Find supremum(S), infimum(S), max(s), min(S) I have little trouble with the following examples of sets to find its minimum, maximum, supremum and infimum. Sometimes it gets tricky if the example is not trivial. I know that here are many examples like that but I want an answer for those two.  For this examples I will show my way of thinking, correct me please:


*

*$ x =  \big\{x:x=\frac{n-1}{n},n \in N^+\big\}  $

*$S=[0,1)\in R $
I'm using the following definitions that supremum is the smallest upper bound $x \in R$, and s is the element of a set S , 
$\forall \epsilon >0 , \exists s \in S:s>x-\epsilon ,  $
respectively the infimum is the greatest lower bound 
$\forall \epsilon >0 , \exists s \in S:s<x+\epsilon ,  $


*

*for $ n=1,2,3,4... , x=0,\frac{1}{2}, \frac{2}{3}, \frac{3}{4}... $ the values of x will never reach 1


So from my point of view, $min(S)=0$ and $ Inf (S)=0$ but $max(S)=\nexists $  and I think that $sup(S)$ does not exist, why I think that?
From the definition, let us pick a number $s $ that would be greater than least upper bound minus a very small number but we don't know what is the smallest upper bound because the set is infinite 
2.$min(S)=0$ and $max(S)=\nexists$
$Inf (S)=0$ but what with $Sup(S)=?$
If we pick a number $s=1 \notin S $ that for sure $1>x-\epsilon $ but the element s is not in the set so how to prove that for example $0.9999999>1-0.0000001$ , easily they gonna be equal but is it possible the left-hand side of the inequality to be grater or simply $Sup(S)=\nexists$
 A: The statement "Every non-empty set $S \subseteq \mathbb{R}$ which is bounded from above has a least upper bound" is often stated as an axiom when defining the set $\mathbb{R}$ of real numbers. This supremum property is equivalent to completeness, i.e., every Cauchy sequence is convergent. Regardless, once this axiom is accepted, then every non-empty set $S \subseteq \mathbb{R}$ which is bounded from above has a least upper bound.
Turning to the task at hand. With $S = \{ 1 - \frac{1}{n} \: : \: n \in \mathbb{N} \}$ it is clear that $r= 1$ is an upper bound. We claim that $r=1$ is the least upper bound. To that end, let $\epsilon > 0$ be given. We must show that $t = 1 -\epsilon$ is not an upper bound. This means finding $n \in \mathbb{N}$ such that such that $1 - \epsilon < 1 - \frac{1}{n}$ or equivalently $\epsilon^{-1} < n$. This is easily done. We can choose $n = \lceil \epsilon^{-1} \rceil + 1$. In short, $r=1$ is an upper bound for $S$ and any number strictly smaller than $r=1$ is not an upper bound for $S$. It follows that $r=1$ is the least upper bound for $S$, i.e., $\sup S = 1$.
