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Question. Let $(a_n)$ be an increasing sequence and $(b_n)$ be a decreasing sequence and assume that $(a_n)\le (b_n)$ for all $n \in \mathbb{N}$. Show that $\lim a_n \le \lim b_n$, and thereby deduce the Nested Interval Property from the Monotone Convergence Theorem.

Proof. It has been given that $(a_n)$ and $(b_n)$ are both monotone. By the Monotone Convergence Theorem (MCT), we only have to show that $(a_n)$ and $(b_n)$ are bounded to show they are convergent.

Define the set $A:=\{ a_n: n \in \mathbb{N}\}$ and $B:=\{b_n: n\in \mathbb{N}\}$. By the definition of suprema, $b_n \le \sup B$ and thus, $a_n \le b_n \le \sup B$ for all $n \in \mathbb{N}$. Thus, we've shown $A$ is convergent. Similarly by the definition of infimum, $\inf A \le a_n$ which implies $\inf A \le a_n \le b_n$ for all $n \in \mathbb{N}$ and we've shown $(b_n)$ is convergent.

And hence by the Order Limit Theorem, we can finally conclude that $\lim a_n \le \lim b_n$.

Finally define the sets $I_n := [a_n, b_n]$ for all $n \in \mathbb{N}$. It can be easily seen that $I_1 \subseteq I_2 \subseteq I_3 \subseteq I_4 \subseteq \dots$. To show that Nested Interval Property holds, we have to show there exists at least one element in the set $\displaystyle\cap_{n=1}^{\infty} I_n$. Consider $x:=\lim a_n$ and $y:= \lim b_n $. Since, $a_n \le x \le y\le b_n $ for all $n \in \mathbb{N}$, we can conclude that $x,y \in \cap_{n=1}^{\infty} I_n$. This completes the proof.$\blacksquare $

Is my proof correct? Will it still hold true if $a_n \le b_n$ is not given? I've been self studying analysis so kindly let me know how can I improve my proof writing.

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    $\begingroup$ The inequality $a_n\leq b_n$ is essential in order to show that the sequence are bounded. And your proof is correct (almost). You just add a statement that $b_n\leq b_1$ and hence $\sup B$ exists. Similarly $a_1\leq a_n$ so $\inf A$ exists. And this is typically how nested interval principle is established. $\endgroup$ – Paramanand Singh Dec 9 '17 at 7:53
  • $\begingroup$ By the way you should try to do the reverse. Prove monotone convergence theorem assuming that the nested interval property holds. Proving that various forms of completeness of real numbers are equivalent is central to understanding the essence of completeness. $\endgroup$ – Paramanand Singh Dec 9 '17 at 7:59

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