Prove that given $A\subseteq\mathbb{R}$, if has a maximum $a_0$, then $a_0$ must be the supremum.
In order to prove this, I have written the following:
Let $A\subseteq \mathbb{R}$, nonempty set because $a_{0}$ $\in A$ for hypotesis. Then, since $a_{0}$ $\in A$ is the largest element of the set $A$ (i.e. for each element $A$ of A we have that $ a\leq a_{0}$) $A$ is bounded above. Then, A has a supremum. Let $s$ be the least upper bound of $A,$ so we have that for each element $a$ of $A$, $a\leq s$, for being $s$ an upper bound. In particular, $a_0\leq s$, which implies that $s=a_0$.
Then, we have that $\sup(A)=\max(A)=a_0$, which is exactly what we wanted to prove.
Observation: this proof is true if $s \in A$
I'm not sure if my proof is correct. I will be thankful if someone tells me.
Best Answer
To be frank this result is typical of many theorems in analysis which are as trivial as they can be and your approach is unnecessarily complicated and based on the unfounded assumption that things in analysis are in general non-trivial.
The result is an immediate consequence of the definitions of maximum and supremum. Let $a_{0}$ be the maximum of set $A$. Then by definition of maximum we have $a\leq a_{0}$ for all $a\in A$. Moreover if we take any number $b<a_{0}$ then it's just repeating the same thing that we have $a_{0}\in A$ and $a_{0}>b$. Thus $a_{0}$ is the supremum of $A$.
There is no need to show that $A$ is bounded above and invoke deep completeness property to justify the existence of $\sup A$. You can see that the above argument (and therefore the result in question) holds in any ordered field (eg $\mathbb{Q}$).
It is rather surprising that other answers here also make use of the completeness property unnecessarily.