Thread: 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 $\displaystyle \epsilon >0$, the interval $\displaystyle (sup S-\epsilon, sup S)$ contains infinitely many elements of S.

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 $\displaystyle \epsilon >0$, the interval $\displaystyle (sup S-\epsilon, sup S)$ contains infinitely many elements of S.

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

CB

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

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 $\displaystyle x_1, x_2, ...., x_n$ is in $\displaystyle (supS- \epsilon, supS)$. So, $\displaystyle supS - \epsilon <x_1, x_2,...,x_n < sup S$ 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 $\displaystyle S$ and less than $\displaystyle {\text{sup}}(S)$ which is a contradiction etc...