Prove that $\sup \{ \varepsilon x:\, x\in A\}=\varepsilon \sup A$ Since this is an introductory course to real analysis I am looking for the most simple and direct proof. Based mostly on definitions. 
Let $\varepsilon$ be a positive real number. If $A$ is a non-empty bounded subset of $\mathbb{R}$ and $B = \{\varepsilon x : x \in A\}$ then $\sup(B) = \varepsilon \sup(A)$.
Since $A$ is a non-empty subset if follows that $\sup(A)$ exists as a finite supremum. Which by definition means that $x \leq \sup(A)$ for all $x\in A$ and by the second multiplicative property since $\varepsilon$ is positive it follows that $\varepsilon x \leq \varepsilon \sup(A)$ which means that $\sup(B) \leq \varepsilon \sup(A)$.
The hard part is to show that $\sup(B) \geq \varepsilon \sup(A)$ I can't think of a way to go about that.
So maybe a one shot deal... it's true always that $$\varepsilon x = \varepsilon x \space \space  \forall x \in A, \space \varepsilon \gt 0 \in \mathbb{R}$$
It's also true by algebraic manipulation and associativity that $(\varepsilon x )= ((\varepsilon)(x))$ by associativity. So it follows that $\sup(B) = ((\varepsilon) (\sup(A))$
To me I don't see any issue with showing $\varepsilon x = \varepsilon x$ can lead to the conclusion of the statement.  

If $S$ is a subset of $\mathbb{R}$ with a minimum, $m$, then $\epsilon m \leqslant \epsilon x$ for all $x \in S$, i.e. the set $\epsilon S$ has the minimum $\epsilon m$.
If $S$ is the set of all upper bounds of $A$, then $\epsilon S$ is the set of all upper bounds of $\epsilon A$, and the result follows.
