Understanding Analysis Exercise 1.4.2 Let $A \subseteq \mathbb{R}$ be nonempty and bounded above, and let $s \in \mathbb{R}$ have the property that for all $n \in \mathbb{N}$, $s + \frac{1}{n}$ is an upper bound for $A$ And $s - \frac{1}{n}$ is not an upper bound for $A$. Show that $s = \sup A$.
I have had a go myself. This is my attempt:
Consider $I_n = [s - \frac{1}{n}, s + \frac{1}{n}].$ Notice that we have that $I_1 \supseteq I_2 \supseteq I_3 \supseteq ...$ and so on. Thus, the Nested Interval Property says that $\bigcap_{n=1}^{\infty}I_n \neq \emptyset$ . I claim that $s \in \bigcap_{n=1}^\infty I_n$. Let $B = \{s - \frac{1}{n}: n \in \mathbb{N}\}$ and $C = \{s + \frac{1}{n}: n \in \mathbb{N}\}$. Note that $B \subseteq \mathbb{R}$ is nonempty and bounded above by $C$, so by the Axiom Of Completeness, $\exists \alpha \in \mathbb{R}$ such that $\alpha = \sup B$. We have that $s - \frac{1}{n} \leq s \leq s + \frac{1}{n}, \forall n \in \mathbb{N} \Rightarrow s \in I_n, \forall n \in \mathbb{N}$. As a result, $s \in \bigcap_{n=1}^\infty I_n$. Which then follows that $s = \alpha = \sup B$.
Earlier in the text, a proof was presented for the Nested Interval Property. Within the proof, it was shown that the infinite intersection is never empty because the supremum of the set of all lower endpoints is an element of the intersection.
In other words, if we had $I_n = [a_n, b_n]$ such that $I_1 \supseteq I_2 \supseteq I_3 \supseteq ...$ then we define $A = \{a_n : n \in \mathbb{N}\}$ where $s = \sup A$. We know that $a_n \leq s$ because $s$ is supposed to be an upper bound and $s \leq b_n$ because $s$ is the least upper bound. Thus, $s = \sup A \in \bigcap_{n=1}^\infty I_n \Rightarrow \bigcap_{n=1}^\infty I_n \neq \emptyset$. Do I need to clarify this in my attempt or would it be already clear?
Thanks!
 A: You are making life hard for yourself.
Since $s+{1 \over n} $ is an upper bound, you have $\sup A \le s+ {1 \over n}$ for all $n$ and hence $\sup A \le s$.
Since $s-{1 \over n}$ is not an upper bound we have $s-{1 \over n} \le \sup A$ for all $n$.
A: $s=\text{sup}A$ iff $s$ is a u.b. of $A$ and if $t<s$ then there exists $u$ in $A$ such that $u>t$.
Proving $s$ is u.b. of $A$
If $s$ is not u.b. then there exists $v$ in $A$ s.t. $v>s$. But I can find an $n$ (by A.P.) s.t. $\frac1n<v-s\implies s+\frac1n<v$ which contradicts one of our assumption.
Proving the second part
SPS $t<s$. Therefore by A.P. there exists an $n$ s.t. $\frac1n<s-t\implies t<s-\frac1n$. But $s-\frac1n$ is not u.b. (by our assumption) so there exists $u$ in $A$ s.t. $t<s-\frac1n<u$.
And we are done.
A: You write:

…then we define $A = \{a_n : n \in \mathbb{N}\}$ where $s = \sup A$.

This is bad.
You are supposed to work with the given $A$ and not redefine it.
If you make a statement about $A$, is it about the new $A$ or the old $A$?
And then you just declare the property you need!
If you want to refine your set into a more convenient one along the way, use a different notation.
Perhaps the first variation would be $A'$ and the second one $A''$.
If you take such a route, remember that you have to prove that $\sup(A)=\sup(A')=\sup(A'')$ or relevant inequality versions thereof.
In your current proof you have not provided any clear connection between the suprema of the original and the redefined set.
In other words, you never seem to give a proper justification for why the given $s$ is the supremum of the given $A$.
Your proof is not only overly complicated, but also seems to fail at proving the claim.
I would even argue that using too heavy tools is wrong as it gives the impression that such tools are needed for the proof; this can leave false impressions that impede future theory building.
You specifically requested feedback on your solution, so I did that.
The suggestion in the answer by copper.hat is an excellent one if you want a better proof.
