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Let $S\subset\mathbb{R}$ be a non-empty bounded above set. Then there exists a monotone increasing sequence $\{x_n\}\subset S$ such that $$\lim_{n\to\infty}x_n=\sup S.$$

I'm struggling with creating a formal proof for this problem. I understand that this has to be true, but I'm unsure of how to go about proving it.

I said that $\sup S = C$ and $x_n= C - y/n$ where $y$ is dependent on the size of $S$, but this is just a bunch of jumbled ideas and I can't really think of a way to formalize a proof. Any idea, even totally different from mine would be appreciated.

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Set $s^* = \sup S$, if $s^* \in S$ take $x_n = s^*$ for every $n$ and the proof is done. So assume $s^* \notin S$. By definition of the supremum there is an element $x_1 \in S$ such that $s^*-x_1 \leq 1$. As $x_1 \in S$, $x_1 \neq s^*$ and $0<s^*-x_1$. Again by definition of the supremum there is some $x_2 \in S$ such that $0 < s^*-x_2 \leq \min\{s^*-x_1, 1\}$. It is clear that $x_1 \leq x_2$. By induction, given $x_k \in S$ you can always find $x_{k+1} \in S$ such that $0 < s^*-x_{k+1} \leq \min\{s^*-x_k, 1/k\}$, it follows then that $x_k \leq x_{k+1}$ and $|x_k-s^*|\leq 1/k$ for every $k$. Monotonicity and convergence are then obvious :).

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    $\begingroup$ Something more is required to ensure the sequence is monotone. $\endgroup$ – egreg Jun 2 '14 at 22:22
  • $\begingroup$ Yeah I saw after posting the answer and adapted it :) $\endgroup$ – Surb Jun 2 '14 at 22:26
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Let $C=\sup S$. If $C\in S$, the required sequence is $C,C,C,\ldots$.

Suppose $C\not\in S$. $C-1$ is not an upper bound of $S$, so there is some $x_1\in S$ with $C-1<x_1<C$. For each positive integer $n$, having chosen $x_n$, since neither $C-\frac{1}{n}$ nor $x_n$ is an upper bound of $S$, there is some $x_{n+1}\in S$ with $\max (C-\frac{1}{n+1},x_n)<x_{n+1}<C$. The required sequence is then $x_1,x_2,x_3,\ldots$.

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  • $\begingroup$ how do we know that 'C-1' is in S? Or more importantly that x1 is in S? $\endgroup$ – Matt Jun 2 '14 at 22:59
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Here is a general idea that you can implement to write down a formal proof. Draw some pictures for better understanding.

Let $c=\sup S$, then pick $x_1 \in \left(c-1,c\right)$. Now pick $x_2$ such that $x_2 \neq x_1$ and $x_2 \in \left(c-r_1,c \right)$, where $r_1=c-x_1$. Now pick $x_3$ such that $x_3 \neq x_1,x_2$ and $x_3 \in \left(c-r_2,c \right)$ , where $r_2=c-x_2$.

This approach takes care of your sequence being "monotone".

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    $\begingroup$ This doesn't work for $S=\{0\}$, for instance. $\endgroup$ – egreg Jun 2 '14 at 22:23

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