ACE 328/Chapter 5

Sequences and Convergence

Convergence of sequences in topological and metric spaces. Subsequences, accumulation points, and the connection between sequences and closed sets.

Sequences are the fundamental tool for probing the fine structure of a space. In a metric space they recover open, closed, and compact sets; in a general topological space they still encode a great deal of information, though the correspondence is sometimes imperfect (a failure we document below). This chapter develops convergence of sequences in topological and metric spaces, treats uniqueness of limits via Hausdorffness, revisits subsequences and accumulation points, proves the Bolzano-Weierstrass theorem, and characterizes closed sets via sequential closure.

Convergence in Topological Spaces

DefinitionConvergent Sequence in a Topological Space

Let $(X, \tau)$ be a topological space and let $(x_n)_{n \geq 1}$ be a sequence in $X$ . We say that $(x_n)$ converges to $x \in X$ , and write $x_n \to x$ , if for every open neighbourhood $U$ of $x$ there exists $N \geq 1$ such that $x_n \in U$ for all $n \geq N$ . The point $x$ is then called a limit of $(x_n)$ .

Remark.

Intuition: A sequence converges to $x$ if it is eventually inside every open neighbourhood of $x$ , no matter how small. In a general topological space, "small" is measured through the system of open sets, not through a distance. A single sequence can, however, converge to more than one limit unless the topology is sufficiently separated — see the next section.

ExampleConvergence in the Trivial Topology

Let $X = \{a, b\}$ with $\tau = \{\emptyset, X\}$ (the trivial topology). Then every sequence in $X$ converges to both $a$ and $b$ : the only nonempty open set is $X$ itself, and every sequence is eventually (in fact always) in $X$ . This shows that uniqueness of limits can fail in general topological spaces.

Convergence in Metric Spaces

DefinitionConvergent Sequence in a Metric Space

Let $(X, d)$ be a metric space and let $(x_n)_{n \geq 1} \subseteq X$ . We say that $x_n \to x$ if for every $\varepsilon > 0$ there exists $N \geq 1$ such that $d(x_n, x) < \varepsilon \quad \text{for all } n \geq N.$ Equivalently, $d(x_n, x) \to 0$ as a sequence of real numbers.

Remark.

Intuition: In a metric space, convergence reduces to "the distance from $x_n$ to $x$ shrinks to zero." This matches the topological definition because the open balls $B_\varepsilon(x)$ form a neighbourhood base at $x$ : every open neighbourhood of $x$ contains some ball $B_\varepsilon(x)$ , and every ball is an open neighbourhood.

TheoremMetric and Topological Convergence Agree

Let $(X, d)$ be a metric space with the induced metric topology $\tau_d$ . A sequence $(x_n) \subseteq X$ converges to $x$ in the metric sense iff $x_n \to x$ in $(X, \tau_d)$ .

TheoremConvergence in Euclidean Space is Componentwise

Let $(x_n)$ be a sequence in $\mathbb{R}^d$ with $x_n = (x_{n,1}, \dots, x_{n,d})$ , and let $L = (L_1, \dots, L_d) \in \mathbb{R}^d$ . Then $x_n \to L$ (in the Euclidean metric) iff $x_{n,i} \to L_i$ in $\mathbb{R}$ for every $i = 1, \dots, d$ .

Basic Properties

TheoremConvergent Sequences are Bounded

Let $(X, d)$ be a metric space. If $x_n \to x$ , then $(x_n)$ is bounded (i.e. contained in some ball $B_R(x_0)$ ).

TheoremAlgebra of Limits in the Real Numbers

If $a_n \to a$ and $b_n \to b$ in $\mathbb{R}$ , then $a_n + b_n \to a + b$ , $a_n b_n \to ab$ , and if $b \neq 0$ , $a_n / b_n \to a/b$ for $n$ sufficiently large.

Uniqueness of Limits: Hausdorff Spaces

DefinitionHausdorff Space

A topological space $(X, \tau)$ is Hausdorff (or $T_2$ ) if for any two distinct points $x, y \in X$ , there exist disjoint open sets $U, V$ with $x \in U$ , $y \in V$ .

