Sequence within a non-empty, bounded above set

**sakuraxkisu** · January 30th 2013, 11:35 AM

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

**Plato** · January 30th 2013, 11:57 AM

Originally Posted by 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 $\sigma=\sup(S)$ . Two cases:
1) $\sigma\in S$ , what is a constant sequence that works?

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

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

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