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??
Sorry I can't follow your argument.
What you need to do is;
1 Define the set T as the image of S after multiplying each point by eps.
2 Show that if B is an upper bound of S, then C=epsB is an upper bound of T.
3 Show that T is non-empty and has an upper bound, so supT exists.
4 Show that supT = C, for in supT < C, then there would be an element y in T which corresponds to some element x in S such that x=y/eps < B, so we could find another value between y/eps and B which maps into T but then would be bigger than C. Hence C=supT.
Hope that helps.