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.
Best Answer
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.
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:
Can you finish it from here?