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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??

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

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