Remark.

Intuition: Hausdorffness says that points can be separated by open neighbourhoods. This is the natural minimum condition under which the notion of "the limit" of a sequence is well defined. Every metric space is Hausdorff: given $x \neq y$ , the balls $B_{d(x,y)/2}(x)$ and $B_{d(x,y)/2}(y)$ are disjoint by the triangle inequality.

TheoremUniqueness of Limits in Hausdorff Spaces

In a Hausdorff space $(X, \tau)$ , a convergent sequence has exactly one limit.

CorollaryUniqueness in Metric Spaces

In a metric space, limits of convergent sequences are unique.

Subsequences and Accumulation Points of Sequences

DefinitionSubsequence

Let $(x_n)_{n \geq 1}$ be a sequence in $X$ . A subsequence is a sequence of the form $(x_{n_k})_{k \geq 1}$ where $n_1 < n_2 < n_3 < \cdots$ is a strictly increasing sequence of positive integers.

TheoremSubsequences of Convergent Sequences

If $x_n \to x$ in $(X, \tau)$ , then every subsequence $(x_{n_k})$ also converges to $x$ .

DefinitionAccumulation Point of a Sequence

A point $a \in (X, \tau)$ is an accumulation point (or cluster point) of a sequence $(x_n)$ if for every open neighbourhood $U$ of $a$ and every $N \geq 1$ , there exists $n \geq N$ with $x_n \in U$ . That is, the sequence enters $U$ infinitely often.

Remark.

Intuition: A limit requires "eventually in every neighbourhood"; an accumulation point only requires "infinitely often in every neighbourhood." Every limit is an accumulation point; in a Hausdorff space, a sequence has at most one limit but may have many accumulation points (e.g. $((-1)^n)$ has accumulation points $+1$ and $-1$ ).

TheoremSubsequential Characterization of Accumulation Points

In a metric space $(X, d)$ , a point $a$ is an accumulation point of $(x_n)$ if and only if there is a subsequence $(x_{n_k})$ with $x_{n_k} \to a$ .

The Bolzano-Weierstrass Theorem

TheoremBolzano-Weierstrass on the Real Line

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

TheoremBolzano-Weierstrass in Euclidean Space

Every bounded sequence in $\mathbb{R}^d$ has a convergent subsequence.

Remark.

Intuition: In $\mathbb{R}^d$ , "bounded" gives enough compactness for a convergent subsequence. The theorem is essentially a manifestation of the fact that closed bounded subsets of $\mathbb{R}^d$ are compact (Heine-Borel). In general metric spaces, boundedness alone is not enough — one needs total boundedness, and the result becomes: in a complete metric space, a sequence has a convergent subsequence iff it is totally bounded.

CorollaryCompactness and Sequential Compactness in Euclidean Space

A subset $K \subseteq \mathbb{R}^d$ is compact iff every sequence in $K$ has a subsequence converging to a point of $K$ .

Sequences and Closed Sets

We now characterize closed sets and closures via sequences, a central link between topology and analysis.

DefinitionSequential Closure

Let $(X, \tau)$ be a topological space and $A \subseteq X$ . The sequential closure of $A$ is $\mathrm{scl}(A) = \{x \in X : \text{there exists } (a_n) \subseteq A \text{ with } a_n \to x\}.$

TheoremSequential Closure is Contained in the Closure

For any $A \subseteq X$ , $\mathrm{scl}(A) \subseteq \overline{A}$ .

The reverse inclusion $\overline{A} \subseteq \mathrm{scl}(A)$ can fail in general topological spaces (one needs first-countability, which we will not discuss in detail), but holds in metric spaces.

TheoremClosure via Sequences in Metric Spaces

Let $(X, d)$ be a metric space and $A \subseteq X$ . Then $\overline{A} = \mathrm{scl}(A) = \{x \in X : \exists (a_n) \subseteq A \text{ with } a_n \to x\}.$

CorollarySequential Characterization of Closed Sets in Metric Spaces

A subset $A$ of a metric space $(X, d)$ is closed iff for every sequence $(a_n) \subseteq A$ with $a_n \to x$ , we have $x \in A$ .

