Proof of Bolzano-Weierstrass

Turning Boundedness into a Convergent Subsequence

The Bolzano-Weierstrass Theorem, stated in the previous tutorial, says that every bounded real sequence has a convergent subsequence. To prove it, we need to turn the broad information that all terms lie in one bounded interval into progressively tighter control over selected terms. Repeated bisection does this: at each stage, retain a half-interval that contains terms at infinitely many indices. The intervals become narrower, while the infinitely many available indices allow us to choose a strictly increasing sequence of them.

The proof depends on two distinct ideas. First, if finitely many sets together contain every nonnegative integer, at least one of those sets must be infinite. Second, nested closed intervals whose lengths tend to zero determine a single real number. We will prove the interval fact using the least-upper-bound property of the real numbers, then use it to complete the subsequence construction.

A Nested-Interval Lemma

Proof. Since the intervals are nested, their left endpoints are nondecreasing: \(u_k\leq u_{k+1}\). Also, every left endpoint is at most \(v_0\), because \(u_k\in I_k\subseteq I_0\). The set \(\{u_k:k\in\mathbb{N}_0\}\) is therefore nonempty and bounded above. By completeness of the real numbers, it has a supremum; set \(L=\sup\{u_k:k\in\mathbb{N}_0\}\).

Fix an index \(m\). We have \(u_m\leq L\), since \(u_m\) belongs to the set whose supremum is \(L\). We also claim that \(L\leq v_m\). For any \(k\geq m\), nesting gives \(u_k\in I_k\subseteq I_m\), so \(u_k\leq v_m\). For \(k<m\), the nondecreasing property gives \(u_k\leq u_m\leq v_m\). Thus \(v_m\) is an upper bound for every left endpoint, and the definition of supremum gives \(L\leq v_m\). We have proved \(u_m\leq L\leq v_m\), so \(L\in I_m\). Since \(m\) was arbitrary, \(L\) belongs to every interval.

To prove uniqueness, suppose \(L'\) also belongs to every \(I_k\). Then \(L,L'\in[u_k,v_k]\), so \(|L-L'|\leq v_k-u_k\) for every \(k\). The right side tends to zero. If \(|L-L'|>0\), then eventually \(v_k-u_k<|L-L'|\), a contradiction. Hence \(|L-L'|=0\), and \(L=L'\). \(\square\)

The lemma is a completeness statement in a form suited to subsequences. The nested intervals do not merely get small: completeness ensures that their common location is an actual real number. The vanishing lengths then make every point chosen from the \(k\)-th interval close to that number.

The Bisection Construction

We now prove the theorem. Start with a bounded sequence \((a_n)_{n=0}^{\infty}\). By the definition of boundedness, there is a \(C\geq0\) such that \(|a_n|\leq C\) for every \(n\). Consequently \(a_n\in[-C,C]\). More generally, the argument works whenever all terms lie in a closed bounded interval \([A,B]\), so use that notation in the construction.

Proof. Let \(I_0=[A,B]\) contain every term of the sequence. We construct intervals \(I_k\) and infinite index sets \(E_k\) recursively. At stage \(k\), let \(I_k=[u_k,v_k]\), and suppose infinitely many indices \(n\) satisfy \(a_n\in I_k\). Bisect \(I_k\) at its midpoint \(m_k=(u_k+v_k)/2\), obtaining the two closed intervals \([u_k,m_k]\) and \([m_k,v_k]\). They cover \(I_k\), including the midpoint. At least one of them contains \(a_n\) for infinitely many indices: otherwise each half would contain terms at only finitely many indices, and their union would also contain terms at only finitely many indices, contrary to the choice of \(I_k\). Choose such a half as \(I_{k+1}\).

This recursive choice begins at \(I_0\), which contains every term and therefore contains terms at infinitely many indices. At every stage it retains an interval containing terms at infinitely many indices. By construction, \(I_{k+1}\subseteq I_k\), and its length is half the length of \(I_k\). Thus

This formula also holds if \(A=B\): the interval has length zero, and its two halves coincide with it. In all cases, the lengths tend to zero. The Nested Intervals with Vanishing Length Lemma gives a real number \(L\) that belongs to every \(I_k\).

For each \(k\), define \(E_k=\{n\in\mathbb{N}_0:a_n\in I_k\}\). Each \(E_k\) is infinite. Choose \(n_0\in E_0\). Having chosen \(n_k\), choose \(n_{k+1}\in E_{k+1}\) with \(n_{k+1}>n_k\). This is possible because every infinite subset of \(\mathbb{N}_0\) is unbounded: if it were bounded, it would be contained in a finite set of nonnegative integers and would itself be finite. The resulting indices are strictly increasing, so \((a_{n_k})\) is a subsequence.

Both \(a_{n_k}\) and \(L\) belong to \(I_k=[u_k,v_k]\). The distance between two points in this interval is at most its length, giving

Since the right side tends to zero, \(a_{n_k}\to L\). For example, given \(\varepsilon>0\), choose \(K\) large enough that \((B-A)/2^K<\varepsilon\). For every \(k\geq K\), the displayed bound gives \(|a_{n_k}-L|\leq(B-A)/2^k\leq(B-A)/2^K<\varepsilon\). Thus the subsequence converges to \(L\), proving the theorem. \(\square\)

Worked Examples: What the Construction Gives

Why the Details of the Proof Matter

Two steps are easy to blur together but serve different purposes. Keeping an interval with infinitely many term indices makes it possible to continue the construction and later choose an index beyond the one already selected. Making the interval lengths tend to zero makes the selected terms converge to a single point. Either feature by itself is insufficient for this proof: infinitely many terms in one fixed interval do not force their values to approach one number, and shrinking intervals that contain only finitely many terms may not allow a subsequence to be built.

The closed halves overlap at their common endpoint. An index whose term equals that endpoint can belong to both associated index sets. This causes no problem: the halves still cover the parent interval, and if both halves contained terms at only finitely many indices, their union could contain terms at only finitely many indices. The same reasoning also covers a degenerate initial interval \([A,A]\).

Finally, the limit \(L\) is obtained from the nested intervals, not assumed in advance. Completeness supplies it through the supremum of the left endpoints. Once \(L\) is known to lie in every interval, the length estimate gives the epsilon-N convergence proof. This is the central technique: use boundedness to repeatedly localize infinitely many terms, and use completeness to turn that localization into a limit.