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Let $a_n_1^infty$ and $b_n_1^infty$ be two sequences in $mathbbR$ such that $forall n in mathbbN$, 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 $forall m,n in mathbbN$ it is true that $a_m leq a_n$ and that there is a number $r in mathbbR$ 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 $forall m,n in mathbbN, a_n leq a_max(m,n) leq b_max(m,n) leq b_n implies a_m leq b_n$. Take $r = a_max(m,n) text or r = b_max(m,n) implies a_m leq r leq b_n$.

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

Since you have already proved that $a_nleq b_nforall nin mathbb N$ it follows that $rleq b_nforall n$

Hence $a_nleq rleq b_n$

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 $Aleq B$. Suppose not. Then we have $A>B$. Let $varepsilon=fracA-B2$. Since $A$ is the limit of $(a_n)$, there exists $MinmathbbZ$ such that for all $ngeq M$, we have $|a_n-A|<varepsilon$. Similarly, since $B$ is the limit of $(b_n)$, there exists $NinmathbbZ$ such that for all $ngeq N$, we have $|b_n-N|<varepsilon$. Choose $K=maxM,N$. We notice that for all $ngeq K$, we have $a_n>b_n$, which contradicts $a_nleq b_n$ for all $ngeq 1$.

Finally, for all $ngeq 1$, we have $a_n < A leq B < b_n$.

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 $m,n in mathbbN$, we have without loss of generality $mle n$. Then $a_m le a_n le b_n$ by your hypothesis, take $r=a_n$ and you are done.