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Let S ⊂ R be a nonempty bounded set. Then there exist monotone sequences ${x_n}$ and ${y_n}$ such that $x_n$, $y_n$ ∈ $S$ and $\sup S = \lim_{n→∞} x_n$ and $\inf S = \lim_{n→∞} y_n$

How can I prove this ..?

2 Answers 2

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Let $x:=\sup S\in \mathbb R$. Then for all $n\in\mathbb N$, there is an $x_n\in S$ such that $x_n+1/n\geq x$ (otherwise $x$ could not be the supremum). We also know that $x_n\leq x$ by definition of supremum. Therefore $|x_n-x|\leq 1/n$. Can you finish it off from here?

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    This shows the limit characterization of supreuma but it doesn't guarantee that the sequence is montone. – 2017-02-28
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Let $x = \sup S$. We're going to have two cases. If $x$ is the maximum then let $a_n = x$ and we have a monotone sequence that converges to $x$. Otherwise consider the following:

Then by the epsilon characterization of supremum given an $n$ we know $\exists a_n$ such that $a_n \in (x - 1/n, x)$. Then let $a_{n+1} \in (\max(a_n, x - 1/n), x)$. This must converge to $x$ because $x-1/n$ converges to $x$. We also know that $A$ is a monotone sequence because given $a_n$ we know $a_{n+1} > a_n$ because it is in the interval $(\max(a_n, x - 1/n), x)$.

To construct this sequence for the infimum you just need to use the minimum and use the epsilon characterization of infimum instead. It's basically the same proof.

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    so second case $a_{n}$ is maximum not supremum ? – 2017-02-28
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    I don't understand $a_{n+1}∈(max(a_{n},x−1/n),x)$ – 2017-02-28
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    how did you get that part.. ? I understand $a_{n}∈(x−1/n,x)$. – 2017-02-28
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    $a_{n}∈(x−1/n,x)$ means $x-1/n < a_n < x$. $x-1/n < a_n < a_{n+1} < x $ This is what I thought. but this doesn't seem like $a_{n+1}∈(max(a_{n},x−1/n),x)$ – 2017-02-28
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    The idea of $(\max(a_n, x - 1/n), x)$ is for every $a_{n+1}$ we want an interval greater than $a_n$ and less than the supremum. If we did just $(x - 1/n, x)$ then we couldn't guarantee that $A$ was monotone and if we did just $(a_n, x)$ we couldn't guarantee convergence to the supremum. – 2017-02-28
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    To me, $(max(a_{n},x−1/n),x)$ should be $(max(x-1/n,a_{n}),x)$ because from $a_{n}$'s intervals, $a_{n}$ should be greater than $x-1/n$. can you explain about this..? Thank you so much. – 2017-02-28
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    Those two things are the same. The order in which things are in a max statement doesn't matter. Let me write it a different way. Let $b_{n+1} = \max(a_n, x-1/n)$. Then we want to pick an $a_n \in (b_{n+1}, x)$. We calculate the max bit before finding the actual interval. – 2017-02-28
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    I understand ! Thank you so much – 2017-02-28