# Proof of $\sup \epsilon x = \epsilon \sup x$

Suppose S is a non empty set of real numbers, and suppose S is bounded above, and that $\epsilon > 0$, Prove $\sup \epsilon x$ = $\epsilon\sup x$

My take so far:

$sup S = B$, then $B$ is an upper bound of $S$, $x \le B$ Let $T = {x:x \in S}$ and since $\epsilon > 0, \epsilon x \le \epsilon B$. Thus $T$ is bounded above by $\epsilon B$ and $\sup T = C$. Now we have to show that $C = \epsilon B$. Since $\epsilon B$ is an upper bound for $T$ and $C$ is $\sup=T$, $C \le \epsilon B$

and from here, I can't figure it out how to prove $\epsilon B \le C$? Because I'm assuming I have to show that $C = \epsilon B$ in order to finish this proof. Am I correct??

• Use a \ before sup so that it renders as a command as so: $\sup$. – Cameron Williams Jan 28 '16 at 3:16
• Sorry I can't follow you argument. What you need to do is : – user247608 Jan 28 '16 at 4:13
• I think you meant $T=\{\varepsilon x: x \in S\}$. – B. Freitas Jan 28 '16 at 4:40