Reading my calculus notes I found this proof of the Bolzano-Weierstrass Theorem:
Let $(x_n)$ be a bounded sequence. Choose $a,b\in\mathbb{R}$ such that $x_n\in I_0:=[a,b]$ for all $n\in\mathbb{N}$. Split $I_0$ in halves, say $I'=[a,(a+b)/2]$ and $I''=[(a+b)/2,b]$.
Now, at least one of this intervals contain infinitely many ${x_n}'s$. Name such interval $I_1$ and choose $n_1>1$ such that ${x_n}_1\in I_1$. Note that $|I_1|=\frac{b-a}{2}$.
Suppose that the closed intervals $I_0\supset I_1\supset …\supset I_m$ and natural numbers $n_1<n_2<…<n_m$ have been chosen in a way that, for every $0\leq k\leq m$ $$|I_k|=\frac{b-a}{2^k} ,$$
${x_n}_k\in I_k, x_n\in I_k$for infinitely many $n's$.
To choose $I_{m+1},$ split $I_m=[a_m,b_m]$ in halves, say $I^*=[a_m,(a_m+b_m)/2]$ and $I^{**}=[(a_m+b_m)/2,b_m]$. At least one of this halves contain infinitely many ${x_n}'s$. Name such interval $I_{m+1}$ and choose $n_{m+1}>n_m$ such that ${x_n}_{m+1}\in I_{m+1}$.
Since $|I_{m+1}|=\frac{|I_m|}{2}=\frac{b-a}{2^{m+1}}$, by induction it follows that there is a sequence of non-empty, bounded, and closed nested intervals $(I_k)_{k\in\mathbb{N}}$ that satisfy $$|I_k|=\frac{b-a}{2^k},\forall k\in\mathbb{N}.$$
By the nested intervals property, there is $x\in\bigcap^\infty_{k=1}I_k$.
Thus, $0\leq |{x_n}_k-x|\leq |I_k|\leq \frac{b-a}{2^k}$. Note that $\frac{b-a}{2^k}\to 0$. Using the squeeze theorem we conclude that ${x_n}_k\to x$.
My question is: What allows me to choose the intervals $I_0\supset I_1\supset …\supset I_m$ and the natural numbers $n_1<n_2<…<n_m$? Is it like an induction hypothesis? I think so, but I'm not sure.
Best Answer
It is a construction proof. Nothing allows you to pick those intervals, and indexes. You are constructing them. However you are guaranteed the existence of indexes with the said property($n_1<n_2<...<n_m$) because you always pick the closed interval with infinite points in the sequence.