alternate statement of Bolzano-Weierstrass theorem

Theorem.

Every bounded, infinite set of real numbers has a limit point.

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

[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 $x-\epsilon<a_{N}\leq x$ . 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

x-\epsilon<a_{n}\leq a_{m}\leq x\leq b_{m}=a_{m}+\dfrac{l}{2^{m}}<x+\epsilon% \text{.}

(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
Canonical name	AlternateStatementOfBolzanoWeierstrassTheorem
Date of creation	2013-03-22 16:40:13
Last modified on	2013-03-22 16:40:13
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	7
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 26A06
Related topic	BolzanoWeierstrassTheorem
Related topic	LimitPoint
Related topic	Bounded
Related topic	Infinite