# Nested Interval Property for open intervals and a alternative proof for NIP.

I was wondering, where does the proof go wrong when applying it to open intervals. I know there are counterexamples but I cannot seem to find any mistakes when you substitute closed intervals with open intervals.(The proof: Proving nested interval theorem from least upper bound property of real numbers) Also, i was wodering if this proof is also valid:

Let $I_{n}=[a_n,b_n]$, if $a_n=b_n$ for some $n$ then the intersection is not empty since $a_n$ will be in it. If $a_n\not=b_n$ for all $n$ then we know that for any $n$, $a_n<b_n$. We know that $a_n$ and $b_n$ are real numbers, therefore, from density of $\mathbb Q$ in $\mathbb R$ we have a real number $x$ that fits between $a_n$ and $b_n$. We can repeat the process infinitely many times. Therefore for any $n$ there is a an infinite amount of rational numbers in the interval. Therefore the intersection is non-empty.

Also a more intuitive proof which i am not sure could be counted as proof. Since the intervals are nested we ha e two cases

1. They "decrease" to the point where they stay the same till infinity. Say they decrease to [a,b]. After that they stay [a,b] forever. Hence the intersection is non empty since [a,b] is in it.

2. They decrease infinitely many times to the point where [a,a] where you cannot decrease more. Hence a is in the intersection of all of the sets.

You've asked two questions. The first is, essentially, "why doesn't the known proof work for open intervals?"

Here is the relevant statement from the accepted answer to the question you link to:

Then $s≥a_i$ for all $i \in I$ by definition. If $i \in I$, then $b_i$ is also an upper bound for $S$, hence $s≤b_i$.

But $s \le b_i$ does not tell you that $s$ is in the open interval $(a_i, b_i)$. It might equal $b_i$. Just think about all the intervals $(-1/i, 0)$.