alternate statement of Bolzano-Weierstrass theoremAlternateStatementOfBolzanoWeierstrassTheorem2007-02-05 20:39:092007-02-06 08:08:00Theoremlimit pointcluster pointaccumulation pointsupremumcompleteleast upper boundcompleteness% this is the default PlanetMath preamble. as your knowledge
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\begin{thm}
Every bounded, infinite set of real numbers has a limit point.
\end{thm}
\begin{proof}
Let $S\subset\mathbb{R}$ be bounded and infinite. Since $S$ is bounded there exist $a,b\in\mathbb{R}$, with $a<b$, such that $S\subset[a,b]$. Let $b-a=l$ and denote the midpoint of the interval $[a,b]$ by $m$. Note that at least one of $[a,m],[m,b]$ must contain infinitely many points of $S$; select an interval satisfying this condition, denoting its left endpoint by $a_1$ and its right endpoint by $b_1$. Continuing this process inductively, for each $n\in\mathbb{N}$, we have an interval $[a_n,b_n]$ satisfying
\begin{equation}
[a_n,b_n]\subset[a_{n-1},b_{n-1}]\subset\cdots\subset[a_1,b_1]\subset[a,b]\text{,}
\end{equation}
where, for each $i\in\mathbb{N}$ such that $1\leq i\leq n$, the interval $[a_i,b_i]$ contains infinitely many points of $S$ and is of length $l/2^i$. Next we note that the set $A=\{a_1,a_2\ldots,a_n\}$ is contained in $[a,b]$, hence is bounded, and as such, has a supremum which we denote by $x$. Now, given $\epsilon>0$, there exists $N\in\mathbb{N}$ such that $x-\epsilon<a_N\leq x$. Furthermore, for every $m\geq N$, we have $x-\epsilon<a_N\leq a_m\leq x$. In particular, if we select $m\geq N$ such that $l/2^m<\epsilon$, then we have
\begin{equation}
x-\epsilon<a_n\leq a_m\leq x\leq b_m=a_m+\dfrac{l}{2^m}<x+\epsilon\text{.}
\end{equation}
Since $[a_m,b_m]\subset(x-\epsilon,x+\epsilon)$ contains infinitely many points of $S$, we may conclude that $x$ is a limit point of $S$.
\end{proof}