# Bounded subset of R

• Sep 7th 2010, 10:47 PM
jackie
Bounded subset of R
I'm having trouble with this problem. Could someone give me a hand?
Let S be a nonempty subset of R that is bounded above. Show that if sup S is not in S, then for every $\epsilon >0$, the interval $(sup S-\epsilon, sup S)$ contains infinitely many elements of S.
• Sep 8th 2010, 12:09 AM
CaptainBlack
Quote:

Originally Posted by jackie
I'm having trouble with this problem. Could someone give me a hand?
Let S be a nonempty subset of R that is bounded above. Show that if sup S is not in S, then for every $\epsilon >0$, the interval $(sup S-\epsilon, sup S)$ contains infinitely many elements of S.

Suppose for some $\epsilon >0$ that $(sup S-\epsilon, sup S)$ contains only finitely many points of $S$ (that it is non-empty should be obvious).

CB
• Sep 8th 2010, 06:14 AM
jackie
Thanks a lot for your help, CB. I actually tried to assume the contrary as you suggested, but I got stuck. I have $x_1, x_2, ...., x_n$ is in $(supS- \epsilon, supS)$. So, $supS - \epsilon I think I should be able to show that this lead to sup S is in S, but I don't know how.
• Sep 8th 2010, 10:05 AM
CaptainBlack
Quote:

Originally Posted by jackie
Thanks a lot for your help, CB. I actually tried to assume the contrary as you suggested, but I got stuck. I have $x_1, x_2, ...., x_n$ is in $(supS- \epsilon, supS)$. So, $supS - \epsilon I think I should be able to show that this lead to sup S is in S, but I don't know how.

A finite subset of the reals has a largest element, which will be an upper bound for $S$ and less than ${\text{sup}}(S)$ which is a contradiction etc...