Subsequences, Limit Superior, and Bolzano–Weierstrass

The monotone convergence theorem handles bounded sequences that are also monotone. A general bounded sequence, like ${(- 1)^{n}}$ , need not converge, but its terms accumulate and some of its subsequences converge. The limit superior and inferior are the highest and lowest values the terms cluster near, and the Bolzano–Weierstrass theorem extracts a convergent subsequence from boundedness alone. The latter is among the most used results in the subject.

Subsequences

A subsequence keeps some of the terms in their original order. For example ${1/ (3 i)}_{i = 1}^{\infty}$ is the subsequence of ${1/ n}$ obtained by taking $n_{i} = 3 i$ . Two constraints are essential: the chosen indices must come from the original sequence, and they must strictly increase, so order is preserved. A useful fact proved by induction is that $n_{i} \geq i$ for every $i$ , since the indices are distinct naturals in increasing order.

A convergent subsequence does not force the whole sequence to converge: $0, 1, 0, 1, \dots$ diverges, though its even-indexed subsequence converges to $1$ and its odd-indexed subsequence to $0$ . The two subsequential limits disagree; quantifying that disagreement leads to the limit superior and inferior.

Limit superior and limit inferior

Fix a bounded sequence ${x_{n}}$ . Every tail ${x_{k} : k \geq n}$ is a bounded set, so it has a supremum and an infimum. Tracking these as $n$ grows produces two new, monotone sequences.

Passing from the $n$ -tail to the $(n + 1)$ -tail removes one element, so the supremum can only stay the same or drop and the infimum can only stay the same or rise:

a_{n + 1} \leq a_{n}, b_{n + 1} \geq b_{n} .

Thus ${a_{n}}$ is decreasing and ${b_{n}}$ increasing, and both are bounded (they lie between the bounds of the original sequence). By the monotone convergence theorem both limits exist, so for a bounded sequence the limit superior and inferior always exist, whether or not the sequence converges. The theorem also identifies them as

n \to \infty lim sup x_{n} = n inf a_{n}, n \to \infty lim inf x_{n} = n sup b_{n} .

Since $b_{n} \leq a_{n}$ for every $n$ , the order lemma gives $lim inf x_{n} \leq lim sup x_{n}$ .² Note that ${a_{n}}$ and ${b_{n}}$ are generally not subsequences of ${x_{n}}$ and need not consist of its values.

A bounded sequence that does not converge. Its terms cluster near two levels: the limit superior above and the limit inferior below.

A fully oscillating sequence makes the tail suprema and infima concrete.

Worked example. Let $x_{n} = (- 1)^{n} (1 + \frac{1}{n})$ . Even terms $1 + \frac{1}{n}$ decrease toward $1$ ; odd terms $- (1 + \frac{1}{n})$ increase toward $- 1$ . The tail supremum $a_{n} = sup {x_{k} : k \geq n}$ is the largest even term in the tail, and the tail infimum $b_{n}$ is the most negative odd term:

$n$	$x_{n}$	$a_{n} = sup_{k \geq n} x_{k}$	$b_{n} = inf_{k \geq n} x_{k}$
$1$	$- 2$	$\frac{3}{2}$	$- 2$
$2$	$\frac{3}{2}$	$\frac{3}{2}$	$- \frac{4}{3}$
$3$	$- \frac{4}{3}$	$\frac{5}{4}$	$- \frac{4}{3}$
$4$	$\frac{5}{4}$	$\frac{5}{4}$	$- \frac{6}{5}$
$5$	$- \frac{6}{5}$	$\frac{7}{6}$	$- \frac{6}{5}$
$6$	$\frac{7}{6}$	$\frac{7}{6}$	$- \frac{8}{7}$

The sequence ${a_{n}}$ decreases to $1$ and ${b_{n}}$ increases to $- 1$ , so $lim sup x_{n} = 1$ and $lim inf x_{n} = - 1$ . They differ, and the sequence diverges.

The tail suprema a_n (decreasing envelope) and tail infima b_n (increasing envelope) squeeze the terms; their limits are the upper and lower limits.

Convergence via limit superior and inferior

The two quantities exist for any bounded sequence, and their coincidence detects convergence.

The limit superior and inferior are, respectively, the largest and smallest subsequential limits. Each is attained.

Every subsequential limit is trapped between $lim inf$ and $lim sup$ , and both bounds are achieved, so these two numbers are the exact range of accumulation of the sequence.

The values of a bounded sequence pile up near its accumulation points on the line; between them the terms are sparse. Here two clusters form.

The Bolzano–Weierstrass theorem

The existence of a subsequence attaining the limit superior immediately yields the following theorem.

That one-line proof uses the full limit-superior development. A direct construction by repeated bisection is more transparent and generalizes to $R^{n}$ .

Bolzano–Weierstrass by bisection. Each step keeps a half-interval holding infinitely many terms; the nested intervals shrink to the subsequence's limit.

The same bisection is the root-finding method behind the intermediate value theorem and the standard route to compactness in metric spaces.

Unbounded sequences

Allowing the extended values $\pm \infty$ removes the boundedness restriction. A sequence diverges to infinity if it eventually exceeds every threshold. For an arbitrary sequence one defines

n \to \infty lim sup x_{n} = n inf a_{n}, n \to \infty lim inf x_{n} = n sup b_{n},

now permitting $a_{n} = \infty$ or $b_{n} = - \infty$ . A monotone increasing sequence then always has a limit in the extended sense: either a finite supremum or $\infty$ . For instance, with $x_{n} = 0$ for odd $n$ and $x_{n} = n$ for even $n$ , every tail supremum is $\infty$ and every tail infimum is $0$ , so $lim sup x_{n} = \infty$ while $lim inf x_{n} = 0$ and the ordinary limit does not exist.

Summary

Object	Definition	Always exists?
subsequence ${x_{n_{i}}}$	terms at $n_{1} < n_{2} < \dots$	yes
$lim sup x_{n}$	$lim_{n} sup {x_{k} : k \geq n}$	yes (extended)
$lim inf x_{n}$	$lim_{n} inf {x_{k} : k \geq n}$	yes (extended)
convergent subsequence	Bolzano–Weierstrass	if bounded

For a bounded sequence, $lim inf x_{n} \leq lim sup x_{n}$ , with equality exactly when the sequence converges; the two are the smallest and largest subsequential limits, each attained. Bolzano–Weierstrass extracts a convergent subsequence from any bounded sequence, and its bisection proof is a template used again for compactness.

Lebl, §2.1, Proposition 2.1.17. ↩
Lebl, §2.3, Definition 2.3.1 and Proposition 2.3.2. ↩
Lebl, §2.3, Proposition 2.3.5. ↩
Lebl, §2.3, Exercise 2.3.5. ↩
Lebl, §2.3, Theorem 2.3.4 and Proposition 2.3.6. ↩
Lebl, §2.3, Theorem 2.3.8. ↩
Lebl, §2.3, alternate proof of the Bolzano–Weierstrass theorem. ↩