On the real line, continuity is a property of [functions](/page/Function) between subsets of $\mathbb{R}$ — a single, familiar metric space. But many of the most important objects in analysis are not real numbers: they are *functions*, *[sequences](/page/Sequence)*, or *operators*, living in spaces where "distance" has a very different meaning. The space $C([0,1])$ of continuous functions on $[0,1]$, equipped with the supremum metric $d(f, g) = \sup_{x \in [0,1]} |f(x) - g(x)|$, is the natural habitat of approximation theory and differential equations. The space $\ell^2$ of square-summable sequences, with metric $d(x, y) = (\sum |x_n - y_n|^2)^{1/2}$, underpins Fourier analysis and quantum mechanics. In these spaces, the $\varepsilon$-$\delta$ definition of continuity carries over verbatim from [Continuity (Real Analysis)](/page/Continuity%20(Real%20Analysis)) — one simply replaces the absolute value $|x - a|$ with the metric $d_X(x, a)$ — but the *consequences* of continuity change dramatically because the geometry of the spaces is fundamentally different.

This page develops continuity for functions $f: (X, d_X) \to (Y, d_Y)$ between metric spaces. The key themes are: the sequential characterisation and why it works (first countability), the equivalence with the [topological](/page/Topology) definition (preimages of [open sets](/page/Open%20Set)), uniform continuity as a metric concept, Lipschitz continuity and its role in fixed-point theory, and the interaction between continuity and completeness. The fully general topological treatment is on the parent page [Continuity](/page/Continuity).

## The $\varepsilon$-$\delta$ Definition in Metric Spaces

[definition:Continuous Map Between Metric Spaces] Let $(X, d_X)$ and $(Y, d_Y)$ be [metric spaces](/page/Metric%20Space). A function $f: X \to Y$ is **continuous at** $a \in X$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that \begin{align*} d_X(x, a) < \delta \implies d_Y(f(x), f(a)) < \varepsilon. \end{align*} The function $f$ is **continuous** if it is continuous at every point of $X$. [/definition]

This is identical in form to the definition on $\mathbb{R}$, but the metrics $d_X$ and $d_Y$ may be very different from the absolute value. The same function can be continuous with respect to one metric on $X$ and discontinuous with respect to another — continuity is a property of $f$ *together with* the metrics on domain and codomain.

[example:Continuity Depends On The Metric] Consider the identity function $\operatorname{id}: (\mathbb{R}, d_{\text{disc}}) \to (\mathbb{R}, d_{\text{Eucl}})$, where $d_{\text{disc}}(x, y) = 0$ if $x = y$ and $1$ otherwise (the discrete metric), and $d_{\text{Eucl}}(x, y) = |x - y|$. This map is continuous: for any $a$ and any $\varepsilon > 0$, take $\delta = 1/2$; then $d_{\text{disc}}(x, a) < 1/2$ forces $x = a$, so $d_{\text{Eucl}}(f(x), f(a)) = 0 < \varepsilon$. In the discrete metric, every function is continuous — every point is isolated. The reverse map $\operatorname{id}: (\mathbb{R}, d_{\text{Eucl}}) \to (\mathbb{R}, d_{\text{disc}})$ is *not* continuous at any point: for $\varepsilon = 1/2$, any $\delta > 0$ allows $x \neq a$ with $|x - a| < \delta$, giving $d_{\text{disc}}(x, a) = 1 \geq \varepsilon$. This pair shows that continuity is not a property of the underlying function alone — it depends essentially on the metrics. [/example]

## The Sequential Characterisation

In any metric space, continuity is equivalent to preservation of convergent sequences. This is the same characterisation as on $\mathbb{R}$, and it works for the same reason: metric spaces are *first-countable* (every point has a countable neighbourhood base — the balls $B(a, 1/n)$).

A function $f: X \to Y$ between metric spaces is continuous at $a$ if and only if $f(x_n) \to f(a)$ whenever $x_n \to a$.

The forward direction is the same as on $\mathbb{R}$. The reverse direction constructs the "bad sequence" $x_n$ with $d_X(x_n, a) < 1/n$ but $d_Y(f(x_n), f(a)) \geq \varepsilon_0$, using the fact that the countable family $B(a, 1/n)$ exhausts all neighbourhoods of $a$.

This characterisation *fails* in general topological spaces that are not first-countable. In the uncountable product $\mathbb{R}^{\mathbb{R}}$ with the product topology, sequential continuity is strictly weaker than continuity. This failure is one reason why the topological definition (preimages of open sets) is needed for full generality — see the parent page [Continuity](/page/Continuity).

