# Set bounded above

• Mar 29th 2011, 03:26 PM
Connected
Set bounded above
Let $A$ be bounded above, then prove that $\sup(c+A)=c+\sup A,$ $c\in \mathbb R.$

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

I think I'm wrong. :(
• Mar 29th 2011, 03:44 PM
hatsoff
Here's what you do: First show that $\sup(c+A)\leq c+\sup A$. You've pretty much done that already. Now you need to show that $c+\sup A\leq\sup(c+A)$.

There are a lot of ways to do that. One way is by contradiction: Suppose $c+\sup A>\sup (c+A)$. Then $\sup A>\sup(c+A)-c$, which means that there is some number $x\in A$ with $x>\sup (c+A)-c\geq c+x-c=x$, a contradiction.
• Mar 29th 2011, 03:45 PM
measureman
Quote:

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

Let $a\in A,$ since $A$ is bounded above, there's supremum, hence $a\le\sup A\implies a+c\le c+ \sup A$ then $\sup(a+c)\le c+\sup A,$ 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, 04:41 PM
Connected
Quote:

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

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

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