Finding Convergence Inside a Bounded Sequence
The Cauchy criterion proved that a real sequence converges when its terms eventually become arbitrarily close to one another. But a bounded sequence need not have that property: its terms may keep moving between separated parts of the real line. The Bolzano-Weierstrass Theorem addresses this by allowing us to keep only selected terms. It guarantees that boundedness is enough to find a convergent subsequence.
The proof uses a selection procedure. Start with a closed interval containing every term. Bisect it and keep a half that contains infinitely many terms; repeat this choice inside each retained interval. The intervals become shorter, while each still contains infinitely many terms. We can therefore select terms with increasing indices from these intervals, producing a Cauchy subsequence. The Cauchy Criterion for Real Sequences then gives convergence.
A subsequence does not need to include every term, and the gaps between selected indices may grow. The theorem guarantees a convergent selection, not convergence of the entire original sequence.
Extracting a Cauchy Subsequence
Proof. Let \((a_n)\) be bounded. There are real numbers \(A\leq B\) such that \(A\leq a_n\leq B\) for every positive integer \(n\). If \(A=B\), then \(a_n=A\) for every \(n\), and the original sequence is Cauchy. Suppose henceforth that \(A<B\), and set \(I_0=[A,B]\).
We construct closed intervals \(I_k\) and infinite sets of indices \(S_k\), for \(k\geq0\). Begin with \(S_0=\{n:a_n\in I_0\}\), which is the set of all positive integers and is infinite. Suppose \(I_k\) has been chosen and \(S_k=\{n:a_n\in I_k\}\) is infinite. Bisect \(I_k\) into its two closed halves. Every term indexed by \(S_k\) belongs to at least one of those halves. If each half contained terms with only finitely many indices from \(S_k\), their union would contain terms with only finitely many such indices, contradicting that \(S_k\) is infinite. Thus at least one half contains terms with infinitely many indices from \(S_k\). Choose such a half as \(I_{k+1}\), and let \(S_{k+1}=\{n:a_n\in I_{k+1}\}\). This ensures \(S_{k+1}\) is infinite and \(I_{k+1}\subseteq I_k\).
At each bisection the interval length is halved. Consequently, if \(L_k\) denotes the length of \(I_k\), then \(L_k=(B-A)/2^k\), which tends to zero. Choose \(n_1\in S_1\). Once \(n_k\in S_k\) has been chosen, choose \(n_{k+1}\in S_{k+1}\) with \(n_{k+1}>n_k\). This is possible because \(S_{k+1}\) is an infinite subset of the positive integers and therefore cannot be bounded above. The indices increase strictly, and \(a_{n_k}\in I_k\) for each \(k\).
To verify the Cauchy condition, let \(\varepsilon>0\). Choose \(m\) so large that \(L_m<\varepsilon\). If \(p,q\geq m\), then \(I_p\subseteq I_m\) and \(I_q\subseteq I_m\). Hence both \(a_{n_p}\) and \(a_{n_q}\) lie in \(I_m\). Any two points in this interval differ by at most its length, so
Thus \((a_{n_k})\) is Cauchy. This proves the lemma. \(\square\)
The construction relies on two separate facts: bisection makes the interval lengths tend to zero, and the choice of a half with infinitely many terms lets us continue selecting indices indefinitely. Keeping a half that contains many terms would not be enough unless it contains infinitely many; a finite collection could run out before an infinite subsequence had been formed.
The Bolzano-Weierstrass Theorem
Proof. Let \((a_n)\) be a bounded real sequence. By the Lemma (Cauchy Subsequence of a Bounded Sequence), it has a Cauchy subsequence \((a_{n_k})\). The Cauchy Criterion for Real Sequences states that every Cauchy real sequence converges. Applying it to \((a_{n_k})\), there is a real number \(L\) such that \(a_{n_k}\to L\). Since \((a_{n_k})\) is a subsequence of the original sequence, this is the required convergent subsequence. \(\square\)
The proof separates the extraction from the convergence. Interval bisection produces terms that are close to one another, and completeness of the real numbers, expressed by the Cauchy Criterion for Real Sequences, turns that Cauchy subsequence into a convergent one. The theorem does not assert that the original sequence converges or that the limit of a selected subsequence is unique.