## Equivalence with the Topological Definition

A function $f: X \to Y$ between metric spaces is continuous if and only if $f^{-1}(V)$ is open in $X$ for every open $V \subseteq Y$.

The proof is a two-line translation between balls and open sets. Forward: if $V \subseteq Y$ is open and $x \in f^{-1}(V)$, then $f(x) \in V$, so there exists $\varepsilon > 0$ with $B_Y(f(x), \varepsilon) \subseteq V$. By $\varepsilon$-$\delta$ continuity, there exists $\delta > 0$ with $f(B_X(x, \delta)) \subseteq B_Y(f(x), \varepsilon) \subseteq V$, so $B_X(x, \delta) \subseteq f^{-1}(V)$, proving $f^{-1}(V)$ is open. Reverse: for any $a$ and $\varepsilon > 0$, the ball $B_Y(f(a), \varepsilon)$ is open in $Y$, so $f^{-1}(B_Y(f(a), \varepsilon))$ is open in $X$ and contains $a$, hence contains some ball $B_X(a, \delta)$.

This equivalence is the bridge between the analytic ($\varepsilon$-$\delta$) and topological (open preimage) perspectives. The closed-set dual — $f$ is continuous iff $f^{-1}(F)$ is closed for every closed $F$ — is equally useful, especially for showing that zero [sets](/page/Set), sublevel sets, and level sets of continuous functions are closed.

[example:Closed Sets Via Continuity] In any metric space $(X, d)$, the closed ball $\overline{B}(a, r) = \{x \in X : d(x, a) \leq r\}$ is closed, because it equals $f^{-1}((-\infty, r])$ where $f(x) = d(x, a)$ is continuous (in fact 1-Lipschitz). The sphere $S(a, r) = \{x : d(x, a) = r\}$ is closed as $f^{-1}(\{r\})$. More generally, for any nonempty $A \subseteq X$, the distance function $d_A(x) = \inf_{a \in A} d(x, a)$ is 1-Lipschitz (hence continuous), and the "closed $r$-neighbourhood" $\{x : d_A(x) \leq r\}$ is closed for every $r \geq 0$. [/example]

## Continuity in Function Spaces

The most important new examples at the metric space level are function spaces, where convergence in the metric corresponds to a specific mode of convergence for functions.

[example:The Supremum Metric And Uniform Convergence] On the space $C([a, b])$ of continuous functions $f: [a, b] \to \mathbb{R}$, the supremum metric $d_\infty(f, g) = \sup_{x \in [a,b]} |f(x) - g(x)| = \|f - g\|_\infty$ turns convergence in the metric into *[uniform convergence](/page/Uniform%20Convergence)*: $f_n \to f$ in $d_\infty$ means $\sup_x |f_n(x) - f(x)| \to 0$, which is exactly uniform convergence on $[a, b]$. A key consequence: the **evaluation functional** $\operatorname{ev}_t: C([a, b]) \to \mathbb{R}$ defined by $\operatorname{ev}_t(f) = f(t)$ is continuous (in fact 1-Lipschitz), since $|\operatorname{ev}_t(f) - \operatorname{ev}_t(g)| = |f(t) - g(t)| \leq \|f - g\|_\infty$. The **integral functional** $I: C([a, b]) \to \mathbb{R}$ defined by $I(f) = \int_a^b f(x) \, dx$ is also continuous: $|I(f) - I(g)| \leq (b - a)\|f - g\|_\infty$, so $I$ is $(b-a)$-Lipschitz. The [uniform limit theorem](/theorems/258) — the uniform limit of continuous functions is continuous — is equivalent to saying that $C([a, b])$ is a *closed subset* of the space of all bounded functions $B([a, b])$ under the supremum metric. Completeness of $C([a, b])$ (every Cauchy sequence of continuous functions converges uniformly to a continuous function) is the statement that $C([a, b])$ is a complete metric space — a [Banach space](/page/Banach%20Space). [/example]

[example:The $L^p$ Metric And Convergence In Mean] On [integrable](/page/Integral) functions, the $L^p$ metric $d_p(f, g) = (\int |f - g|^p)^{1/p}$ captures convergence in $p$-th mean. This is a weaker notion than uniform convergence: $f_n \to f$ in $L^p$ allows pointwise discrepancies on sets of small measure, as long as the integrated $p$-th power of the difference tends to zero. The identity map $\operatorname{id}: (C([a,b]), d_\infty) \to (C([a,b]), d_p)$ is continuous (uniform convergence implies $L^p$ convergence), but the reverse map is not: a sequence can converge in $L^p$ without converging uniformly. The choice of metric on the same underlying set of functions fundamentally changes which operations are continuous. [/example]