Nested Interval Property and the intersection of infinite sequences Given sequences such that $A_n = \{n,n+1,\cdots \}$, then it can be shown that $\bigcap\limits_{n=1}^\infty A_n = \emptyset$
Now, according to nested interval property if $\mathbf{I}_n = [a_b,b_n] = \{x \in \Re: a_n\leq x\leq b_n\}$, then $\bigcap\limits_{n=1}^\infty \mathbf{I}_n \neq \emptyset$
The above two statements looks very similar, but the results are just opposite. From what I can see is, $A_n$ is a countable infinite set whereas $\mathbf{I}_n$ is an uncountable infinite set.  Is that the only difference, if it is true ?. Or is there anything more than that which relates two statements above ?
 A: There is a theorem which generalizes the nested interval property:

Theorem: (Nested Compact Sets Property) Let $K_1\supseteq K_2\supseteq K_3\dots$ be a nested sequence of nonepmty compact sets in a Hausdorff topological space. Then $\bigcap_n K_n$ is nonempty.

Proof: If $\bigcap_n K_n=\varnothing$ , then $\{K_1\setminus K_n\}_{n\ge 2}$ is an open cover of $K_1$ with no finite subcover. $\square$
However, the assumption of compactness is necessary. In a general topological space, a nested sequence of sets $A_n$ can have null intersection. Your sets $A_n$ are a typical counter-example. Note that the $A_n$ are not compact. 
In summary, the commonality to your two examples is nested nonempty sets, and the difference is compactness.
A: First of all note that both $\{A_n\}$ and $\{I_n\}$ are countable so this is not the problem.
The difference is in that $A_n$ is not a bounded closed interval, so we can not  apply the nested interval theorem. 
The nested interval theorem is about nested bounded closed intervals not arbitrary sets. 
A: First note that there are countably many $A_{n}$ and $I_{n}$ so the problem is not the countability or non-countability of the intersection. In other words the set containing all the $A_{n}$ and the set containing all the $I_{n}$ are both countable being indexed by the countable set $\mathbb{N}$.
Secondly it is true that the sets $I_{n}$ are uncountable whilst the sets $A_{n}$ are countable. This is a different statement from the above, for here we are talking about what the sets $I_{n}$ and $A_{n}$ contain, not what the sets that contain them contain.
However, the fact that the sets $I_{n}$ are uncountable whilst the sets $A_{n}$ are countable is not the reason for the discrepancy.
In fact it is possible to have a countable collection of countable sets $J_{n} = \{x \in \mathbb{Q} : 1 \leq x \leq 1 + \frac{1}{n}\}$ such that the intersection is non-empty, in this case equal to the set containing the number 1.
Conversely it is possible to have a collection of uncountable sets $B_{n} = \{x \in \mathbb{R} : x \geq n\}$ such that the intersection is empty.
Thus the crucial difference between your sets $A_{n}$ and $I_{n}$ is that the $I_{n}$ are ${bounded}$ and nested whilst the $A_{n}$ are ${unbounded}$ and nested.
It is this difference that is responsible for the different results for their respective intersections.
