# Simple supremum question

• Sep 28th 2009, 03:28 PM
cgiulz
Simple supremum question
If $\displaystyle A$ and $\displaystyle B$ are bounded and non-empty subsets of $\displaystyle \mathbb{R}$, and $\displaystyle c \in \mathbb{R}$, let

$\displaystyle cA =$ {$\displaystyle ca: a \in A$}.

Prove the following:
if $\displaystyle c > 0$, supremum $\displaystyle cA = c$ supremum $\displaystyle A$.

[SOLVED]
• Sep 28th 2009, 06:59 PM
Danneedshelp
Quote:

Originally Posted by cgiulz
If $\displaystyle A$ and $\displaystyle B$ are bounded and non-empty subsets of $\displaystyle \mathbb{R}$. Prove the following:

if $\displaystyle c > 0$, supremum $\displaystyle cA = c$ supremum $\displaystyle A$.

Let $\displaystyle \alpha=sup(A)$. Since $\displaystyle \alpha$ is an upper bound for $\displaystyle A$, $\displaystyle c\alpha\geq\\ca$ for all $\displaystyle a\in{A}$. Therefore, $\displaystyle c\alpha$ is an upper bound for $\displaystyle A$. Moreover, if we let $\displaystyle \beta$ be any other upper bound for $\displaystyle cA$, we have that $\displaystyle \frac{\beta}{c}$ is an upper bound for $\displaystyle A$. Thus, $\displaystyle \alpha\leq\frac{\beta}{c}$ $\displaystyle \Leftrightarrow\\c\alpha\leq\beta$. So, $\displaystyle c\alpha$ satisfies the definition of $\displaystyle sup(cA)$.

That is my stab at it.
• Sep 29th 2009, 12:57 PM
cgiulz
Wow, I left out a piece sorry! (Itwasntme)