Proof verification of $a\sup S=\sup(aS)$ I'm trying to show that $a\sup S=\sup aS$ where $a>0$ and $aS:=\{as:s\in S\}$. 
My attempt:
Let $u$ be an arbitary upper bound for the set $S$ then by definition we have that $\forall s\in S\Rightarrow s\leq u$ so $as\leq au$ so $au$ is clearly an upper bound for the set $aS$. To show that this is the supremum of $aS$ by the archimedean property we have that $\forall s\in S$ and $\varepsilon_0>0\Rightarrow |s-\sup S|<\varepsilon_0$, multiplying both sides by $a>0$ and taking $\varepsilon_0=\frac{\varepsilon}{a}>0$ we obtain $$a|s-\sup S|=|as-a\sup S|<a\cdot\varepsilon_0=a\cdot\frac{\varepsilon}{a}=\varepsilon$$
This shows that $a\sup S=\sup(aS)$

There is another way to do this. 
Let $s \in S$, then by the definition of the supremum, we have $s \leq \sup S$.
 But this is same as saying $$as \leq a\sup S, \; a > 0.$$
Now $as \in aS$, so by the definition of $\sup$, we get $as \leq \sup(aS) \leq a\sup S$. 
Conversely, for $as\in aS$ we have $as \leq \sup(aS)$; dividing both sides by $a$ and repeating the argument as above gives the other inequality.
