Analysis proof involving bounded intervals This proof is somewhat similar to the question asked here: Help with real analysis proof involving supremum. But it's the other way around.
Question: Let $S$ be a non-empty subset of the real numbers, bounded above. Show that if $u = sup S$, then for every natural number $n$, the number $u − 1/n$ is not an upper bound of $S$, but the number $u + 1/n$ is an upper bound of S.
Here is my approach.
First, I write the conditions for my proof to satisfy. They are, for every natural number $n$ the following two must satisfy,
$$
(I)\hspace{1cm} s\leq u+1/n,\forall s\in S \\
(II)\hspace{1cm}\exists s \in S \ni s\geq u-1/n
$$
Now we prove this by induction. First, we have to establish the base case for $n=1$.
We can easily show that, when $n=1$
$u+1/n = u+1$
Since $u+1>u$, it is clearly an upper bound for $S$. (I) holds.
and
$u-1/n = u-1$. 
since $u-1 < u$, it is clearly not an upper bound for $S$. Which means (II) holds.
$\\\\\\$
Now let's assume that the conditions (I) and (II) are true for some integer $m$.
i.e:
$ s\leq u+1/m,\forall s\in S$ and
$\exists s \in S \ni s\geq u-1/m$
but the case $m+1$ implies, $u+1/m > u+1/(m+1)$
It can be proven using the following process.
$u+1/m > u+1/(m+1) \Leftrightarrow 1/m > 1/(m+1) \Leftrightarrow m+1 > m \Leftrightarrow 1>0 $
Now the problem is this does not guarantee $ s\leq u+1/(m+1),\forall s\in S$.
Note that $u+1/(m+1)$ fall between $u$ and $u+1/m$.
The same is true for $u-1/(m+1)$ situation. 
Any insights into this would be appreciated. 
 A: *

*Let $T$ be the set of all upper bounds for $S.$ By definition, $u=\sup S$ is the least member of $T.$ If $n\in \Bbb N$ then $u-1/n<u.$ So if $u-1/n$ were an upper bound for $S$ then $u-1/n$ would be a member of $T$ that's less than the least member ($u$) of $T$, which is absurd.

*By definition of $T,$  we have, for any $v,v',$ that $$v'>v\in T \implies \forall s\in S\,(v'>v\ge s)$$ $$ \implies \forall s\in S\,(v'\ge s)$$ $$\implies v'\in T.$$ In particular $n\in \Bbb N\implies u+1/n>u\in T\implies u+1/n\in T.$
A: Note:  $u - \frac 1n < u = \sup S$.  And note that $u + \frac 1n > u = \sup S$.
That is IT!  You are done.
The definintion of $u = \sup S$ is:


*

*$u\ge x$ for all $x \in S$.


So $u+ \frac 1n > u \ge x$ for all $x \in S$.  So $u+\frac 1n$ is an upper bound.  That's all there is to it.


*If $w < u$ then $w$ is not an upper bound of $S$.


So $u-\frac 1n < u$.  So $u-\frac 1n$ is not an upper bound of $S$.   End of story.
A: Supremum means the least upper bound. You are given that u is the least upper bound of the set S. $u+1/n>u$ for all natural numbers $n$ since $1/n>0$. So, it is greater than the least upper bound, meaning it is an upper bound.
Similarly, $u-1/n<u$ since $1/n<0$ for all natural numbers $n$. Since it is smaller than the least upper bound, it is not an upper bound.
