Nested interval principle I am asked to prove the nested interval principle by using the axiom of completeness. That is, for a decreasing sequence of nested closed intervals $I_1, I_2, I_3,...$, there exists exactly one $x ∈ R$ such
that $∀n ∈ N : x ∈ I_n $. 
The way I understand the question, I need to prove that the intersection of all these nested intervals is a singleton set. Is my intuition correct? If so, I think I know how to prove this. Please do not give any hints for the proof if my guess is correct. Thanks in advance.
 A: It is better to understand exactly what is at play at each stage.
Let $I_n=[a_n,b_n]$. Then $I_1 \supseteq I_2 \supseteq \cdots$ implies
$$
a_1 \le a_2 \le \cdots a_n \le \cdots \le b_n \le \cdots \le b_2 \le b_1
$$
Thus, $A=\{a_1,a_2,\dots\}$ is a non empty set that is bounded above and $B=\{b_1,b_2,\dots\}$ is a non empty set that is bounded below. Let $a=\sup A$ and $b=\inf B$, using completeness. Let $I=[a,b]$. Then $I$ is non empty and $I \subseteq \bigcap I_n$, which proves the nested interval principle. (Actually, $I=\bigcap I_n$, but this not needed at this point.) 
Now, if $b_n-a_n \to 0$, then $I=\bigcap I_n$ is a singleton.
A: Your intuition is correct - the intersection of decreasing (to length $0$) nested intervals is a singleton.
Also, you might consider this additional intuitive explanation of what is happening:
A 'small' closed interval $[a_i,b_i]$ can be viewed as an approximation of some real number. Indeed a scientist may even announce that the 'actual' number $x$ satisfies
$\quad a_i \le x \le b_i$
As these nested approximations 'telescope down', and you 'keep going',
you 'pick out' or 'point out' the real number number $x$ of interest (it can't be 'missing in action').
A fun exercise is to see if you can come up with a definition of nested decreasing open intervals so that the intersection is also guaranteed to be a singleton.
