Bolzano-Weierstrass

Bolzano-Weierstrass

Let $(a_n{)}_{n\in \mathbb{N}}\subset \mathbb{C}$ be a bounded sequence.
Then, $(a_n{)}_{n\in \mathbb{N}}$ has at least one convergent subsequence.
That is, $(a_n{)}_{n\in \mathbb{N}}$ has at least one accumulation point.

Bounded sequences can’t escape to infinity, so the terms must “pile up” somewhere. Bolzano-Weierstrass says you can always find a convergent subsequence hiding inside.

The converse is false

A sequence with a convergent subsequence need not be bounded. E.g. $a_{2n}=n$, $a_{2n-1}=0$: the subsequence $(a_1,a_3,a_5,\dots )=(0,0,0,\dots )$ converges, but the sequence is unbounded.

Proof idea (interval halving)

Assume $0\le a_n\le 1$ (any bounded sequence can be rescaled). Split $[0,1]$ into $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$. At least one half contains infinitely many terms. Pick that half and split again. Repeat.

This produces nested intervals $[a_1,b_1]\supset [a_2,b_2]\supset \dots$ with lengths $b_k-a_k=\frac{1}{2^k}\to 0$, each containing infinitely many terms. The left endpoints $a_k$ form a non-decreasing bounded sequence, the right endpoints $b_k$ a non-increasing bounded sequence, and $b_k-a_k\to 0$, so both converge to the same point $a$ (sandwich rule). From each interval we can pick one term of the sequence, giving a subsequence converging to $a$.

Why boundedness is needed

Without boundedness, terms can spread out forever. $a_n=n$ has no convergent subsequence: every subsequence is also unbounded.