Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 2 down vote accepted

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 $\sup S$.

share|improve this answer
1  
Shouldn't the second inequality be $a\le\sup S$? Great answer, btw, +1. –  Dejan Govc Apr 5 '12 at 1:20
    
@DejanGovc I have edited that. –  fpqc Apr 6 '12 at 14:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.