Supremum and infimum of $\{ \frac{|x|}{1+|x|}:x\in \mathbb{R} \}$ and $\{ x+x^{-1}:0<x \leq 2 \}$

Question

I want to find out if these sets have a supremum and/or infimum:

$\{ \frac{|x|}{1+|x|}:x\in \mathbb{R} \}$

and

$\{ x+x^{-1}:0<x \leq 2 \}$

Regarding the first one I would say that all elements are non-negative and since $0 \in \mathbb{R}$ it follows that $0$ is minimum as well as infimum. But is that enough?

Regarding the second one we can rewrite $x^{-1}$ as $\frac{1}{x}$ and since x can't be $0$ we could use $2$ and get $2,5$ as a supremum. Is that correct/and or enough reasoning?

Answer 1

You are right about infimum of the first set. The supremum of this set is $1$ because every element of the set is $<1$ and taking $x=1,2,3,..$ and letting $x \to \infty$ we get $1$ in the limit. This supremum is not a maximuum.

For the second set your answer is not correct. We can see that $x+\frac 1x \geq 2$ for all $x>0$ using the inequality $(x-1)^{2} \geq 0$. Hence $2$ is a lower bound for this set. The value $2$ is attained when $x=1$ so the infimum and the minimum are $2$. The set is not bounded above: take $x =\frac 1 n$ to see this.

Answer 2

1. Absolutely. More than that, for any set if it has a minimum, it is also infimum.

2. Not correct. For small x > 0 the value can be any large, there is no supremmum then. On the other hand it has also minimum, so it has infimum.

Answer 3

If $S_1$ is your first set, then $0=\min S_1$ because $0\in S_1$ (you did not state this) and because $(\forall s\in S_1):0\leqslant s$. On the other hand, every element of $S_1$ is smallar than $1$ and therefore $1$ is an upper bound of $S_1$. In fact, $1=\sup S_1$ (and $S_1$ has no maximum) since $\lim_{x\to\infty}\frac{\lvert x\rvert}{1+\lvert x\rvert}=1$.

And if $S_2$ is your second set, then $S_2$ has no upper bounds and therefore it has supremum nor maximum. On the other hand, $\min S_2=2$. Can you justify it?

Answer 4

for $$A=\{ \frac{|x|}{1+|x|}:x\in \mathbb{R} \}$$

We have $\inf (A) = 0$ and $\sup (A) =1$

For $$B= \{ x+x^{-1}:0<x \leq 2 \}$$

$\inf (B) = 2$ and $\sup (B)$ does not exist because it is not bounded above.