# Nested intervals theorem - a special case on open intervals

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.

• A relevant comment: math.stackexchange.com/questions/1370695/… Jun 12, 2020 at 9:54
• 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 $n$. Jun 12, 2020 at 9:55

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 $$\bigcap_{n=1}^\infty(a_n, b_n) \supseteq \bigcap_{n=1}^\infty[a_{n+1}, b_{n+1}] \ne \varnothing.$$
Hints: Let $$A=\sup_n a_n =\lim a_n$$. Since $$(a_n)$$ is strictly increasing it follows that $$a_n for all $$n$$. Now $$a_{n+k} 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 $$b_{n+1}. This implies that $$b_{n+1} for all $$m$$ sufficiently large. But then $$a_m when $$m$$ is large enough. This contradiction shows that $$A for al $$n$$. Hence $$A \in (a_n,b_n)$$ for all $$n$$.
Our assumption is that $$a_{n} < a_{n+1},a_nb_{n+1}\tag{1}$$ Then by density of reals we can find numbers $$a'_n, b'_n$$ such that $$a_nb'_n>b_{n+1}\tag{2}$$ Then the above inequalities lead to $$a'_nb'_{n+1}\tag{3}$$ 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.
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 $$n \in \mathbb N$$, $$a_m \lt b_n$$.
• Welcome to MSE. Why do you use $$...$$ for each and every mathematical expression, when $...$ would be enough in almost every case? Nov 10, 2020 at 7:33