"Show that if a set $A$ of integers is bounded above, then $A$ has a largest element."
I've written the following proof. Is it valid?
Let $A$ be a nonempty set of integers which is bounded above. Then by the least-upper-bound axiom, the set $A$ has a supremum, which we shall denote by $\sup{A}$. By definition, $a≤\sup{A}$ for every $a\in A$. Consider the number $\sup{A}-1$. Since $\sup{A}-1<\sup{A}$, there exists an integer $m\in A$ such that $\sup{A}-1<m$. Thus $m≤\sup{A}<m+1$. Suppose that there is an integer $q\in A$ such that $m<q$. Then $q≥m+1>\sup{A}$, so that $q>\sup{A}≥q$, which implies $q>q$, a contradiction. Hence $m≥a$ for all $a\in A$. Thus $m$ is an upper bound for $A$. Therefore, we have shown that $A$ has a greatest element. Moreover, since $\sup{A}$ is the least upper bound, we must have $\sup{A}≤m$. But since $m\in A$, we also have $m≤\sup{A}$. Therefore, $m=\sup{A}$. $\blacklozenge$