let $A\subseteq \mathbb{R}$ be a non-empty set and $s\in \mathbb{R}$ and upper bound of $A$.
So $s$ is the supremum of $A$ $\iff$ $\forall \epsilon>0$ there is $x\in A$ so $s-\epsilon<x\leq s$.
Prove that $s\in \mathbb{R}$ upper bound of $A$ is a needed condition:
If I say that $A=\mathbb{N}$ so $s-\epsilon<x\leq s$ holds but $s\in \mathbb{R}$ is a upper bound of $A$ don't?
let $A⊆ℝ$ be a non-empty set and $s∈ℝ$. The condition $∀ϵ>0$ there is $x∈A$ so $s−ϵ<x≤s$ is equivalent to that there is a monotone increasing sequence contained by $A$ which tends to $s$. For every $a\in A$ the constant sequence $a,a,\ldots$ is such a sequence. One can prove, that the supremum of a set is unique, so if $A$ has more than one element, the condition which states $s$ is an upper bound is required.