Bolzano-Weierstrass and Bounded Sequences

What Boundedness Reveals About Subsequences

The Bolzano-Weierstrass Theorem states that every bounded real sequence has a convergent subsequence. That existence result does not, by itself, describe the possible subsequential limits or tell us whether the whole sequence converges. We can obtain more information by looking at the terms remaining after each index: the largest possible tail value is recorded by a supremum, and the smallest by an infimum.

These tail bounds move in only one direction as the sequence progresses. Their limits give the upper and lower limits of the original sequence. We will show that both are subsequential limits, that no subsequential limit can lie outside the interval they determine, and that the original sequence converges exactly when the two limits agree. The argument uses the Bolzano-Weierstrass Theorem and the Monotone Convergence Theorem established earlier in this course.

Tail Suprema, Tail Infima, and Their Limits

Let \((a_n)_{n=0}^{\infty}\) be bounded. For each \(n\), consider the tail \(\{a_k:k\geq n\}\). It is nonempty and bounded, so it has a supremum and an infimum by completeness of the real numbers. These bounds depend on \(n\):

The tail beginning at \(n+1\) is contained in the tail beginning at \(n\). Consequently, \(s_{n+1}\leq s_n\), while \(i_{n+1}\geq i_n\). Also, every \(i_n\) and \(s_n\) lies between any global lower and upper bounds for the sequence. Thus \((s_n)\) is nonincreasing and bounded, and \((i_n)\) is nondecreasing and bounded. The Monotone Convergence Theorem gives finite limits for both.

Write \(U=\limsup_{n\to\infty}a_n\) and \(I=\liminf_{n\to\infty}a_n\). Since \(i_n\leq s_n\) for every \(n\), order preservation for limits gives \(I\leq U\). These numbers describe the extreme values that can persist arbitrarily far out in the sequence. A large term occurring only among the first few terms can affect an early tail supremum, but it cannot keep the tail suprema high forever.

The Tail Limits Are Subsequential Limits

Proof. We first construct a subsequence converging to \(U\). For each \(j\in\mathbb{N}_0\), set \(m_0=0\) and, once \(n_{j-1}\) has been chosen, set \(m_j=n_{j-1}+1\) for \(j\geq1\). By the Approximation Property of the Supremum and Infimum, there is an index \(n_j\geq m_j\) such that

Indeed, \(s_{m_j}\) is the supremum of the nonempty tail \(\{a_k:k\geq m_j\}\), so some term in that tail is within \(1/(j+1)\) of the supremum from below; every term in the tail is at most the supremum. Since \(n_j\geq m_j=n_{j-1}+1\) when \(j\geq1\), the chosen indices are strictly increasing. They therefore define a subsequence.

As \(m_j\to\infty\), the convergence \(s_n\to U\) gives \(s_{m_j}\to U\). The displayed inequalities show that \(0\leq s_{m_j}-a_{n_j}<1/(j+1)\). Hence

The right side tends to zero, so \(a_{n_j}\to U\).

For the lower limit, choose indices \(q_j\geq r_j\), where \(r_0=0\) and \(r_j=q_{j-1}+1\) for \(j\geq1\), using the approximation property of the infimum to ensure

These indices are strictly increasing. Since \(r_j\to\infty\), we have \(i_{r_j}\to I\), and

Thus \(a_{q_j}\to I\) as well. Both limits are subsequential limits, as claimed. \(\square\)

This result sharpens the existence guarantee in the Bolzano-Weierstrass Theorem: the upper and lower limits are not merely bounds but are actually approached along subsequences. In particular, if \(I=U\), the two constructed subsequences approach the same number. The next theorem explains why that equality forces the entire sequence to converge.

Convergence and Equality of the Tail Limits

Proof. Suppose first that \(a_n\to L\). Let \(\varepsilon>0\). There is an \(N\) such that \(|a_k-L|<\varepsilon\) whenever \(k\geq N\). For every \(n\geq N\), all terms in the tail beginning at \(n\) satisfy \(L-\varepsilon<a_k<L+\varepsilon\). It follows that \(i_n\geq L-\varepsilon\) and \(s_n\leq L+\varepsilon\). The tail is nonempty, and each of its terms also lies in this interval, so its infimum cannot exceed \(L+\varepsilon\) and its supremum cannot be below \(L-\varepsilon\). Therefore

Taking limits gives \(I=L=U\), since \(\varepsilon\) was arbitrary.

Conversely, suppose \(I=U=L\). For every \(n\), the definition of infimum and supremum gives \(i_n\leq a_n\leq s_n\). We already know \(i_n\to L\) and \(s_n\to L\). The Squeeze Theorem therefore gives \(a_n\to L\). This proves both directions. \(\square\)

The theorem gives a useful test for convergence that does not begin by guessing a limit. Instead, it compares the long-term upper and lower bounds of the tails. If they coincide, every term is eventually trapped between quantities approaching the same number. If they differ, the theorem also shows that the sequence cannot converge.

Worked Examples

How to Interpret the Bounds

For any subsequence that converges to a finite number \(x\), the eventual bounds \(i_n\leq a_n\leq s_n\) imply \(I\leq x\leq U\). In fact, this follows by applying order preservation for limits along the subsequence: its indices tend to infinity, so the corresponding tail bounds still approach \(I\) and \(U\). Thus the upper and lower limits bound every subsequential limit, and the theorem above shows that both endpoints themselves occur as subsequential limits.

A common pitfall is to treat boundedness as if it guaranteed convergence. The sequence in the first worked example is bounded, yet its even and odd terms approach different numbers. Bolzano-Weierstrass guarantees at least one convergent subsequence; the tail-limit approach gives more: it identifies two extreme subsequential limits and tells us whether they coincide. When they do, the squeeze argument establishes convergence of the whole sequence. When they do not, the two distinct subsequential limits rule out convergence.