Worked Applications
Worked Example: A Bounded Sequence with Two Different Subsequence Limits
Define \(a_n=(-1)^n(1+1/n)\). Since \(n\geq1\), we have \(0<1+1/n\leq2\), so \(|a_n|\leq2\) and the sequence is bounded. For even indices \(n=2k\),
For odd indices \(n=2k+1\),
The reciprocal terms tend to zero, so these calculations give two subsequences with different limits. In particular, the full sequence cannot converge: if it converged, every subsequence would converge to the same limit. The example shows why the Bolzano-Weierstrass conclusion is about finding at least one convergent subsequence, not about forcing the original sequence to converge.
Worked Example: A Sequence with Values in a Finite Set
Let \(a_n=1\) when \(n\) is a perfect square, and let \(a_n=0\) otherwise. The sequence is bounded because every term is either \(0\) or \(1\). For the indices \(n_k=k^2+1\), where \(k\geq1\), the index is not a square: \(k^2<k^2+1<(k+1)^2\), since \((k+1)^2-(k^2+1)=2k>0\). Thus \(a_{n_k}=0\) for every \(k\), giving a constant subsequence converging to \(0\).
There is also a subsequence indexed by \(m_k=k^2\), for which \(a_{m_k}=1\) for every \(k\), so that subsequence converges to \(1\). Both index sequences increase strictly. This example illustrates that a bounded sequence may have several subsequential limits; the theorem guarantees existence, not uniqueness.
Worked Example: Fractional Parts of a Sequence
For a real number \(x\), let \(\{x\}=x-\lfloor x\rfloor\) denote its fractional part, where \(\lfloor x\rfloor\) is the greatest integer not exceeding \(x\). The defining property of the floor gives \(0\leq\{x\}<1\). Consider \(a_n=\{n\sqrt{2}\}\). Every term lies in \([0,1)\), so \((a_n)\) is bounded. The Bolzano-Weierstrass Theorem therefore guarantees indices \(n_1<n_2<\cdots\) and a real number \(L\) such that \(a_{n_k}\to L\).
Because \(0\leq a_{n_k}<1\) for every \(k\), order is preserved under limits and gives \(0\leq L\leq1\). This application does not identify the indices or the value of \(L\); it establishes their existence from boundedness alone. The theorem is useful precisely when explicit subsequence calculations are not readily available.
Worked Example: Why Boundedness Is Necessary
The sequence \(a_n=n\) is unbounded. No subsequence can converge to a real number. Indeed, for any subsequence indices \(n_1<n_2<\cdots\), strict increase among positive integers implies \(n_k\geq k\). Given any real \(M\), choose \(K\) with \(K>M\). Then for every \(k\geq K\),
Thus the subsequence eventually exceeds every real bound, which is incompatible with convergence to a real limit. The boundedness hypothesis in Bolzano-Weierstrass cannot simply be omitted.
What the Theorem Does—and Does Not—Say
A common mistake is to replace “has a convergent subsequence” with “converges.” These statements differ: a subsequence can discard terms that prevent the full sequence from settling near one number. The first worked example makes the distinction explicit. When using the theorem, keep track of the selected indices and state the conclusion as convergence of that subsequence.
The theorem is also a key compactness tool. It converts a bound on all terms into the existence of a convergent selection, without requiring monotonicity or a candidate limit. Its proof reveals where the real-number structure matters: intervals can be bisected indefinitely, their lengths can be made arbitrarily small, and Cauchy sequences of real numbers converge. In the next tutorial, this subsequence property will be connected with the Heine-Borel Theorem.
Check Your Understanding
Use the interval construction and the theorem to answer each question.
- Why must the half-interval selected at each stage contain infinitely many terms rather than merely one term?
- How does the length of the interval after \(k\) bisections depend on its original length?
- Why does the selected sequence \((a_{n_k})\) satisfy the Cauchy condition?
- What additional result turns the Cauchy subsequence into a convergent subsequence?
- Does the Bolzano-Weierstrass Theorem imply that every bounded sequence converges? Explain.