# alternate statement of Bolzano-Weierstrass theorem

###### Proof.

Let $S\subset\mathbb{R}$ be bounded and infinite. Since $S$ is bounded there exist $a,b\in\mathbb{R}$, with $a, 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

 $[a_{n},b_{n}]\subset[a_{n-1},b_{n-1}]\subset\cdots\subset[a_{1},b_{1}]\subset[% a,b]\text{,}$ (1)

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. Furthermore, for every $m\geq N$, we have $x-\epsilon. In particular, if we select $m\geq N$ such that $l/2^{m}<\epsilon$, then we have

 $x-\epsilon (2)

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 $S$. ∎

Title alternate statement of Bolzano-Weierstrass theorem AlternateStatementOfBolzanoWeierstrassTheorem 2013-03-22 16:40:13 2013-03-22 16:40:13 mathcam (2727) mathcam (2727) 7 mathcam (2727) Theorem msc 26A06 BolzanoWeierstrassTheorem LimitPoint Bounded Infinite