# Set bounded above

• Mar 29th 2011, 02:26 PM
Connected
Set bounded above
Let $\displaystyle A$ be bounded above, then prove that $\displaystyle \sup(c+A)=c+\sup A,$ $\displaystyle c\in \mathbb R.$

Let $\displaystyle a\in A,$ since $\displaystyle A$ is bounded above, there's supremum, hence $\displaystyle a\le\sup A\implies a+c\le c+ \sup A$ then $\displaystyle \sup(a+c)\le c+\sup A,$ I'm on the right track? How do I finish?

I think I'm wrong. :(
• Mar 29th 2011, 02:44 PM
hatsoff
Here's what you do: First show that $\displaystyle \sup(c+A)\leq c+\sup A$. You've pretty much done that already. Now you need to show that $\displaystyle c+\sup A\leq\sup(c+A)$.

There are a lot of ways to do that. One way is by contradiction: Suppose $\displaystyle c+\sup A>\sup (c+A)$. Then $\displaystyle \sup A>\sup(c+A)-c$, which means that there is some number $\displaystyle x\in A$ with $\displaystyle x>\sup (c+A)-c\geq c+x-c=x$, a contradiction.
• Mar 29th 2011, 02:45 PM
measureman
Quote:

Originally Posted by Connected
Let $\displaystyle A$ be bounded above, then prove that $\displaystyle \sup(c+A)=c+\sup A,$ $\displaystyle c\in \mathbb R.$

Let $\displaystyle a\in A,$ since $\displaystyle A$ is bounded above, there's supremum, hence $\displaystyle a\le\sup A\implies a+c\le c+ \sup A$ then $\displaystyle \sup(a+c)\le c+\sup A,$ I'm on the right track? How do I finish?

I think I'm wrong. :(

How did you go from a + c <= c + sup A to sup(a + c) <= c + sup A?
• Mar 29th 2011, 03:41 PM
Connected
Quote:

Originally Posted by hatsoff
Here's what you do: First show that $\displaystyle \sup(c+A)\leq c+\sup A$. You've pretty much done that already

Just a thing: I determined that $\displaystyle \sup(c+a)\leq c+\sup A,$ can I change $\displaystyle a$ for $\displaystyle A$ ?

The reason I took $\displaystyle a\in A$ is because I didn't want to say $\displaystyle A\le \sup A,$ since $\displaystyle A$ is a set and it's supremum is a scalar, so that doesn't make sense.