---
title: Subsequences, Limit Superior, and Bolzano–Weierstrass
module: Sequences and Series
moduleNumber: 2
lessonNumber: 3
order: 203
summary: >
  A bounded sequence need not converge, but it always has convergent
  subsequences, and its terms cluster between two extreme values. The limit
  superior and inferior are the limits of the tail suprema and infima; they
  always exist for a bounded sequence, coincide exactly when it converges, and
  are its largest and smallest subsequential limits. Bolzano–Weierstrass
  extracts a convergent subsequence from boundedness alone.
topics: [Sequences and Series]
sources:
  - book: Lebl
    ref: "§2.1 Subsequences; §2.3 Limit superior, limit inferior, and Bolzano–Weierstrass"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

The [monotone convergence theorem](/real-analysis/sequences-series/limit-laws-monotone)
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

> **Definition (Subsequence).** Let $\{x_n\}_{n=1}^\infty$ be a sequence and let
> $\{n_i\}_{i=1}^\infty$ be a strictly increasing sequence of natural numbers,
> $n_1 < n_2 < n_3 < \cdots$. The sequence $\{x_{n_i}\}_{i=1}^\infty$ is a
> **subsequence** of $\{x_n\}$.

A subsequence keeps some of the terms in their original order. For example
$\{1/(3i)\}_{i=1}^\infty$ is the subsequence of $\{1/n\}$ obtained by taking
$n_i = 3i$. 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 $n_i \ge i$ for every $i$, since the
indices are distinct naturals in increasing order.

> **Proposition.** If $\{x_n\}$ converges to $x$, then every subsequence
> converges to $x$.

> **Proof.** Given $\epsilon > 0$, take $M$ with $|x_n - x| < \epsilon$ for
> $n \ge M$. Since $n_i \ge i$, any $i \ge M$ gives $n_i \ge M$, hence
> $|x_{n_i} - x| < \epsilon$.[^lebl-sub] $\square$

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 $\{x_k : k \ge n\}$ is a bounded
set, so it has a supremum and an infimum. Tracking these as $n$ grows produces
two new, monotone sequences.

> **Definition (Limit superior and inferior).** For a bounded sequence
> $\{x_n\}$, set $a_n = \sup\{x_k : k \ge n\}$ and $b_n = \inf\{x_k : k \ge
> n\}$. Define
> $$
> \limsup_{n\to\infty} x_n = \lim_{n\to\infty} a_n, \qquad
> \liminf_{n\to\infty} x_n = \lim_{n\to\infty} b_n.
> $$

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:

$$
a_{n+1} \le a_n, \qquad b_{n+1} \ge b_n.
$$

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

$$
\limsup_{n\to\infty} x_n = \inf_n a_n, \qquad \liminf_{n\to\infty} x_n = \sup_n b_n.
$$

Since $b_n \le a_n$ for every $n$, the order lemma gives $\liminf x_n \le
\limsup x_n$.[^lebl-ls] Note that $\{a_n\}$ and $\{b_n\}$ are generally not
subsequences of $\{x_n\}$ and need not consist of its values.

$$
% caption: A bounded sequence that does not converge. Its terms cluster near two
% levels: the limit superior above and the limit inferior below.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.8,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,4.2);
\draw[acc, thick] (0,3.0) -- (9.8,3.0) node[right, black] {upper limit};
\draw[acc, thick] (0,1.0) -- (9.8,1.0) node[right, black] {lower limit};
\foreach \n/\y in {1/0.4, 2/3.6, 3/0.7, 4/3.3, 5/0.85, 6/3.15, 7/0.92, 8/3.08, 9/0.96, 10/3.04}
  \fill[black] (\n*0.9,\y) circle (2.2pt);
\end{tikzpicture}
$$

> **Worked example.** Let $x_n = \frac{n+1}{n}$ for odd $n$ and $x_n = 0$ for
> even $n$. The infimum of every tail is $0$, so $\liminf x_n = 0$. The supremum
> of the $n$-tail is $\frac{n+1}{n}$ or $\frac{n+2}{n+1}$ depending on parity,
> each tending to $1$, so $\limsup x_n = 1$. The sequence does not converge,
> consistent with $\liminf \ne \limsup$.

