[proofplan] Let $m = \inf A$ and let $B = \{\lambda a : a \in A\}$. Because multiplication by the negative number $\lambda$ reverses inequalities, the lower-bound property $m \le a$ for $A$ becomes the upper-bound property $\lambda a \le \lambda m$ for $B$. To prove leastness, take an arbitrary upper bound $u$ of $B$, divide by $\lambda < 0$ to obtain a lower bound $u/\lambda$ of $A$, and then use the greatest-lower-bound property of $m$. [/proofplan]

[step:Define the scaled set and prove it is nonempty and bounded above]Let $m = \inf A$, and define the scaled set \begin{align*} B = \{\lambda a : a \in A\} \subset \mathbb{R}. \end{align*} Since $A$ is nonempty, choose $a_0 \in A$. Then $\lambda a_0 \in B$, so $B$ is nonempty. Since $m = \inf A$, the number $m$ is a lower bound for $A$. Thus, for every $a \in A$, \begin{align*} m \le a. \end{align*} Multiplying this inequality by $\lambda < 0$ reverses the inequality, giving \begin{align*} \lambda a \le \lambda m. \end{align*} Therefore every element of $B$ is at most $\lambda m$, so $\lambda m$ is an upper bound for $B$. Hence $B$ is nonempty and bounded above.[/step]

[guided]We introduce two pieces of notation to separate the order argument from the set-builder notation. Let \begin{align*} m = \inf A, \end{align*} and define \begin{align*} B = \{\lambda a : a \in A\} \subset \mathbb{R}. \end{align*} First we verify that $B$ is a set for which a supremum can be discussed. Since $A$ is nonempty, there exists $a_0 \in A$. By the definition of $B$, the number $\lambda a_0$ belongs to $B$. Hence $B$ is nonempty. Next we show that $B$ is bounded above. Because $m = \inf A$, the number $m$ is a lower bound for $A$. This means that for every $a \in A$, \begin{align*} m \le a. \end{align*} The scalar $\lambda$ is negative. Multiplication by a negative real number reverses the order, so the preceding inequality gives \begin{align*} \lambda a \le \lambda m. \end{align*} This holds for every $a \in A$. Equivalently, every element of the form $\lambda a$ with $a \in A$ is bounded above by $\lambda m$. Since those are exactly the elements of $B$, the number $\lambda m$ is an upper bound for $B$. Thus $B$ is nonempty and bounded above.[/guided]

[step:Show that every upper bound of the scaled set lies above $\lambda m$]Let $u \in \mathbb{R}$ be an arbitrary upper bound for $B$. Then for every $a \in A$, the element $\lambda a$ belongs to $B$, so \begin{align*} \lambda a \le u. \end{align*} Dividing by $\lambda < 0$ reverses the inequality, and therefore \begin{align*} a \ge \frac{u}{\lambda} \end{align*} for every $a \in A$. Hence $u/\lambda$ is a lower bound for $A$. Since $m = \inf A$ is the greatest lower bound of $A$, every lower bound of $A$ is at most $m$. Applying this to the lower bound $u/\lambda$, we obtain \begin{align*} \frac{u}{\lambda} \le m. \end{align*} Multiplying again by $\lambda < 0$ reverses the inequality, giving \begin{align*} u \ge \lambda m. \end{align*} Thus every upper bound $u$ of $B$ satisfies $\lambda m \le u$.[/step]

[guided]To prove that $\lambda m$ is the least upper bound of $B$, it is not enough to know that $\lambda m$ is an upper bound. We must also prove that no smaller number can be an upper bound. So let $u \in \mathbb{R}$ be an arbitrary upper bound for $B$. The statement that $u$ is an upper bound for $B$ means that every element of $B$ is at most $u$. For each $a \in A$, the number $\lambda a$ belongs to $B$, and therefore \begin{align*} \lambda a \le u. \end{align*} Now we want to translate this back into a statement about the original set $A$. Since $\lambda < 0$, dividing by $\lambda$ reverses the inequality. Hence, for every $a \in A$, \begin{align*} a \ge \frac{u}{\lambda}. \end{align*} This says exactly that $u/\lambda$ is a lower bound for $A$. Because $m = \inf A$, the number $m$ is the greatest lower bound of $A$. Thus every lower bound of $A$ is less than or equal to $m$. Applying this to the lower bound $u/\lambda$, we get \begin{align*} \frac{u}{\lambda} \le m. \end{align*} Multiplying this inequality by the negative number $\lambda$ reverses the inequality once more, so \begin{align*} u \ge \lambda m. \end{align*} Therefore every upper bound $u$ of $B$ lies above $\lambda m$.[/guided]

[step:Conclude that $\lambda m$ is the supremum of the scaled set] From the first step, $\lambda m$ is an upper bound for $B$. From the second step, every upper bound $u$ of $B$ satisfies $\lambda m \le u$. Therefore $\lambda m$ is the least upper bound of $B$. Hence \begin{align*} \sup B = \lambda m. \end{align*} Substituting $B = \{\lambda a : a \in A\}$ and $m = \inf A$ gives \begin{align*} \sup\{\lambda a : a \in A\} = \lambda \inf A. \end{align*} This proves the theorem. [/step]