Continuity between Metric Spaces¶

Definition¶

Consider two metric spaces $(X,d_{X})$ and $(Y,d_{Y})\text{,}$ and a function $f\colon E\to Y$ where $E\subseteq X\text{.}$ Let $x_0\in X\text{.}$ Assume there exists an $r>0$ such that $B_{(X,d_{X})}(x_0,r)\subseteq E\text{.}$ We say that $f$ is continuous at $x=x_0$ if, for any $\varepsilon > 0\text{,}$ there exists a $0<\delta \leq r$ such that whenever $d_{X}(x,x_0)<\delta\text{,}$ we have $d_{Y}(f(x),f(x_0))<\varepsilon\text{.}$

We say that $f$ is continuous on $E$ (or simply continuous) if it is continuous at every point of $E\text{.}$ At a point $x_0\in E$ that has no such ball inside $E$ (a boundary point of $E$), the same condition is required only for $x\in E\text{.}$

The condition says that the intersection of the domain with a sufficiently small open metric ball around $x_0$ is mapped into the $\varepsilon$-ball around $f(x_0)$. With the usual real distance $|x-y|$$y|$, this is the familiar definition of continuity.

Interpretation¶

A function is continuous at a point if you can make the outputs as close as you want to the value at that point by taking inputs sufficiently close to it. In metric spaces, "close" is measured by the metrics: $d_{X}$ for the inputs and $d_{Y}$ for the outputs.

On the left is the open ball $B(x_0,\delta)$ in $X$; on the right, its image $f(B(x_0,\delta))$ lies entirely inside the ball $B(f(x_0),\varepsilon)$ in $Y$. This inclusion is the definition restated: every $x$ with $d_{X}(x,x_0)<\delta$ satisfies $d_{Y}(f(x),f(x_0))<\varepsilon$$\epsilon$. The order of choice matters. The tolerance $\varepsilon$ on the outputs comes first, and $\delta$ is chosen in response; a smaller $\varepsilon$ usually forces a smaller $\delta$.

The $\delta$ may depend on both $\varepsilon$ and the point $x_0$. If one $\delta$ works at all points at once, $f$ has the stronger property of uniform continuity, known from real functions. The bound $\delta\le r$ only keeps the $\delta$-ball inside the domain $E$, so that $f(x)$ is defined there.

With $X=Y=\mathbb{R}$ and the distance $|x-y|$$y|$, this is the $\varepsilon$-$\delta$ continuity of Math I. Continuity depends on the metrics: if $X$ carries the discrete metric, every function on $X$ is continuous, because $B(x_0,\tfrac{1}{2})=\{x_0\}$$\{x_0\}$. Descriptions by sequences and by preimages of open sets are given in the equivalent formulation of continuity.

A common slip is to fix $\delta$ first and then look for an $\varepsilon$. That reversed condition holds for every bounded function, including the discontinuous signum function at $0$, where $\varepsilon=2$ works for every $\delta$.

Key Results¶

• If $f(x)$ tends to two different values as $x$ approaches $x_0$ in two different ways (for example along two different lines, or along two sequences converging to $x_0$), then the limit of the function at that point does not exist, so the function is not continuous there. More generally, if for some sequence $x_k\to x_0$ the values $f(x_k)$ do not tend to $f(x_0)\text{,}$ then $f$ is not continuous at $x_0\text{.}$
• A vector-valued function into $\mathbb{R}^n$ with its usual Euclidean metric is continuous if and only if each real component function is continuous.
• Most properties of real-valued functions of one real variable are preserved under continuity. In particular, continuity is preserved under addition, subtraction, multiplication, division (where the denominator is nonzero), as well as taking the maximum, minimum, and composition of functions.

Examples¶

• $f(x,y)=x^2+y^2$ on $\mathbb{R}^2$. At $(0,0)$, if you want $f$ within $\varepsilon$ of $0$, choose $\delta=\sqrt{\varepsilon}$$\sqrt{\epsilon }$. Then $\sqrt{x^2+y^2}<\delta$ gives $x^2+y^2<\varepsilon$$\epsilon$.

Prerequisites¶

• Open Metric Ball
• Continuity

Related Concepts¶

• Uniform Continuity between Metric Spaces

Sources¶

• Math III: Analysis · Lecture 8

Symbols: delta, element of, epsilon, real numbers, square root, subset

Last updated: 20 October 2026 03:02