NestedIntervalTheorem 2007-08-06 08:11:45 2008-04-30 23:35:15 Theorem real number % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} \newtheorem{prop}{Proposition} \begin{prop} If $$[a_1,\,b_1]\;\supseteq [a_2,\,b_2]\;\supseteq [a_3,\,b_3]\;\supseteq\ldots$$ is a sequence of nested closed intervals, then \begin{align*} \bigcap_{n=1}^\infty [a_n,\,b_n] \neq \varnothing. \end{align*} If also\, $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$,\, then the infinite intersection consists of a unique real number. \end{prop} \begin{proof} There are two consequences to nesting of intervals: $[a_m,\,b_m]\subseteq[a_n,\,b_n]$ for $n\le m$: \begin{enumerate} \item first of all, we have the inequality $a_n\le a_m$ for $n\le m$, which means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ 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 $i=1,2,\ldots$. \end{enumerate} Therefore, the limit of the sequence $(a_i)$ exists, and is just the supremum, say $a$ (see proof \PMlinkname{here}{NondecreasingSequenceWithUpperBound}). 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 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 non-empty (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 non-empty. 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 actually coincides with $[a,b]$. Now, since $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$, we have that $b-a=\displaystyle\lim_{n\to\infty}b_n - \displaystyle\lim_{n\to\infty} a_n = \displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$. So $a=b$. This means that the intersection of the nested intervals contains a single point $a$. \end{proof} \textbf{Remark}. This result is called the \emph{nested interval theorem}. It is a restatement of the \emph{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.