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Let $\{a_n\}_1^\infty$ and $\{b_n\}_1^\infty$ be two sequences in $\mathbb{R}$ such that $\forall n \in \mathbb{N}$, it is true that $a_n \leq b_n, a_n \leq a_{n+1}, \text{ and } b_{n+1} \leq b_n$.

We want to show $\forall m,n \in \mathbb{N}$ it is true that $a_m \leq a_n$ and that there is a number $r \in \mathbb{R}$ such that $a_m \leq r \leq b_n$.

I've proceeding as follows:

We have $a_{n} \leq b_{n} \implies a_{n+1} \leq b_{n+1}$ and thus $a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$.

Does this not imply that $a_m \leq a_n$? Even without stating the obvious fact that the sets are upper and lower bounds of each other? It seems then that $r$ would follow..

EDIT: Taking some of the ideas from below I have written a simple proof. Feedback is welcome and appreciated.

Since $a$ is monotonically increasing and $b$ is monotonically decreasing we have $\forall m,n \in \mathbb{N}, a_n \leq a_{\max(m,n)} \leq b_{\max(m,n)} \leq b_n \implies a_m \leq b_n$. Take $r = a_{\max(m,n)} \text{ or } r = b_{\max(m,n)} \implies a_m \leq r \leq b_n$.

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It is clear that $a_n$ is a monotonically increasing sequence bounded above by $b_1$.Hence by Monotone Convergence Theorem $a_n\to r$ (say )

Since you have already proved that $a_n\leq b_n\forall n\in \mathbb N$ it follows that $r\leq b_n\forall n$

Hence $a_n\leq r\leq b_n$

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    $\begingroup$ I've taken this idea and applied it in a slightly different way above. $\endgroup$
    – Chris
    Oct 8 '15 at 14:31
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First we note that $b_1$ is an upper bound for $(a_n)$. Since $(a_n)$ is monotone increasing, the Monotone Convergence Theorem states that $(a_n)$ converges to its supremum $A$. Similarly, $a_1$ is a lower bound for $(b_n)$. Since $(b_n)$ is monotone decreasing, the Monotone Convergence Theorem states that $(b_n)$ converges to its infimum $B$.

We claim that $A\leq B$. Suppose not. Then we have $A>B$. Let $\varepsilon=\frac{A-B}{2}$. Since $A$ is the limit of $(a_n)$, there exists $M\in\mathbb{Z}$ such that for all $n\geq M$, we have $|a_n-A|<\varepsilon$. Similarly, since $B$ is the limit of $(b_n)$, there exists $N\in\mathbb{Z}$ such that for all $n\geq N$, we have $|b_n-N|<\varepsilon$. Choose $K=\max\{M,N\}$. We notice that for all $n\geq K$, we have $a_n>b_n$, which contradicts $a_n\leq b_n$ for all $n\geq 1$.

Finally, for all $n\geq 1$, we have $a_n < A \leq B < b_n$.

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You are asked to show $a_m \le r \le b_n$. Notice that the indices must be different, the result you give has the same index (i.e. $ a_n \le r \le b_n$). Given $m,n \in \mathbb{N}$, we have without loss of generality $m\le n$. Then $a_m \le a_n \le b_n$ by your hypothesis, take $r=a_n$ and you are done.

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  • $\begingroup$ In my edit above I show something very similar written differently. $\endgroup$
    – Chris
    Oct 8 '15 at 14:31

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