Remark.

Intuition: In metric spaces, closed sets are "closed under taking limits of sequences." This is the usual working definition of closedness in real analysis and typically the easiest to check.

Sequential Continuity Revisited

We can now restate continuity in metric spaces in its cleanest form.

TheoremSequential Continuity in Metric Spaces

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and $f : X \to Y$ . Then $f$ is continuous iff for every convergent sequence $x_n \to x$ in $X$ , $f(x_n) \to f(x)$ in $Y$ .

Limits Superior and Inferior

For bounded sequences in $\mathbb{R}$ , two useful quantities capture the asymptotic oscillation behaviour.

DefinitionLimit Superior and Limit Inferior

Let $(a_n)$ be a bounded sequence of real numbers. Define $\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k, \qquad \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_k.$ The sequences $(\sup_{k \geq n} a_k)$ and $(\inf_{k \geq n} a_k)$ are monotone (decreasing and increasing respectively) and bounded, so these limits exist.

Remark.

Intuition: $\limsup a_n$ is the largest accumulation point of $(a_n)$ ; $\liminf a_n$ is the smallest. The sequence converges iff they are equal, and the common value is the limit.

TheoremConvergence via lim sup and lim inf

Let $(a_n)$ be a bounded real sequence. Then $a_n \to L$ iff $\limsup a_n = \liminf a_n = L$ .

Theoremlim sup as the Greatest Subsequential Limit

Let $(a_n)$ be a bounded real sequence. Then $\limsup a_n$ is the greatest accumulation point of $(a_n)$ , and $\liminf a_n$ is the smallest.

ExampleOscillating Sequences

(1) $a_n = (-1)^n$ : $\limsup = 1$ , $\liminf = -1$ ; the sequence does not converge.

(2) $a_n = (-1)^n (1 + 1/n)$ : $\limsup = 1$ , $\liminf = -1$ ; again divergent.

(3) $a_n = \sin(n)$ : the set of accumulation points is exactly $[-1, 1]$ (a nontrivial equidistribution argument), so $\limsup = 1$ , $\liminf = -1$ .

Cauchy Sequences (Preview)

Cauchy sequences — sequences whose terms become arbitrarily close to each other — play a central role in the following chapter on completeness. We record the definition here because it is a property of sequences, and we note the basic relationship with convergence.

DefinitionCauchy Sequence

A sequence $(x_n)$ in a metric space $(X, d)$ is a Cauchy sequence if for every $\varepsilon > 0$ there exists $N \geq 1$ such that $d(x_n, x_m) < \varepsilon$ for all $n, m \geq N$ .

TheoremConvergent Implies Cauchy

In any metric space, every convergent sequence is Cauchy.

The converse — every Cauchy sequence converges — is the completeness property, studied in detail next.

TheoremCauchy Sequences are Bounded

Every Cauchy sequence in a metric space is bounded.

TheoremCauchy Sequence with a Convergent Subsequence Converges

Let $(x_n)$ be a Cauchy sequence in a metric space $(X, d)$ . If a subsequence $(x_{n_k})$ converges to $L$ , then the full sequence converges to $L$ .

This last result is the key technical observation that drives the next chapter: in $\mathbb{R}^d$ , a Cauchy sequence is bounded, hence has a convergent subsequence by Bolzano-Weierstrass, hence converges. That argument establishes completeness of $\mathbb{R}^d$ .

Sequences of Functions: Pointwise vs Uniform Convergence

We now move from sequences of points to sequences of functions, recording the two basic notions of convergence and the central facts that uniform convergence preserves continuity and commutes with integrals.

DefinitionPointwise and Uniform Convergence

Let $I \subseteq \mathbb{R}$ and $f_n, f : I \to \mathbb{R}$ .

$f_n \to f$ pointwise if $f_n(x) \to f(x)$ for every $x \in I$ .
$f_n \to f$ uniformly if $\sup_{x \in I} |f_n(x) - f(x)| \to 0$ as $n \to \infty$ . Equivalently: for every $\varepsilon > 0$ there exists $N$ such that for all $n \geq N$ and all $x \in I$ , $|f_n(x) - f(x)| < \varepsilon$ .

