Bounded subset of R

**jackie** · September 7th 2010, 10:47 PM

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.

**CaptainBlack** · September 8th 2010, 12:09 AM

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

**jackie** · September 8th 2010, 06:14 AM

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, $supS - \epsilon <x_1, x_2,...,x_n < sup S$ I think I should be able to show that this lead to sup S is in S, but I don't know how.

**CaptainBlack** · September 8th 2010, 10:05 AM

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

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