I found an interesting problem, but I couldn't find someone who solved it like I did, so I am asking if I've gone wrong somewhere.
Let $S \subseteq R$ be nonempty. Prove that if a number $u$ in $R$ has the properties:
(i) for every $n \in N$ the number $u - \frac{1}{n} $ is not an upper bound of S, and
(ii) for every number $n \in N$ the number $u + \frac{1}{n}$ In is an upper bound of S,
then u = sup S.
Now notice that this is an implication, $(i) \land(ii) \implies u =$ sup S. So I used the contrapositive to prove it:
$ u \neq$ sup S $ \implies \lnot(i) \lor \lnot(ii)$
This is what I did:
Notice that (i) is the same as saying: for every $n \in N$, there exits an $s \in S$ such that $ u - \frac{1}{n} < s$.
The negation of (i) is : there exits a natural number $n$ such that for all $s \in S$ we have $u-\frac{1}{n} \geq s $.
Simply solve the inequality $u-\frac{1}{n} \geq $ sup S for $n$, we get $ n \geq \frac{1}{u-sup S}$.
And since $ u \neq$ sup S the donaminator is not zero. The Archimedian property garantees the existence of this $n$.
As for (ii), the negation is: there exits a natural $n$ and an $s \in S$ such that $ u + \frac{1}{n} \leq s$ .
Since $u$ is not the upper bound, there exists an $s' \in S$ such that $s' > u$. So we solve the inequality $ u + \frac{1}{n} \leq s'$ for $n$, we obtain $n \geq \frac{1}{s'-u}$.
By the Archimedean property such $n$ exits.
And I am done with the proof.