Bolzano-Weierstrass Theorem
The Bolzano-Weierstrass theorem states the following:
Suppose {$S$} is bounded and infinite ({$S \neq \emptyset$}) Then {$S^* \neq \emptyset$}. In other words, Every Bounded Infinte Set Has a Point of Accumulation.
We can prove this as follows: Since {$S$} is bounded, we can certainly say that {$ S \subset [a,b] $}. Let {$ c $} be the midpoint. Consider {$ S \cap [a,c] $}, and ${ S \cap [c,b] $}. Consider also that {$ S = S \cap [a,c] \cup S \cap [c,b] $}.
The key remark is that one of these last two sets must be infinite, as S is an infinite sequence. The union of two finitely indexed sets is a finite set, so at least one must be infinite.
For the sake of argument, let’s assume {$ [a,c] $} is infinte. {$d$} is the midpoint of {$ [a,c] $} Consider {$S \cap [a,d] $} and {$S \cap [d,c] $}. Because {$S$} is infinite, one of these two splits also must be infinite.
Continue this process ad infinitum. Let’s formalize this:
The process generates closed intervals {$I_1 \superset I_2 \superset I_3, $} … etc, such that {$\forall n$}, ${I_n \cap S$}, {$I_n \rightarrow 0$}.
Now, apply the nested interval principle. We get that the intersection of all ${I_n}$ is a single point *, let’s call it {$x$}. ${x \in I_n \forall n$}.
The claim is that {$x \in S^*$}. We must show that {$N_r^*(x) \cap S =\neq 0 \forall r > 0$}.
Somewhere in the interval is x. Consider the neighborhood around it. The neighborhood surrounding it must contain {$I_n$} for some {$n$}, since {$I_n$} tends to length zero. Because all {$I_n \subset S$}, we have satisfied this. QED.
The BWT is now going to be used to prove the last step of ht emain theory in this class. Summary of BWT: Bounded infinite sets have points of accumulation.