[proofplan] We prove the result by unpacking the definitions. A maximum is already an element of the set and dominates every element of the set, so it is an upper bound. To prove it is the least upper bound, we compare it with an arbitrary upper bound $u$; since $m \in A$, the upper-bound property of $u$ forces $m \le u$. [/proofplan] [step:Use maximal membership and domination to make $m$ an upper bound] Since $m$ is a maximum of $A$, by definition $m \in A$ and for every $a \in A$ one has $a \le m$. Therefore $m$ is an upper bound of $A$. [guided] We start by recording exactly what the word maximum contributes. The hypothesis that $m$ is a maximum of $A$ has two parts: first, $m$ is itself an element of $A$; second, every element of $A$ is below $m$ in the order $\le$. Thus \begin{align*} m \in A \end{align*} and, for every $a \in A$, \begin{align*} a \le m. \end{align*} The second displayed condition is precisely the definition that $m$ is an upper bound of $A$. The membership condition will be used in the next step, where we compare $m$ with an arbitrary upper bound. [/guided] [/step] [step:Compare $m$ with an arbitrary upper bound] Let $u \in P$ be an arbitrary upper bound of $A$. By definition of upper bound, every element of $A$ is less than or equal to $u$. Since $m \in A$, applying this property to the element $m$ gives \begin{align*} m \le u. \end{align*} Thus $m$ is less than or equal to every upper bound of $A$. [/step] [step:Conclude that $m$ is the least upper bound] We have shown that $m$ is an upper bound of $A$, and that for every upper bound $u \in P$ of $A$ one has $m \le u$. This is exactly the definition that $m$ is a supremum of $A$. Therefore $m$ is a supremum of $A$. [/step]