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

> **Worked example.** Let $x_n = (-1)^n\bigl(1 + \tfrac{1}{n}\bigr)$. Even terms
> $1 + \tfrac{1}{n}$ decrease toward $1$; odd terms $-\bigl(1 + \tfrac{1}{n}\bigr)$
> increase toward $-1$. The tail supremum $a_n = \sup\{x_k : k \ge n\}$ is the
> largest even term in the tail, and the tail infimum $b_n$ is the most negative
> odd term:
>
> | $n$ | $x_n$ | $a_n = \sup_{k \ge n} x_k$ | $b_n = \inf_{k \ge n} x_k$ |
> | --- | --- | --- | --- |
> | $1$ | $-2$ | $\tfrac{3}{2}$ | $-2$ |
> | $2$ | $\tfrac{3}{2}$ | $\tfrac{3}{2}$ | $-\tfrac{4}{3}$ |
> | $3$ | $-\tfrac{4}{3}$ | $\tfrac{5}{4}$ | $-\tfrac{4}{3}$ |
> | $4$ | $\tfrac{5}{4}$ | $\tfrac{5}{4}$ | $-\tfrac{6}{5}$ |
> | $5$ | $-\tfrac{6}{5}$ | $\tfrac{7}{6}$ | $-\tfrac{6}{5}$ |
> | $6$ | $\tfrac{7}{6}$ | $\tfrac{7}{6}$ | $-\tfrac{8}{7}$ |
>
> The sequence $\{a_n\}$ decreases to $1$ and $\{b_n\}$ increases to $-1$, so
> $\limsup x_n = 1$ and $\liminf x_n = -1$. They differ, and the sequence
> diverges.

$$
% caption: The tail suprema a_n (decreasing envelope) and tail infima b_n
% (increasing envelope) squeeze the terms; their limits are the upper and lower limits.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.8,0) node[right, black] {$n$};
\draw[black] (0,-1.9) -- (0,1.9);
% terms: even up near +1, odd down near -1, shrinking amplitude
\foreach \n/\y in {1/-1.7, 2/1.55, 3/-1.35, 4/1.28, 5/-1.2, 6/1.17, 7/-1.12, 8/1.1}
  \fill[black] (\n*1.1,\y) circle (2.0pt);
% upper envelope a_n (decreasing toward +1)
\draw[black, dashed] (1.1,1.55) -- (2.2,1.55) -- (3.3,1.28) -- (4.4,1.28) -- (5.5,1.17) -- (6.6,1.17) -- (7.7,1.1) -- (8.8,1.1);
% lower envelope b_n (increasing toward -1)
\draw[black, dashed] (1.1,-1.7) -- (2.2,-1.35) -- (3.3,-1.35) -- (4.4,-1.2) -- (5.5,-1.2) -- (6.6,-1.12) -- (7.7,-1.12) -- (8.8,-1.1);
\draw[acc, thick] (0,1.0) -- (9.8,1.0) node[right, black] {upper limit};
\draw[acc, thick] (0,-1.0) -- (9.8,-1.0) node[right, black] {lower limit};
\node[black, anchor=south] at (8.8,1.12) {$a_n$};
\node[black, anchor=north] at (8.8,-1.12) {$b_n$};
\end{tikzpicture}
$$

## Convergence via limit superior and inferior

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

> **Proposition.** A bounded sequence $\{x_n\}$ converges if and only if
> $\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n$. When it converges, the
> common value is the limit.

