# compact nonempty sets contain Sups.

If $E$ iis a compact, nonempty subset of REAL numbers (hence closed), we know that every convergent sequence in $E$ converges in $E$, are there sequences $a_{n}$ and $b_{n}$ in $E$ that converge to $\sup E$ and $\inf E$, respectively?

I would think so since we can just take any collection on points in $E$ that are arbitrary close to $E$, but how would I show this, or is it not true in an arbitrary metric space?

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sorry, I meant in the reals. – The Substitute Apr 5 '12 at 0:48
Yes, since $\sup E\in E$ in this case, you can just take the constant sequence defined by $s_n:=\sup E$ and the same for $\inf$. (And in an arbitrary metric space, $\sup$ and $\inf$ aren't defined, so no in that case.) – Dejan Govc Apr 5 '12 at 0:57
The reason I am taking this sequence to converge to Sup is because I am trying to prove SupE is in E. – The Substitute Apr 5 '12 at 1:08

I think you don't need to talk about compactness or anything. Your question is asking given a bounded subset $S$ of $\Bbb{R}$, is there a sequence of points $a_n$ that live in $S$ that converge to $\sup S$? Note that if $S$ is closed this would force $\sup S \in S$.
By definition of the supremum for all $\epsilon > 0$, there is a point $a \in S$ such that $$\sup S - \epsilon < a \leq \sup S.$$
In particular if we choose $\epsilon = 1/n$, for every $n \in \Bbb{N}$ we get some ball of radius $1/n$ about $\sup S$. Choose a point $a_n$ in this ball. Then you get a sequence of points $a_n$ that live in $S$ and converge to $\sup S$.
Shouldn't the second inequality be $a\le\sup S$? Great answer, btw, +1. – Dejan Govc Apr 5 '12 at 1:20