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Bolzano-Weierstrass Theorem — Definition, Formula & Examples

The Bolzano-Weierstrass Theorem states that every bounded sequence of real numbers contains at least one convergent subsequence. In other words, if the terms of a sequence all stay within some finite interval, you can always extract a subsequence that settles down to a limit.

Let (an)(a_n) be a sequence in R\mathbb{R} such that there exists M>0M > 0 with ∣an∣≤M|a_n| \leq M for all n∈Nn \in \mathbb{N}. Then there exists a subsequence (ank)(a_{n_k}) and a point L∈RL \in \mathbb{R} such that lim⁡k→∞ank=L\lim_{k \to \infty} a_{n_k} = L.

How It Works

The classical proof uses repeated bisection. Start with a closed interval [a,b][a, b] that contains all terms of the sequence. Split it into two halves; at least one half contains infinitely many terms of the sequence. Pick that half and repeat. This produces a nested sequence of closed intervals whose lengths shrink to zero, and by the Nested Intervals Property, they share exactly one common point. A subsequence chosen by picking one term from each interval converges to that point.

Worked Example

Problem: Show that the bounded sequence an=(−1)n(1−1n)a_n = (-1)^n \left(1 - \frac{1}{n}\right) has a convergent subsequence.
Step 1: Verify boundedness: Each term satisfies ∣an∣<1|a_n| < 1 since 1−1/n<11 - 1/n < 1 for all n≥1n \geq 1. The sequence is bounded by M=1M = 1.
∣an∣=1−1n<1|a_n| = 1 - \frac{1}{n} < 1
Step 2: Identify a subsequence: Take the subsequence of even-indexed terms: nk=2kn_k = 2k. Then a2k=(−1)2k(1−1/(2k))=1−1/(2k)a_{2k} = (-1)^{2k}(1 - 1/(2k)) = 1 - 1/(2k).
ank=1−12ka_{n_k} = 1 - \frac{1}{2k}
Step 3: Find the limit: As k→∞k \to \infty, 1/(2k)→01/(2k) \to 0, so the subsequence converges.
lim⁡k→∞ank=1\lim_{k \to \infty} a_{n_k} = 1
Answer: The subsequence (a2k)(a_{2k}) converges to 11, confirming the Bolzano-Weierstrass Theorem. (Similarly, the odd-indexed subsequence converges to −1-1.)

Why It Matters

The Bolzano-Weierstrass Theorem is a cornerstone of real analysis and underpins proofs of the Extreme Value Theorem, the Heine-Borel Theorem, and compactness arguments throughout mathematics. In applied fields like optimization and numerical analysis, it guarantees that bounded iterative algorithms have accumulation points, which is essential for proving convergence results.

Common Mistakes

Mistake: Applying the theorem to unbounded sequences, such as an=na_n = n.
Correction: Boundedness is essential. The sequence an=na_n = n has no convergent subsequence because every subsequence diverges to infinity. Always verify that the sequence is bounded before invoking the theorem.