $|A|$ is also bounded by $M$.

Answer 3

• The first two sentences are irrelevant descriptions. Try to ask yourself, what is the question asking, what do you have to do to answer the question, and execute the plan, the other irrelevant stuffs even if they are correct, should not be included.

• You have guess the answer correctly, but we want to look at the proof.

• You can't take an interval and claim that you have covered every single bounded set. (Hence you have not provided a proof at all).

• Now take an arbitary set $A$ that is bounded. What you should do is think of what does boundedness mean. Your textbook might use a different definition from the rest. Upon checking, a set is bounded if it has an upper bound and a lower bound. That is

$$\exists l_1, u_1 \in \mathbb{R}, \forall x \in A, l_1 \le x \le u_1$$

That's all you know about $A$.

• What do you have to show is $|A|$ is bounded. that is you have to show that

$$\exists l_2, u_2 \in \mathbb{R}, \forall y \in |A|, l_2 \le y \le u_2$$

Here are some claims that I will leave it to you to prove:

• We can pick $l_2 = 0$.
• We can pick $u_2=\max(|l_1|, |u_1|)$

By saying to "prove", I mean write it in general. Do not just verify using a few examples.