# Constant times Supremum

I'm trying to show that $\sup \{ rx \ | \ x \in S \} = r \sup S$ where $S$ is a set of real numbers and $r$ is a real number greater than 0.

I started by letting $b = \sup S$. That would imply that $b \geq x$ for all $x \in S$ and $b \leq y$ for any upper bound $y$.

Since $b \geq x$, that implies $rb \geq rx$. So I've shown that $rb$ is an upper bound of $\{ rx \ | \ x \in S \}$.

Edit: show that $rb$ is the least upper bound of $\{ rx \ | \ x \in S \}$.

Since $b \leq y$, then $rb \leq ry$. This means that $rb$ is less than or equal to any upper bound of $\{rx \ | \ x \in S \}$. So $rb$ must be the least upper bound.

-
This may not be much of a hint, but: Start by writing out exactly what it means to satisfy the definition in terms of the new set $rS$. Suppose $y$ is an upper bound for $rS$... – Jonas Meyer Mar 14 '12 at 5:12
+1 for showing what you've done, but you're almost there. I think the best advice we could give you right now is to keep thinking about it. I would feel bad to give you the answer right now ; you're really not that far away from it. – Patrick Da Silva Mar 14 '12 at 5:15
@JonasMeyer: Could you take a look at my edit? – user26139 Mar 14 '12 at 5:30
@user26139: The edited in solution is not correct. You need to go roughly in the opposite direction, starting with an upper bound for $\{rx|x\in S\}$. I.e., suppose that $y$ is an upper bound for $\{rx|x\in S\}$. Now use this to obtain an upper bound for $S$ and use $b=\sup S$, etc. – Jonas Meyer Mar 14 '12 at 5:34
@JonasMeyer: I see that since $b \leq y$ and $b \geq x$, then $y \geq x$. So $y$ is an upper bound of $S$. But I don't see how that helps. – user26139 Mar 14 '12 at 5:56

To prove that, $t$ is the supremum of a set $S$, you'll have to verify that

• For all $s \in S$, $s \le t$.
• For any upper bound $u$ of $S$, $t \le u$

Hint:

Show that $r \sup S$ satisfies the definition of the $\sup rS$ and observe that $\sup A$ is unique for any set $A$.

Let $x \in rS$. $\quad \bullet$ Note that $x=rs$ for some $s \in S$. Since for all $s \in S$, $s \le b=\sup S$, $sr \le br$ for all $s \in S$. Thus for any $x \in rS$,we must have, $x \le br$. $\quad \bullet$ Let $u$ be an upper bound for $rS$. Then, $x \le u$. This means, $s \le \dfrac u r$. Since $b=\sup S$ is the least upper bound, we must have that, $\dfrac u r \ge b$. So, you have that, $br \le u$. (Important: To make sure you understand the proof, decipher where have I used the fact that $r \gt 0$.)

Let $r \le 0$, what can you say about the $\sup rS$? $\quad$ Hint: $\sup -S=-\inf S$