nested interval theorem
Proposition 1.
If
$$[{a}_{1},{b}_{1}]\supseteq [{a}_{2},{b}_{2}]\supseteq [{a}_{3},{b}_{3}]\supseteq \mathrm{\dots}$$ 
is a sequence of nested closed intervals^{}, then
$\bigcap _{n=1}^{\mathrm{\infty}}}[{a}_{n},{b}_{n}]\ne \mathrm{\varnothing}.$ 
If also $\underset{n\to \mathrm{\infty}}{lim}({b}_{n}{a}_{n})=0$, then the infinite intersection consists of a unique real number.
Proof.
There are two consequences to nesting of intervals: $[{a}_{m},{b}_{m}]\subseteq [{a}_{n},{b}_{n}]$ for $n\le m$:

1.
first of all, we have the inequality^{} ${a}_{n}\le {a}_{m}$ for $n\le m$, which means that the sequence ${a}_{1},{a}_{2},\mathrm{\dots},{a}_{n},\mathrm{\dots}$ is nondecreasing;

2.
in addition, we also have two inequalities: ${a}_{m}\le {b}_{n}$ and ${a}_{n}\le {b}_{m}$. In either case, we have that ${a}_{i}\le {b}_{j}$ for all $i,j$. This means that the sequence ${a}_{1},{a}_{2},\mathrm{\dots},{a}_{n},\mathrm{\dots}$ is bounded from above by all ${b}_{i}$, where $i=1,2,\mathrm{\dots}$.
Therefore, the limit of the sequence $({a}_{i})$ exists, and is just the supremum, say $a$ (see proof here (http://planetmath.org/NondecreasingSequenceWithUpperBound)). Similarly the sequence $({b}_{i})$ is nonincreasing and bounded from below by all ${a}_{i}$, where $i=1,2,\mathrm{\dots}$, and hence has an infimum $b$.
Now, as the supremum of $({a}_{i})$, $a\le {b}_{i}$ for all $i$. But because $b$ is the infimum of $({b}_{i})$, $a\le b$. Therefore, the interval $[a,b]$ is nonempty (containing at least one of $a,b$). Since ${a}_{i}\le a\le b\le {b}_{i}$, every interval $[{a}_{i},{b}_{i}]$ contains the interval $[a,b]$, so their intersection also contains $[a,b]$, hence is nonempty.
If $c$ is a point outside of $[a,b]$, say $$, then there is some ${a}_{i}$, such that $$ (by the definition of the supremum $a$), and hence $c\notin [{a}_{i},{b}_{i}]$. This shows that the intersection actually coincides with $[a,b]$.
Now, since $\underset{n\to \mathrm{\infty}}{lim}({b}_{n}{a}_{n})=0$, we have that $ba=\underset{n\to \mathrm{\infty}}{lim}{b}_{n}\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=\underset{n\to \mathrm{\infty}}{lim}({b}_{n}{a}_{n})=0$. So $a=b$. This means that the intersection of the nested intervals contains a single point $a$. ∎
Remark. This result is called the nested interval theorem. It is a restatement of the finite intersection property for the compact set $[{a}_{1},{b}_{1}]$. The result may also be proven by elementary methods: namely, any number lying in between the supremum of all the ${a}_{n}$ and the infimum of all the ${b}_{n}$ will be in all the nested intervals.
Title  nested interval theorem 

Canonical name  NestedIntervalTheorem 
Date of creation  20130322 17:27:12 
Last modified on  20130322 17:27:12 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  11 
Author  pahio (2872) 
Entry type  Theorem 
Classification  msc 54C30 
Classification  msc 2600 