Bolzano-Weierstrass theorem

Bolzano-Weierstrass theorem#

Below is a rigorous statement and proof of the Bolzano–Weierstrass theorem:

Theorem (Bolzano–Weierstrass theorem)

Every bounded sequence \((x_n)\subseteq\mathbb{R}\) has a convergent subsequence.

Formally: if \((x_n)\) is bounded, there exist a subsequence \((x_{n_k})\) and a limit \( x^*\in\mathbb{R}\) such that:

\[ \lim_{k\to\infty} x_{n_k} = x^*. \]

Proof. Bolzano–Weierstrass theorem.

The proof of the Bolzano–Weierstrass theorem is based on the completeness property of the real numbers, which states that every bounded sequence has a least upper bound (supremum) and greatest lower bound (infimum).

We use the method of repeated bisection intervals (also known as the “nested interval argument”).

Step 1 (boundedness): Let \((x_n)\subseteq\mathbb{R}\) be a bounded sequence. Then there exist real numbers \(m, M\) with \(m\le x_n\le M\) for all \(n\in\mathbb{N}\).

Define the initial interval:

\[ [a_1,b_1] = [m,M]. \]

This interval contains all terms \(x_n\).

Step 2 (interval bisection): We construct nested intervals by repeatedly halving intervals:

  • Divide the initial interval \([a_1,b_1]\) into two equal halves:

\[ [a_1,c_1], \quad [c_1,b_1] \quad \text{where } c_1 = \frac{a_1+b_1}{2}. \]

At least one of these intervals contains infinitely many terms of the sequence (otherwise, each interval would contain only finitely many terms, contradicting the infinite nature of the sequence).

  • Choose the interval (say \([a_2,b_2]\)) containing infinitely many terms. Clearly, the length of this interval is exactly half the original length:

\[ |b_2 - a_2| = \frac{1}{2}(b_1 - a_1). \]

  • Again, divide this chosen interval into two equal subintervals and pick the one containing infinitely many terms.

We continue indefinitely, obtaining a nested sequence of intervals:

\[ [a_1,b_1]\supseteq[a_2,b_2]\supseteq[a_3,b_3]\supseteq\cdots \]

such that:

  • Each interval contains infinitely many terms of \((x_n)\).

  • The length of each interval shrinks by a factor of \(1/2\) at each step, hence:

\[ |b_k - a_k| = \frac{1}{2^{k-1}}(b_1 - a_1)\quad\rightarrow\quad 0\quad\text{as}\quad k\to\infty. \]

Step 3 (intersection of intervals): By the Nested Interval Property, the intersection of these intervals is nonempty and contains exactly one point. Let:

\[ \{x^*\} = \bigcap_{k=1}^{\infty}[a_k,b_k]. \]

This follows from completeness of \(\mathbb{R}\). Such a point \(x^*\) clearly exists and is unique because interval lengths shrink to 0.

Step 4 (construction of the convergent subsequence): We now construct a subsequence \((x_{n_k})\) converging to \(x^*\):

  • From the interval \([a_1,b_1]\), select a term \(x_{n_1}\).

  • From interval \([a_2,b_2]\), select a term \(x_{n_2}\) with \(n_2>n_1\).

  • Continue this way: from each interval \([a_k,b_k]\), choose a term \(x_{n_k}\) with \(n_k>n_{k-1}\). Such a choice is always possible since each interval contains infinitely many terms.

Thus, by construction, the subsequence \((x_{n_k})\) satisfies:

  • \(x_{n_k}\in[a_k,b_k]\), and the intervals \([a_k,b_k]\) shrink to the single point \(x^*\).

Since the intervals shrink to length zero and each contains \(x_{n_k}\), we must have:

\[ \lim_{k\to\infty} x_{n_k} = x^*. \]

Conclusion:

Thus, we have shown rigorously that every bounded sequence in \(\mathbb{R}\) has a convergent subsequence. This completes the proof of the Bolzano–Weierstrass theorem.