Remark.

Intuition: Pointwise convergence asks each input separately; the speed at which $f_n(x) \to f(x)$ may depend on $x$ . Uniform convergence requires a single rate that works for every $x$ simultaneously. Uniform convergence implies pointwise; the converse fails.

ExamplePointwise Convergence Does Not Preserve Continuity

Let $f_n : [0, 1] \to \mathbb{R}$ , $f_n(x) = x^n$ . Then $\lim_{n \to \infty} x^n = \begin{cases} 0 & \text{if } 0 \leq x < 1, \\ 1 & \text{if } x = 1. \end{cases}$ Each $f_n$ is continuous, but the pointwise limit is discontinuous at $x = 1$ . The convergence is not uniform: $\sup_{x \in [0,1]} |f_n(x) - f(x)| = 1$ for every $n$ .

TheoremUniform Convergence Preserves Continuity

Let $f_n : [a, b] \to \mathbb{R}$ be continuous for each $n$ and suppose $f_n \to f$ uniformly. Then $f$ is continuous.

TheoremUniform Convergence and Integration / Differentiation

Let $f_n : [a, b] \to \mathbb{R}$ be continuous.

(1) If $f_n \to f$ uniformly, then $\lim_{n \to \infty} \int_a^b f_n(x)\, dx = \int_a^b f(x)\, dx$ .

(2) If $f_n \in C^1([a, b])$ , $f_n \to f$ uniformly on $[a, b]$ , and $f_n' \to g$ uniformly on $(a, b)$ , then $f$ is differentiable and $f' = g$ .

Remark.

Uniform convergence is generally a sufficient condition to interchange a limit with integration or differentiation. Pointwise convergence alone is not enough — consider "moving bump" examples where pointwise convergence to $0$ is incompatible with integral convergence.

Equicontinuity and the Arzelà-Ascoli Theorem

When can we extract a uniformly convergent subsequence from a sequence of continuous functions? Boundedness alone is not enough — the analogue of Bolzano-Weierstrass fails in $C([a,b])$ because the space is infinite-dimensional. The right hypothesis is equicontinuity.

DefinitionEquibounded and Equicontinuous

Let $(f_n)_{n \geq 1} \subseteq C([a, b])$ .

$(f_n)$ is equibounded (or uniformly bounded) if there exists $C > 0$ such that $\sup_{x \in [a, b]} |f_n(x)| \leq C \quad \text{for all } n \geq 1.$ The bound $C$ does not depend on $n$ .
$(f_n)$ is equicontinuous if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $\forall s, t \in [a, b] \text{ with } |s - t| < \delta \text{ and } \forall n \geq 1, \quad |f_n(s) - f_n(t)| < \varepsilon.$ The same $\delta$ works for every function in the sequence.

Remark.

Intuition: Equiboundedness is uniform bound across the family. Equicontinuity is uniform continuity that is uniform across the family — the modulus of continuity does not depend on $n$ . A typical sufficient condition for equicontinuity is a uniform Lipschitz bound: if $|f_n'(x)| \leq M$ for all $n$ and $x$ , then $|f_n(s) - f_n(t)| \leq M|s - t|$ for all $n$ , so we can take $\delta = \varepsilon / M$ .

TheoremArzelà-Ascoli Theorem

Let $(f_n)_{n \geq 1} \subseteq C([a, b])$ be equibounded and equicontinuous. Then there exists a subsequence $(f_{n_k})$ that converges uniformly on $[a, b]$ (in particular to a continuous function).

Remark.

Intuition: Arzelà-Ascoli is the infinite-dimensional analogue of Bolzano-Weierstrass for function spaces. It is the workhorse for compactness arguments in $C([a, b])$ and underlies existence proofs for solutions to ODEs (Peano's theorem) and many problems in calculus of variations.

Remark.

Historical note: Cesare Arzelà (1847-1912) and Giulio Ascoli (1843-1896) were Italian mathematicians who developed this theorem in the late 19th century. Arzelà was a professor at the University of Bologna; three years after his death his students erected a monument to him that still stands in the mathematics department.