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Prove If $S \subset R$ is a nonempty set, bounded from above, then for every $\epsilon >0$ there exists $ x\in S$ such that $(\sup S) - \epsilon < x \le \sup S$

How can I prove for this.. ?

I always use this proposition, but I dont know how to prove.

1 Answers 1

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If $ \epsilon >0$ , the $\sup S - \epsilon$ is not an upper bound of $S$.

Your turn !

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    I know that $\sup S - \epsilon$ is not an uppper bound of $S$ Since sup S is least upper bound, which means every element of S should be less than or equal to S. Also, I know that $\sup S - \epsilon \le x $ but I dont know how to prove that proposition.. – 2017-02-28
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    Suppose that for all $x \in S$: $x \le \sup S - \epsilon$. Then $\sup S - \epsilon$ would be an upper bound of $S$. But this is nonsense ! Hence, there is $x \in S$ with $x>\sup S - \epsilon$. – 2017-02-28
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    is this proof by contradiction? Should I have to prove that $x \le \sup S $? If I have to, can I use the same way that you used.. ? – 2017-02-28
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    ex) Suppose that for all $ x \in S: x > \sup S$ Then x is upper bound of S. if x is in S and x is upper bound, then x should be $\sup S$ by definition of supremum. Therefore $x = \sup S$, so the assumption $x > \sup S $ is contradiction. – 2017-02-28
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    Is this right..? – 2017-02-28
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    @Kwangi Yu Why you have to prove that $x\leq \sup S$? Notice that $\sup S$ is an upper bound, in particular, of set $S$, and that $x\in S$. By the way, I already made an answer to the link I mentioned. – 2017-02-28