I am aware that the Nested Interval Property (NIP) does not hold for open intervals. The standard way to convince someone of this fact is by providing the standard counterexample $ \bigcap_{n=1}^{\infty}(0,\frac{1}{n}) $.
What I'm wresting with is whether NIP ever holds for open intervals. It turns out that, if in $I_n=(a_n,b_n)$, $(a_n)$ is a strictly increasing sequence and $(b_n)$ is a strictly decreasing one, then NIP holds!
My reference for the above observation: Nested intervals theorem - a special case on open intervals .
Question: Are there any cases other cases where the NIP holds for open intervals?
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Dhiraj Rao
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1Yes. Another case is when the length $b_n-a_n$ does not converge to $0$. – GEdgar May 31 '23 at 12:23
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Consider the nested intervals $(-\frac1n , \frac1n)$ which all contain $0$ so the intersection must be non-empty. – CyclotomicField May 31 '23 at 12:40