Let $S$ be a nonempty bounded subset of $\mathbb{R}$. Define the set $kS = \{ks : s \in S\}$. We wish to prove that if $k \geq 0$, then $\sup(kS) = k\cdot \sup S$.
I'm pretty sure the upper half of the proof is fine, but it's the lower half that attempts to show that $k\cdot \sup S \leq \sup(kS)$ with which I am concerned. I begin the lower half by, "However, the following argument(?)…".
Proof
Let $k \geq 0$ be an arbitrary constant. Since $S$ is bounded above, the completeness axiom entails the existence of the least upper bound $\sup S$. Hence, the following inequality is readily established for all $s \in S$
$$s \leq \sup S$$
Since $k \geq 0$ we can multiply the above inequality by $k$
$$ks \leq k\cdot \sup S$$
So the set $kS$ is bounded above. Further, we know that $kS \neq \emptyset$, because if we choose any $s \in S$ where $S \neq \emptyset$, then $ks \in kS$ by definition of $kS$. Hence, $kS$ has the least upper bound $\sup(kS)$.
The second inequality shows that $k \cdot \sup S$ is an upper bound of $kS$, so we must have the inequality
$$ \sup(kS) \leq k \cdot \sup S$$
(since $\sup(kS)$ is the smallest upper bound of $kS$ and $k \cdot \sup S$ is an upper bound of $kS$).
However, the following argument(?) shows that $k \cdot \sup S \leq \sup(kS)$, thereby establishing the fact that $\sup(kS) = k \cdot \sup S$.
Since $\sup S$ is the least upper bound of $S$, the number $\sup S – \epsilon$ for $\epsilon > 0$ is not an upper bound of $S$. Therefore, there exists a number $s' \in S$ such that
$$\sup S – \epsilon < s'$$
Since $k\geq 0$, we can multiply the above inequality by $k$ to construct the following inequality
$$k \cdot \sup S – k \cdot \epsilon \leq k \cdot s'$$
Due to the fact that, for every $k\cdot s \in kS$, $k\cdot s \leq \sup kS$, it follows that $k \cdot s' \leq \sup kS$. Thus by transitivity, we conclude that
$$k \cdot \sup S – k \cdot \epsilon \leq \sup(kS)$$
Adding $k\cdot \epsilon$ to both sides of the above inequality, we have
$$k \cdot \sup S \leq \sup(kS) + k\cdot \epsilon$$
Because the above inequality is true for every $\epsilon > 0$, we infer
$$k \cdot \sup S \leq \sup(kS)$$
as desired.
Therefore, $k \cdot \sup S = \sup(kS)$.
Best Answer
Assume $k> 0$. we have from the definition of $\sup$, there exists $x\in S$ such that $$\sup (S) -x \leq \frac{\epsilon}{k}$$ $$k \sup (S) - k x \leq \epsilon$$ therefore $$k \sup (S) \leq k x + \epsilon \leq \sup (k S) + \epsilon.$$
For the other inequality, again from the definition of $\sup$, there exists $kx\in kS$ such that $$\sup (kS) - kx \leq \epsilon$$ $$\sup (kS) \leq kx + \epsilon$$ therefore $$\sup(kS) \leq kx + \epsilon \leq k\sup(S) + \epsilon.$$
For $k=0$, we have $0= 0 \sup\{S\} = \sup \{0\} = 0$.