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I learned the nested intervals theorem in the class: If $I_n\ (n\in\Bbb N)$ is a sequence of bounded closed intervals, i.e. $[a_n,b_n]$, then $\bigcap I_n\ (n\in\Bbb N)\neq\varnothing$. In the proof of the theorem, we used another theorem: monotone non-decreasing sequence (in this case $a_n$) is convergent, if bounded above.

And we discussed in the class that the theorem does not hold for open intervals, i.e. $(a_n,b_n)$. We made a counterexample $I_n=(0,\frac{1}{n}),\ n\in\Bbb N$, where $\bigcap I_n=\varnothing$. It also makes sense to me. There is no real staying in the intersection of nested intervals, because the candidate $0$ is ruled out by the open intervals.

Then, the professor gave us an extended question: What if we have nested $I_n$ as open intervals $(a_n,b_n)$, but this time, let $a_n$ be a strictly increasing sequence ($\forall n\in\Bbb N,\ a_n\lt a_{n+1}$) and $b_n$ a strictly decreasing sequence? It seems to me that there would be a real finally staying in the nested interval. However, I cannot prove it. Appreciated if anyone can provide me a hint. Thanks.

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  • $\begingroup$ A relevant comment: math.stackexchange.com/questions/1370695/… $\endgroup$ Jun 12, 2020 at 9:54
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    $\begingroup$ Try to show that the closed interval $[\sup a_n , \inf b_n]$ is contained in the intersection. By the way, you need the additional hypothesis that $a_n < b_n$ for all $n$. $\endgroup$ Jun 12, 2020 at 9:55

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For all $n,$ $[a_n, b_n] \supset (a_n, b_n) \supset [a_{n+1}, b_{n+1}],$ therefore $([a_n, b_n])_{n\geqslant1}$ is a nested sequence of closed intervals, therefore $$ \bigcap_{n=1}^\infty(a_n, b_n) \supseteq \bigcap_{n=1}^\infty[a_{n+1}, b_{n+1}] \ne \varnothing. $$

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  • $\begingroup$ Way better and simpler than my answer +1. This is a clear evidence that simplicity often eludes us. $\endgroup$
    – Paramanand Singh
    Jun 13, 2020 at 1:28
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Hints: Let $A=\sup_n a_n =\lim a_n$. Since $(a_n)$ is strictly increasing it follows that $a_n <A$ for all $n$. Now $a_{n+k} <a_n <b_n$ for all $n,k$ so $A=\lim_k a_{n+k} \leq b_n$ for all $n$. Suppose $A=b_n$ for some $n$. Then $b_{n+1}<b_n=A(=\lim a_n)$. This implies that $b_{n+1}<a_m$ for all $m$ sufficiently large. But then $a_m <b_m <b_{n+1} <a_m$ when $m$ is large enough. This contradiction shows that $A<b_n$ for al $n$. Hence $A \in (a_n,b_n)$ for all $n$.

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Our assumption is that $$a_{n} < a_{n+1},a_n<b_n,b_n>b_{n+1}\tag{1}$$ Then by density of reals we can find numbers $a'_n, b'_n$ such that $$a_n<a'_n<a_{n+1},b_n>b'_n>b_{n+1}\tag{2}$$ Then the above inequalities lead to $$a'_n<a'_{n+1},a'_n<b'_n,b'_n>b'_{n+1}\tag{3}$$ Thus the sequence of closed intervals $[a'_n, b'_n] $ is nested and has a non-empty intersection. So there is a real number $x$ such that $x\in[a'_n, b'_n]\subset (a_n, b_n) $ for all $n$. So the intervals $(a_n, b_n) $ have a non-empty intersection.

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Let $$A = \lim_{n\to \infty} {a_n}$$$$B=\lim_{n\to \infty}{b_n}$$For all n, $ a_n \lt A \le B \lt b_n $according to Sandwich Theorem, thus $[A,B] \subset I_n$ for all n. So $$\bigcap_{n=1}^\infty{I_n} = [A,B] \neq \emptyset$$ Besides, for every $m \in \mathbb N$ and every $ n \in \mathbb N$, $a_m \lt b_n$.

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  • $\begingroup$ Welcome to MSE. Why do you use $$...$$ for each and every mathematical expression, when $...$ would be enough in almost every case? $\endgroup$ Nov 10, 2020 at 7:33

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