| Version 3 |
Version 2 |
| If |
If |
| $$[a_1,\,b_1]\;\supseteq [a_2,\,b_2]\;\supseteq [a_3,\,b_3]\;\supseteq\ldots$$ |
$$[a_1,\,b_1]\;\supseteq [a_2,\,b_2]\;\supseteq [a_3,\,b_3]\;\supseteq\ldots$$ |
| is a sequence of nested closed intervals, then |
is a sequence of nested closed intervals, then |
| \begin{align*} |
\begin{align*} |
| \bigcap_{n=1}^\infty [a_n,\,b_n] \neq \varnothing. |
\bigcap_{n=1}^\infty [a_n,\,b_n] \neq \varnothing. |
| \end{align*} |
\end{align*} |
| If also\, $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$,\, then the infinite |
If also\, $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$,\, then the infinite |
| intersection consists of a unique real number. |
intersection consists of a unique real number. |
|
|
| This result is called the \emph{nested interval theorem}. |
This result is called the \emph{nested interval theorem}. |
| It is a restatement of the \emph{finite intersection property} |
It is a restatement of the \emph{finite intersection property} |
| for the compact set $[a_1, b_1]$. |
for the compact set $[a_1, b_1]$. |
| The result may also be proven by elementary methods: |
The result may also be proven by elementary methods: |
| namely, any number lying in between the supremum of all the $a_n$ |
namely, any number lying in between the supremum of all the $a_n$ |
| and the infimum of all the $b_n$ |
and the infimum of all the $b_n$ |
|
will be in all the nested intervals.
|
will be in the infinite intersection.
|
