alternate statement of Bolzano-Weierstrass theorem

Theorem.

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

Proof.

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

$$[a_n,b_n]\subset [a_{n-1},b_{n-1}]\subset \mathrm{\cdots }\subset [a_1,b_1]\subset [a,b]\text{,}$$(1)

where, for each $i\in \mathbb{N}$ such that $1\le i\le n$, 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 $A=\{a_1,a_2\mathrm{\dots },a_n\}$ is contained in $[a,b]$, hence is bounded, and as such, has a supremum which we denote by $x$. Now, given $ϵ>0$, there exists $N\in \mathbb{N}$ such that . Furthermore, for every $m\ge N$, we have . In particular, if we select $m\ge N$ such that , then we have

(2)

Since $[a_m,b_m]\subset (x-ϵ,x+ϵ)$ contains infinitely many points of $S$, we may conclude that $x$ is a limit point of $S$.∎