Bolzano–Weierstrass Theorem — Mathematics Knowledge System

Overview

The Bolzano–Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. The sequence itself need not converge — it may oscillate forever, like aₙ = (-1)ⁿ — but boundedness alone guarantees that some subsequence settles down. The standard proof uses repeated bisection: halve an interval containing infinitely many terms, keep the half that still contains infinitely many terms, and iterate, producing nested intervals whose lengths shrink to 0 and which (by completeness of ℝ) pin down a single point that a subsequence converges to. The theorem is foundational for compactness arguments in ℝⁿ and underlies proofs of the Extreme Value Theorem and existence results throughout analysis.

Intuition

Think of a bounded sequence as infinitely many darts all landing inside a fixed finite interval. With infinitely many darts crammed into a finite region, they must pile up densely somewhere — some sub-region, however small, has to contain infinitely many of them (a pigeonhole-style argument). Repeatedly zooming into the sub-region that still holds infinitely many darts (bisection) traps a single accumulation point, and picking one dart from each successive sub-region builds a subsequence converging to it. The theorem is really a statement about compactness: bounded closed sets in ℝ cannot let infinitely many points 'escape to infinity' or 'escape through a gap,' so they must have an accumulation point.

Formal Definition

Definition

For a sequence of real numbers:

(a_n) \text{ bounded} \implies \exists \text{ subsequence } (a_{n_k}) \text{ that converges}

Bolzano–Weierstrass theorem

\text{A set } K \subseteq \mathbb{R}^n \text{ is compact} \iff K \text{ is closed and bounded (Heine–Borel)}

Heine–Borel characterization, the higher-dimensional generalization

\text{every sequence in a compact set has a subsequence converging to a point of that set}

Sequential compactness (equivalent formulation)

Worked Examples

The sequence is -1,1,-1,1,... which is bounded (between -1 and 1) but never converges as a whole sequence.
$a_n = (-1)^n$
The subsequence of even-indexed terms is constant: a₂=1, a₄=1, a₆=1, ..., which trivially converges to 1.
$a_{2k} = 1 \to 1$

Answer: The subsequence (a₂ₖ) = 1,1,1,... converges to 1, illustrating Bolzano–Weierstrass even though (aₙ) itself diverges.

Practice Problems

Difficulty 4/10

Find two different convergent subsequences of aₙ = sin(nπ/2), and state their limits.

Difficulty 4/10

Bolzano–Weierstrass guarantees a convergent subsequence for any sequence that is:

MonotoneBoundedCauchyContinuous

Difficulty 7/10

Use Bolzano–Weierstrass to prove that every Cauchy sequence of real numbers converges (an alternative route to completeness).

Quiz

The Bolzano–Weierstrass theorem states that every bounded sequence of real numbers:

ConvergesHas a convergent subsequenceIs monotoneIs Cauchy

The proof of Bolzano–Weierstrass by repeated bisection ultimately relies on which property of ℝ?

The Archimedean property aloneCompleteness (nested interval property)Countability of ℝThe distributive law

Summary

Every bounded sequence of real numbers has a convergent subsequence, even if the full sequence diverges (e.g. (-1)ⁿ).
The standard proof uses repeated bisection, producing nested shrinking intervals that pin down an accumulation point via completeness.
The theorem generalizes to compact sets via the Heine–Borel characterization and underlies existence proofs like the Extreme Value Theorem.

References

WebsiteWikipedia — Bolzano–Weierstrass theorem