Prove that $\sup(aS)=a\sup (S)$

Question

Let $S$ be a nonempty bounded set in $\mathbb{R}$ Let $a>0$ and let $aS=\{as: s\in \mathbb{R}\}$. Prove that $\sup(aS)=a\sup S$

I did my proof a different way then what the book did but I am not clear if my proof is still correct. The book proved $\sup(aS) \geq a\sup(S)$ and $\sup(aS) \leq a\sup(S)$. My proof goes like assume $a>0$ and set $S$ is bounded above by the the completeness axiom $\sup(S)$ exists. Let $x=\sup(S)$. Then $x \leq m$ for all $m\in S$. Thus $ax$ is an upper bound for set $aS$. By definition if we let $v$ be another upper bound for set $S$ then $x \leq v$. It follows that $ax \leq av$. Since $av$ is an arbitrary upper bound it follows $\sup(aS)=ax=a\sup(S)$. I just need to know if my proof is correct? If not I'm going to redo it the other way.

I just realised that even though the statement to prove is the same as the one on the link above, this doesn't really qualify as a duplicate. — Git Gud, Jun 08 '13 at 05:59

score 4 · Answer 1 · answered Jun 08 '13 at 05:56

You inadvertently got an inequality backwards: you mean that $x\ge m$ for all $m\in S$.

Your proof goes a bit astray after you show that $ax$ is an upper bound for $aS$, but the idea is sound. Let me rephrase the start of it.

$S$ is bounded, so let $x=\sup S$. Then $x\ge m$ for each $m\in S$, and $a>0$, so $ax\ge am$ for each $m\in S$. It follows that $ax$ is an upper bound for $aS$.

Up to here you’re fine, but then you go astray a little. You need to show that $ax$ is the least upper bound of $aS$, so you need to compare it with an arbitrary upper bound of $aS$. In other words, instead of starting with another upper bound $v$ of $S$, you should proceed as follows:

Now suppose that $v$ is any upper bound of $aS$. Then $v\ge am$ for each $m\in S$, and $a^{-1}>0$, so $a^{-1}v\ge m$ for each $m\in S$. But then $a^{-1}v$ is an upper bound for $S$, so ...

Can you finish it from here?

yes since $a^{-1}v$ is an upper bound for $S$ then $a^{-1}v \geq sup(S)=x$. It follows that $ v \geq ax=asup(S)$. It follows that $ax=sup(aS)=asup(S)$. — user60887, Jun 08 '13 at 16:42

score 1 · Answer 2 · answered Jun 08 '13 at 05:53

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Your proof that $ax$ is an upper bound for the set $aS$ is correct (but not very rigorous). The second half though is a bit lacking. What you have to do is show that if $t$ is any upper bound of $aS$, then $as\le t$. That is not quite what you showed. There is no need to redo the proof in the other way you mention, as it is quite easy to correct the little mistake and make the proof 100% good.

answered Jun 08 '13 at 05:53

Ittay Weiss

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oh ok i think ill use contradiction for that part. – user60887 Jun 08 '13 at 05:56

Prove that $\sup(aS)=a\sup (S)$

2 Answers2