why only $\sup(S) $ can satisfy $M - \epsilon <= x$

Question

in real analysis, for an elment to be the superior of the set it have to be:

1. bigger than all other elements (for every element $x$ in set, $x \le M$)
2. for any positive epsilon $\epsilon > 0$, there exist an elment of the set that is bigger than M, $M - \epsilon \le x$

the first condition of being an upper bound is totally understandable, but the second one, i mean any upper bound can satisfy it, the all the upper bounds are $\sup(S) $

Answer 1

Often the definition of the least upper bound is given as follows. Let $S$ be a set bounded above. We say $\sup S$ is the least upper bound of $S$ if:

1. $\sup S$ is an upper bound for $S$;
2. If $\ell$ is an upper bound for $S$, then $\ell\geq\sup S$.

From this, we can prove that for any $\varepsilon>0$, there exists $s\in S$ such that $\sup S-\varepsilon<s$. Suppose there exists some $\varepsilon>0$ such that for all $s\in S$, $\sup S-\varepsilon\geq s$. Then $\sup S-\varepsilon$ must also be an upper bound for $S$, since $s$ is an arbitrary element of $S$. But this implies that $\sup S$ is not the least upper bound of $S$, since $\sup S-\varepsilon<\sup S$ for all $\varepsilon>0$, which is a contradiction.

Now try to prove that if $\ell$ is an upper bound for $S$ such that for all $\varepsilon>0$, there exists $s\in S$ such that $\ell-\varepsilon<s$, then $\ell=\sup s$, and you'll see that only $\sup S$ can satisfy the property given above.