Sequence within a non-empty, bounded above set

Let S be a non-empty subset of R (real numbers) that is bounded above. Show that there exists a sequence (xn, n is a natural number), contained in S (that is, xn is an element of S for all n in the set of natural numbers) and which is convergent with limit equal to sup S.

Any help would be greatly appreciated :)

Re: Sequence within a non-empty, bounded above set

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**sakuraxkisu** Let S be a non-empty subset of R (real numbers) that is bounded above. Show that there exists a sequence (xn, n is a natural number), contained in S (that is, xn is an element of S for all n in the set of natural numbers) and which is convergent with limit equal to sup S.

Suppose that $\displaystyle \sigma=\sup(S)$. Two cases:

1) $\displaystyle \sigma\in S$, what is a constant sequence that works?

2) $\displaystyle \sigma \notin S$ and $\displaystyle \exists s_1\in S$ such that $\displaystyle s<\sigma$. You explain why!

If $\displaystyle n>1$ then $\displaystyle \exists s_n\in S$ such that $\displaystyle \sigma-\tfrac{1}{n}\le s_n<\sigma$. You explain why!