Below is my attempt to prove the topological version of the Bolzano-Weierstrass Theorem. Is it an effective proof? I'd appreciate any comments on it. The book gave a hint to use a nested sequence of half-intervals. The idea is pretty intuitive…I'll explain it if anyone would like me to.
Bolzano-Weierstrass Theorem: "Every bounded, infinite subset of $\mathbb{R}$ has a limit point."
"Let $A$ be a bounded, infinite subset of $\mathbb{R}$. Then since $A$ is bounded, it is a subset of some closed interval $[a,b]$. Take a sequence of half-intervals of $[a,b]$, $\{[a_n,b_n]\}_{n=1}^\infty$ where $[a_1,b_1]=[a,b]$. By Cantor's Nested Intervals Theorem $\displaystyle\bigcap_{n=1}^\infty [a_n,b_n]$ is nonempty and since $\mbox{diam}([a_n,b_n]) \to 0$ as $n \to \infty$, the intersection contains exactly one element, say $p$. Since $p \in \displaystyle\bigcap_{n=1}^\infty [a_n,b_n]$ and every $[a_n,b_n]$ contains infinitely many elements of $A$*, so does $(a_n,b_n)$. Since $\mbox{diam}(a_n,b_n) \to 0$ as $n \to \infty$, every open set containing $p$ also contains some $(a_n,b_n)$, so $p$ is a limit point of $A$."
*Intuitively, if we cram an infinite number of points into a bounded interval, they will be 'dense' in that interval (I haven't formally learned what 'dense' means yet), but how do we prove it?
Best Answer
You’re missing one crucial aspect of dividing the intervals in two at each stage. Let’s say that you currently have $[a_n,b_n]$, and its intersection with $A$ is infinite. Let $c_n$ be the midpoint of $[a_n,b_n]$. Then at least one of the sets $A\cap[a_n,c_n]$ and $A\cap[c_n,b_n]$ must be infinite. If $A\cap[a_n,c_n]$ is infinite, let $a_{n+1}=a_n$ and $b_{n+1}=c_n$; otherwise, let $a_{n+1}=c_n$ and $b_{n+1}=b_n$. Now you know that $A\cap[a_{n+1},b_{n+1}]$ is infinite, and therefore so is $(a_{n+1},b_{n+1})$.
Without some such argument, you can’t in fact be sure that every $[a_n,b_n]$ has an infinite intersection with $A$; in fact, if you make one bad choice, all of your smaller intervals will contain only finitely many points of $A$.
The rest of your argument is fine, however.