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Revision difference : nested interval theorem |
| Version 5 |
Version 4 |
| 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. |
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| \begin{proof} There are two consequences to nesting of intervals: $[a_n,\,b_n]\supseteq [a_m,\,b_m]$ for $m\ge n$: |
\begin{proof} There are two consequences to nesting of intervals: $[a_n,\,b_n]\supseteq [a_m,\,b_m]$ for $m\ge n$: |
| \begin{enumerate} |
\begin{enumerate} |
| \item first of all, we have the inequality $a_m\le a_n$ for $m\le n$, which means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ is nondecreasing; |
\item first of all, we have the inequality $a_m\le a_n$ for $m\le n$, which means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ is nondecreasing; |
| \item 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 $a_1, a_2, \ldots, a_n, \ldots$ is bounded from above by all $b_i$, where $i=1,2,\ldots$. |
\item 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 $a_1, a_2, \ldots, a_n, \ldots$ is bounded from above by all $b_i$, where $i=1,2,\ldots$. |
| \end{enumerate} |
\end{enumerate} |
| Therefore, the limit of the sequence $(a_i)$ exists (see proof here), and is just the supremum, say $a$. Similarly the sequence $(b_i)$ is nonincreasing and bounded from below by all $a_i$, where $i=1,2,\ldots$, and hence has an infimum $b$. Now, as an supremum of $(a_i)$, $a\le b_i$ for all $i$. But $b$ is the infimum of $(b_i)$, $a\le b$. Therefore, the interval $[a,b]$ is non-empty (containing at least one of $a,b$). Since every interval $[a_i,b_i]$ contains the interval $[a,b]$, their intersection also contains $[a,b]$, hence non-empty. |
Therefore, the limit of the sequence $(a_i)$ exists (see proof here), and is just the supremum, say $a$. Similarly the sequence $(b_i)$ is nonincreasing and bounded from below by all $a_i$, where $i=1,2,\ldots$, and hence has an infimum $b$. Now, as an supremum of $(a_i)$, $a\le b_i$ for all $i$. But $b$ is the infimum of $(b_i)$, $a\le b$. Therefore, the interval $[a,b]$ is non-empty (containing at least one of $a,b$). Since every interval $[a_i,b_i]$ contains the interval $[a,b]$, their intersection also contains $[a,b]$, hence non-empty. |
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| If $c$ is a point outside of $[a,b]$, say $c<a$, then there is some $a_i$, such that $c<a_i$ (by the definition of the supremum $a$), and hence $c\notin [a_i,b_i]$. This shows that the intersection is in fact equal to $[a,b]$. |
If $c$ is a point outside of $[a,b]$, say $c<a$, then there is some $a_i$, such that $c<a_i$ (by the definition of the supremum $a$), and hence $c\notin [a_i,b_i]$. This shows that the intersection is in fact equal to $[a,b]$. |
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Now, if $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$, then $b-a=\displaystyle\lim_{n\to\infty}b_n - \displaystyle\lim_{n\to\infty} a_n = \displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$. So $a=b$. This means that the intersection of the nested intervals contains a single point $a$.
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Now, if $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$, then $b-a=\displaystyle\lim_{n\to\infty}b_n - \displaystyle\lim_{n\to\infty} a_n = \displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$. So that $a=b$. This means that the intersection of the nested intervals contains a single point $a$.
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| \end{proof} |
\end{proof} |
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| 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 all the nested intervals. |
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