# 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 $k \cdot \sup S \leq \sup(kS) + k\cdot \epsilon$ ? – Jay 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! – Jay 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 $0= 0 \sup\{S\} = \sup \{0\} = 0$.

In order to prove $$k \cdot \sup S \leq \sup(kS)$$, I thought that $$\sup(kS)$$ is the smaller upper bound of $$kS$$ thus $$\sup(kS)\geq ks,\forall s\in S\Rightarrow \frac{\sup(kS)}{k}\geq s,\forall s\in S$$, since $$k\geq 0$$. Hence $$\frac{\sup(kS)}{k}$$ is an upper bound of $$S$$, then $$\sup(S)\leq \frac{\sup(kS)}{k}\Rightarrow k\cdot\sup(S)\leq \sup(kS)$$. And that's what we want to prove. Greeting from Uruguay!

• If you can speak English, please write in English. – Micah Windsor Apr 29 '20 at 22:43
• Also this question was asked and answered almost five years ago, I'm pretty sure he's good without your input. – Micah Windsor Apr 29 '20 at 22:52
• @MicahWindsor It's my understanding that answers here are for the benefit of anyone who searches for this question in the future, not just for the original asker. There is definitely limited utility to answering an older question, but this ultimately has less to do with the original asker and more to do with visibility in internet searches, – Brian Moehring Apr 30 '20 at 0:34
• @BrianMoehring The accepted answer is much neater though – Micah Windsor Apr 30 '20 at 0:37
• @MicahWindsor I agree there, but the accepted answer doesn't utilize the identity of sets $S = \frac{1}{k}(kS)$ which this answer uses to realize the second half of the proof as an immediate consequence of the first half (at least for $k>0$, Javier's answer has an error for $k=0$, but this would be easily remedied). If we were to completely clean up both answers, Javier's would be the one I would have been looking for back when I was learning this topic. – Brian Moehring Apr 30 '20 at 0:48