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I am attempting to show that there is a sequence $x_1, x_2, \ldots$ in $S$ such that $x_n$ converges to $a$, where $a = \sup S$ and $S$ is contained in the set of all reals.

Since $a = \sup S$, given $\epsilon > 0$ and that $x_n$ is in $S$, we know that $a-\epsilon < x_n \leq a$ since $a$ is a least upper bound by def'n of a supremum. I do not know where to go from applying the def'n of supremum.

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 13 '15 at 17:49
  • $\begingroup$ Any additional assumptions about $S$? Nonempty? Bounded from above? $\endgroup$ – Hagen von Eitzen Oct 13 '15 at 17:50
  • $\begingroup$ We can assume that S is bounded from above. $\endgroup$ – Matt B. Oct 14 '15 at 2:27
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Well, we have two cases:

Case 1: $a<\infty$. In that case, take a sequence $x_n$ such that $$\forall n\in\mathbb{N}:a-\frac{1}{n}<x_n<a$$ (there is such $x_n$ for each $n$, due to the definition of supremum). Prove that this sequence converges to $a$.

Case 2: $a=\infty$. In that case, take a sequence $x_n$ such that $$\forall n\in\mathbb{N}:x_n>n$$ and continue similarly

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  • $\begingroup$ You are missing the case 3: $a=-\infty$. Though in that case there of course is no such sequence. $\endgroup$ – Hagen von Eitzen Oct 13 '15 at 17:49
  • $\begingroup$ Since the supremum of a set cannot be $-\infty$, it is irrelevant. $\endgroup$ – Guy Oct 13 '15 at 17:51

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