Can't finish this proof about supremums :(

Question

Let $S$ be a nonempty bounded subset of $\Bbb R$ and let $k \in \Bbb R$. Define $kS = \{ks:s \in S\}$. Prove: If $k \ge 0$, then $sup(kS) = k \cdot supS$.

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Okay, here's my attempt at the proof:

Suppose $k \ge 0$

Case I

If $k = 0$, then $kS = \{0\}$

This implies that $sup(kS) = 0$

Since $k=0$, $k \cdot supS = 0$

So $k \cdot supS = sup(kS)$.

Case II

If $k = 1$, then $kS = \{s:s \in S\} = S$

Since $ k = 1$, $k \cdot supS = sup S$

So, $sup(kS) = supS = k \cdot supS$.

Case III

If $k \gt 1$ …

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And this is where I got stuck.

Answer 1

Let $k>0$.

1. If $x \in kS$, then $x=ks$ for some $s \in S$, hence $x \le k \sup S.$ This gives $\sup (kS) \le k \sup S$.

2. Let $c:= \sup(kS)$ and take $s \in S$.Then $ks \le c$, hence $s \le \frac{c}{k}$. This shows that $sup S \le \frac{c}{k}$, therfore $k \sup S \le c = \sup(kS).$

Answer 2

$k>0;$

1) $a \le \sup A$ , then $ka \le k(\sup A)$, i.e.

$k (\sup A)$ is an upper bound for $kA$,

$\sup (kA) \le k \sup (A)$, since $\sup (kA)$ is the least upper bound for $kA$.

2) $a= (1/k)(ka)\le (1/k)\sup (kA)$.

$(1/k) \sup (kA)$ is an upper bound for $A$,

$\sup (A) \le (1/k)\sup(kA)$, since $\sup (A)$ is the least upper bound for $A$.

Hence $k \sup (A) \le \sup (kA)$.