[Math] Constant times Supremum

real-analysissupremum-and-infimum

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.

Best Answer

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 $r \gt 0$.)

Additional exercise:

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

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