> **Proof.** From $b_n \le x_n \le a_n$: if $\liminf = \limsup$, then $\{a_n\}$
> and $\{b_n\}$ share a limit and the squeeze lemma forces $\{x_n\}$ to it.
> Conversely, if $x_n \to x$, every subsequence converges to $x$; in particular
> the subsequences realizing $\limsup$ and $\liminf$ (constructed next) converge
> to $x$, so both equal $x$.[^lebl-lsc] $\square$

> **Worked example.** Let $x_n = \frac{(-1)^n}{n}$: the terms
> $-1, \tfrac{1}{2}, -\tfrac{1}{3}, \tfrac{1}{4}, \dots$ alternate sign while
> shrinking. The tail supremum $a_n$ is the largest even-indexed term at or past
> $n$, so $a_n = \tfrac{1}{n}$ for even $n$ and $a_n = \tfrac{1}{n+1}$ for odd
> $n$; either way $a_n \to 0$, so $\limsup x_n = 0$. Symmetrically the tail
> infimum is $-\tfrac{1}{n}$ or $-\tfrac{1}{n+1}$, so $\liminf x_n = 0$. The two
> agree, and the criterion confirms $x_n \to 0$.[^lebl-ex235]

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

> **Theorem.** If $\{x_n\}$ is bounded, there is a subsequence converging to
> $\limsup_{n\to\infty} x_n$ and a (possibly different) subsequence converging
> to $\liminf_{n\to\infty} x_n$. Moreover, for any convergent subsequence
> $\{x_{n_k}\}$,
> $$
> \liminf_{n\to\infty} x_n \le \lim_{k\to\infty} x_{n_k} \le \limsup_{n\to\infty} x_n.
> $$

> **Proof (sketch, for the limit superior).** Write $x = \limsup x_n = \lim
> a_n$. Build the subsequence inductively: having chosen $n_{k-1}$, use that
> $a_{n_{k-1}+1}$ is a supremum to find an index $n_k > n_{k-1}$ with
> $a_{n_{k-1}+1} - x_{n_k} < 1/k$. Then $|a_{n_k} - x_{n_k}| < 1/k$, and since
> $a_{n_k} \to x$, the triangle inequality gives $x_{n_k} \to x$.[^lebl-attain]
> $\square$

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

$$
% caption: The values of a bounded sequence pile up near its accumulation points
% on the line; between them the terms are sparse. Here two clusters form.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10,0);
% cluster A near x=2, cluster B near x=8
\foreach \x in {1.4,1.7,1.85,1.93,1.97,2.02,2.06,2.13,2.28,2.55}
  \fill[black] (\x,0) circle (2.0pt);
\foreach \x in {7.4,7.7,7.85,7.93,7.97,8.02,8.06,8.13,8.28,8.55}
  \fill[black] (\x,0) circle (2.0pt);
% sparse strays in the middle
\foreach \x in {3.6,5.0,6.2}
  \fill[black] (\x,0) circle (1.6pt);
\draw[acc, thick] (2.0,-0.55) -- (2.0,0.55);
\draw[acc, thick] (8.0,-0.55) -- (8.0,0.55);
\node[acc, anchor=north] at (2.0,-0.6) {cluster};
\node[acc, anchor=north] at (8.0,-0.6) {cluster};
\end{tikzpicture}
$$

## The Bolzano–Weierstrass theorem

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

> **Theorem (Bolzano–Weierstrass).** Every bounded sequence of real numbers has
> a convergent subsequence.

> **Proof.** By the previous theorem there is a subsequence converging to
> $\limsup_{n\to\infty} x_n$.[^lebl-bw] $\square$

