Is it always true that the maximum of a closed bounded set, $[a,b]$, is also the supremum

Question

This seems true and I've seen some tangential proofs in my real analysis class but I wasn't 100% sure. I think this is the case because there can be no bound smaller than the maximum.

Answer 1

An even more general fact is true and helpful here:

If $A$ is a subset of $\mathbb R$, then if $u$ is an upper bound of $A$ and $u$ is an element of $A$, then $u$ is the supremum of $A$.

This fact can easily be proven using the definition of supremum. Remember, for a bounded set $A$, the supremum of $A$ is the smallest lower bound for $A$. Therefore, to show $s$ is the supremum, we must show that (1) $s$ is an upper bound of $A$ and (2) any value smaller than $s$ is not an upper bound of $A$.

Well, apply this to our case.

1. $u$ is an upper bound of $A$ by definition.
2. If $v<u$, then $v$ is not the upper bound of $A$, because $u\in A$ and $u>v$.

Therefore, $u$ is the supremum of $A$.