Let $S\subseteq \mathbb R$ be nonempty. Prove that a number $u\in \mathbb R$ satisfies $u=\sup S$ if for every $n\in \mathbb N$:
(i): $\Big(u-\dfrac{1}{n}\Big)$ is not an upper bound of $S$
(ii):$\Big(u+\dfrac{1}{n}\Big)$ is an upper bound of $S$
I know that this question has been asked before here but most of the answers use proof by contradiction whereas I require to proof without contradiction. So please help me regarding my approach.
My approach:
By condition (i), we have $\Big(u-\dfrac{1}{n}\Big)\in S$, which implies that for some $s\in S$, $\Big(u-\dfrac{1}{n}\Big)<s$
Let $\dfrac{1}{n}=\epsilon$
$\therefore$ For all $\epsilon>0$, There exists some $s\in S$ such that $(u-\epsilon)<s$
Now if I can prove that $u$ is an upper bound of $S$, I can imply from previous condition that $u=\sup S$. This is where I am stuck.
Please help me to prove that $u$ is an upper bound from (ii).
Also if you have any alternate approach, please provide that too.
Best Answer
For each $\epsilon >0$ there exist a $n$ such that $\epsilon >\frac{1}{n}$......(*)
By first condition, for each $\epsilon >0$there exist a $s$ such that $u-\epsilon < u- \frac{1}{n}<s$.
Also , from second condition $\sup S \leq u+ \frac{1}{n}$ for all $n$. So $\sup S $ is a lower bound for the set $A=\{u+\frac{1}{n} : n \in \mathbb{N} \}$. So, $\sup S \leq \inf A= u$. So $u$ is also an upper bound.
Hence , $u=\sup S$.