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

> **Proof (by bisection).** Since $\{x_n\}$ is bounded, fix $a_1 < b_1$ with
> $a_1 \le x_n \le b_1$ for all $n$. The interval $[a_1, b_1]$ contains $x_n$
> for infinitely many $n$ — indeed for all of them. Suppose $[a_k, b_k]$ has
> been chosen to contain infinitely many terms. Cut it at its midpoint $y =
> \frac{a_k + b_k}{2}$. At least one of the halves $[a_k, y]$, $[y, b_k]$
> contains $x_n$ for infinitely many $n$; take that half as $[a_{k+1}, b_{k+1}]$
> and pick an index $n_{k+1} > n_k$ with $x_{n_{k+1}}$ in it. The interval
> lengths satisfy
> $$
> b_{k} - a_{k} = \frac{b_1 - a_1}{2^{k-1}} \longrightarrow 0,
> $$
> so $\{a_k\}$ (increasing) and $\{b_k\}$ (decreasing) converge to a common
> point $x$, and $a_k \le x_{n_k} \le b_k$ squeezes the subsequence to
> $x$.[^lebl-bw2] $\square$

$$
% caption: Bolzano–Weierstrass by bisection. Each step keeps a half-interval
% holding infinitely many terms; the nested intervals shrink to the subsequence's limit.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% four nested intervals, each half the previous, kept alternately
\draw[black, thick] (0,3) -- (8,3);
\node[black, anchor=east] at (-0.15,3) {stage 1};
\draw[black, thick] (4,2) -- (8,2);
\node[black, anchor=east] at (-0.15,2) {stage 2};
\draw[black, thick] (4,1) -- (6,1);
\node[black, anchor=east] at (-0.15,1) {stage 3};
\draw[black, thick] (5,0) -- (6,0);
\node[black, anchor=east] at (-0.15,0) {stage 4};
% endpoints ticks
\foreach \yy/\xa/\xb in {3/0/8, 2/4/8, 1/4/6, 0/5/6} {
  \draw[black] (\xa,\yy+0.13) -- (\xa,\yy-0.13);
  \draw[black] (\xb,\yy+0.13) -- (\xb,\yy-0.13);
}
% converging point
\draw[acc, dashed] (5.5,3.3) -- (5.5,-0.4);
\fill[acc] (5.5,0) circle (2.2pt);
\node[acc, anchor=north] at (5.5,-0.45) {limit};
\end{tikzpicture}
$$

The same bisection is the root-finding method behind the intermediate value
theorem and the standard route to
[compactness](/real-analysis/metric-spaces/compactness) in metric spaces.

## Unbounded sequences

Allowing the extended values $\pm\infty$ removes the boundedness restriction. A
sequence
[**diverges to infinity**](/real-analysis/sequences-series/sequences-limits)
if it eventually exceeds every threshold. For an
arbitrary sequence one defines

$$
\limsup_{n\to\infty} x_n = \inf_n a_n, \qquad \liminf_{n\to\infty} x_n = \sup_n b_n,
$$

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 $\limsup x_n =
\infty$ while $\liminf x_n = 0$ and the ordinary limit does not exist.

## Summary

| Object | Definition | Always exists? |
| --- | --- | --- |
| subsequence $\{x_{n_i}\}$ | terms at $n_1 < n_2 < \cdots$ | yes |
| $\limsup x_n$ | $\lim_n \sup\{x_k : k \ge n\}$ | yes (extended) |
| $\liminf x_n$ | $\lim_n \inf\{x_k : k \ge n\}$ | yes (extended) |
| convergent subsequence | Bolzano–Weierstrass | if bounded |

For a bounded sequence, $\liminf x_n \le \limsup x_n$, 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-sub]: Lebl, §2.1, Proposition 2.1.17.
[^lebl-ls]: Lebl, §2.3, Definition 2.3.1 and Proposition 2.3.2.
[^lebl-lsc]: Lebl, §2.3, Proposition 2.3.5.
[^lebl-ex235]: Lebl, §2.3, Exercise 2.3.5.
[^lebl-attain]: Lebl, §2.3, Theorem 2.3.4 and Proposition 2.3.6.
[^lebl-bw]: Lebl, §2.3, Theorem 2.3.8.
[^lebl-bw2]: Lebl, §2.3, alternate proof of the Bolzano–Weierstrass theorem.