Proving that $k\cdot \sup S = \sup(kS)$ for $k\geq 0$

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, $k \cdot \sup S = \sup(kS)$.

• It looks good for me. – SomeOne Oct 7 '15 at 22:54
• So there's no problem with the last two inequalities in the proof? The lemma that I'm invoking there is the theorem that states that, Let $x,y \in \mathbb{R}$ such that $x \leq y+ \epsilon$ for every $\epsilon > 0$. Then $x \leq y$. My real concern is with the $k\cdot \epsilon$: Can the lemma be applied to infer $k\cdot \sup S \leq \sup(kS)$ from $k \cdot \sup S \leq \sup(kS) + k\cdot \epsilon$ ? – Jeremiah Dunivin Oct 7 '15 at 22:58
• The condition in the lemma that you used was for every $\epsilon > 0$. We have $\epsilon > 0$ goes through each positive value does that imply $k \epsilon$ go through each positive value given that $k >0$? – SomeOne Oct 7 '15 at 23:04
• hm, good point! That's very obvious now that you've mentioned it. Thank you! – Jeremiah Dunivin Oct 7 '15 at 23:06
• You are welcome. I think there is a minor edit you should have a special case $k = 0$. – SomeOne Oct 7 '15 at 23:08

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 $0= 0 \sup\{S\} = \sup \{0\} = 0$.