Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric).

A sequence in $X$ is a function $x : \mathbb{N} \to X$, written $(x_k)$ with $x_k := x(k)$. As everywhere in this library, $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)) and a sequence is indexed from $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences); an index range copied from a text that starts at $1$ must be shifted before it is used here.

Let $(x_k)$ be a sequence in $X$ and $p \in X$. The function $k \mapsto d(x_k, p)$ is a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and it is nonnegative (Nonnegativity of a metric is a consequence of the other axioms, not an axiom), so $|d(x_k,p)| = d(x_k,p)$ (Absolute value in an ordered field). Define

$x_k \longrightarrow p \text{ in } (X,d) \quad :\Longleftrightarrow \quad d(x_k,p) \longrightarrow 0 \text{ in } \mathbb{R},$

the convergence on the right being that of Limits and Cauchy sequences of reals. Unwound, this says: for every rational $\varepsilon > 0$ there is $K \in \mathbb{N}$ with $d(x_k, p) < \varepsilon$ for every $k \ge K$. We then call $p$ a limit of $(x_k)$, and say $(x_k)$ converges in $(X,d)$ if it has a limit.

Rational and real $\varepsilon$ agree here, as they do on the real line. Limits and Cauchy sequences of reals tests convergence against rational $\varepsilon$ only, and its own remark, restated for sequences in Sequences of reals: bounded, eventually, frequently, tails, subsequences, records that nothing is lost: below any real $\eta > 0$ lies a positive rational (The rationals embed densely in the reals), and the index belonging to that rational serves for $\eta$. So a proof may establish convergence by producing an index for every real $\varepsilon > 0$, and may use a convergence hypothesis at a real $\varepsilon$ by first passing to a rational below it. Both moves are used on this page and are always cited.

Subsequences and subsequential limits. A subsequence of $(x_k)$ is the composite $x \circ n$ for a strictly increasing $n : \mathbb{N} \to \mathbb{N}$, written $(x_{n_j})$, exactly as for sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences); and $p$ is a subsequential limit of $(x_k)$ in $(X,d)$ when some subsequence converges to $p$, which is the metric-space form of Subsequential limit of a real sequence, and the subsequential limit set.

• A limit is a point of $X$, and uniqueness is a theorem. Nothing in the definition rules out two limits; that a sequence in a metric space has at most one is A sequence in a metric space has at most one limit, and its proof is where the separation axiom (M1) is spent. Reading the same definition with a pseudometric (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) that is not a metric, that is one with $p(a,b) = 0$ for some $a \ne b$, limits are genuinely not unique: the constant sequence at $a$ converges to $b$ as well.
• Convergence is defined from the metric but determined by the topology. It can be restated as "every open set containing $p$ contains $x_k$ for all large $k$", which follows from The balls $B(x, 1/n)$, $n \ge 1$, form a countable neighbourhood base at $x$, so every metric space is first countable; so it is unchanged by passing to a topologically equivalent metric (Topologically, uniformly and Lipschitz equivalent metrics on a set). That restatement is not made part of the definition, because the metric form is what every proof on this page uses.
• The relation between convergence and closure is A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed: a point lies in the closure of $A$ exactly when some sequence in $A$ converges to it.