Bolzano-Weierstrass Theorem

Boundedness Guarantees a Convergent Subsequence

A sequence can remain within a fixed range without settling toward a single number. For instance, it may keep moving between different parts of that range. The result from the previous tutorial, Convergent Subsequences, showed how selected terms can have their own limits. The Bolzano-Weierstrass Theorem gives a broad guarantee: boundedness alone is enough to ensure that at least one such selection converges.

The theorem promises existence, not a particular limit or a particular set of indices. A bounded sequence may have several convergent subsequences with different limits, and the theorem does not say that the original sequence converges. The detailed proof of the theorem follows in Proof of Bolzano-Weierstrass. First, a finite-partition argument gives a useful way to understand how boundedness can support subsequence selection.

A Finite-Partition Selection Principle

Suppose all terms of a sequence lie in a closed bounded interval. Divide that interval into finitely many smaller closed intervals. Since the smaller intervals cover the original one, at least one must contain terms at infinitely many indices. This does not yet produce a convergent subsequence, but it identifies a region where infinitely many terms can be selected.

Proof. For \(j=0,1,\ldots,r\), define \(x_j=A+j(B-A)/r\). The \(r\) closed intervals are \([x_{j-1},x_j]\), for \(j=1,\ldots,r\). They cover \([A,B]\), including when \(A=B\). For each \(j\), let \(E_j=\{n\in\mathbb{N}_0:a_n\in[x_{j-1},x_j]\}\). Because the intervals cover \([A,B]\), every index belongs to at least one of the sets \(E_j\). Thus

If every \(E_j\) were finite, their finite union would be finite. That would imply that \(\mathbb{N}_0\) is finite, which is false. Therefore at least one \(E_j\) is infinite, proving the claim. \(\square\)

The intervals in the partition share endpoints, so an index whose term is an endpoint may belong to more than one \(E_j\). This causes no difficulty: the union still covers every index, and a finite union of finite sets is still finite. The argument also covers the degenerate case \(A=B\), when each partition interval is the singleton \([A,A]\).

Using the Theorem Without Confusing the Sequence and Its Subsequence

A convergent subsequence describes the behavior of the original sequence at selected indices only. The theorem is therefore a statement about what can be extracted from a bounded sequence, not a claim that all its terms approach the same number. In particular, the result Subsequences of a Convergent Sequence says that every subsequence of a convergent sequence converges to the same limit; Bolzano-Weierstrass provides a convergent subsequence even when the full sequence is not known to converge.

Bounded Sequences Have Cluster Points

Recall from Cluster Points of Sequences that a cluster point of a sequence is a limit of one of its subsequences. The Bolzano-Weierstrass Theorem can therefore be restated as an existence claim about cluster points.

Proof. Let \((a_n)\) be bounded. By the Bolzano-Weierstrass Theorem, there are strictly increasing indices \(n_0<n_1<n_2<\cdots\) such that the subsequence \((a_{n_k})\) converges to some real number \(L\). By the definition of cluster point, \(L\) is a cluster point of \((a_n)\). \(\square\)

This application is often useful when a problem asks whether a sequence has any cluster points, rather than asking for the indices or the value of a particular subsequential limit. The theorem guarantees at least one; it does not rule out additional cluster points, as the preceding example demonstrates.

When a Unique Cluster Point Forces Convergence

The theorem also supports a useful converse-style criterion. A bounded sequence may have several cluster points and fail to converge. But if a bounded sequence has exactly one cluster point, then it must converge to that point. The reason is that any persistent failure to approach the proposed limit would itself yield another convergent subsequence.

Proof. Suppose, for a contradiction, that \(a_n\) does not converge to \(L\). By the Failure-Witness Criterion, there is an \(\varepsilon_0>0\) such that for every \(N\in\mathbb{N}_0\), some \(n\geq N\) satisfies \(|a_n-L|\geq\varepsilon_0\). We can choose such indices successively to obtain \(n_0<n_1<n_2<\cdots\), with \(|a_{n_k}-L|\geq\varepsilon_0\) for every \(k\): after choosing \(n_k\), use the failure witness with \(N=n_k+1\) to choose \(n_{k+1}\).

The subsequence \((a_{n_k})\) is bounded because all its terms are terms of the bounded sequence \((a_n)\). By Bolzano-Weierstrass, it has a convergent subsequence, say \((a_{n_{k_j}})\to M\). Since this is a subsequence of the original sequence, \(M\) is a cluster point of \((a_n)\). The hypothesis that \(L\) is the only cluster point gives \(M=L\).

On the other hand, \(|a_{n_{k_j}}-L|\geq\varepsilon_0\) for every \(j\). Because \(a_{n_{k_j}}\to M\), the Absolute Values Preserve Limits theorem, applied to the sequence \(a_{n_{k_j}}-L\), gives \(|a_{n_{k_j}}-L|\to|M-L|\). Since every term of this magnitude sequence is at least \(\varepsilon_0\), its limit is at least \(\varepsilon_0\). Hence \(|M-L|\geq\varepsilon_0>0\), contradicting \(M=L\). The contradiction proves \(a_n\to L\). \(\square\)

Boundedness is essential to this argument: it is what lets Bolzano-Weierstrass produce a convergent subsequence from the terms that stay at least \(\varepsilon_0\) away from \(L\). Without that extraction, failure to approach \(L\) would not by itself produce another real cluster point.

What the Theorem Does—and Does Not—Say

The Bolzano-Weierstrass Theorem is an existence result. It does not generally tell us how to write down a convergent subsequence, nor does it specify the subsequential limit. The finite-partition argument suggests how one can repeatedly narrow the range containing infinitely many terms; the proof in the next tutorial develops this idea into a complete construction.

Keep the distinction between three claims clear: a sequence is bounded; it has a convergent subsequence; and it converges. The theorem establishes the second claim from the first. It does not establish the third. When the full sequence does converge, all its subsequences share that limit. When it does not, boundedness still guarantees at least one subsequential limit, and different selections may reveal different cluster points.