Epsilon-Delta Continuity in Metric Spaces

Continuity Measured by Distances

The open-set definition of continuity describes how neighborhoods are carried into neighborhoods. The Sequential Characterization of Continuity, established in “Continuity Between Metric Spaces,” instead tests the images of convergent sequences. The epsilon-delta formulation expresses the same local idea directly in terms of distances: given any required accuracy in the output, we choose an input tolerance that guarantees it.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $f:X\to Y$, and fix $x\in X$. The input tolerance is measured by $d_X$, while the output accuracy is measured by $d_Y$. Keeping these metrics distinct is essential when the spaces are different.

Definition: The function $f:X\to Y$ is epsilon-delta continuous at $x\in X$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that for every $u\in X$, $$ d_X(u,x)<\delta \quad\Longrightarrow\quad d_Y\bigl(f(u),f(x)\bigr)<\varepsilon. $$ The function is epsilon-delta continuous if it is epsilon-delta continuous at every $x\in X$.

The order of the quantifiers matters. First an output tolerance $\varepsilon$ is specified; then we are allowed to choose an input tolerance $\delta$. The choice of $\delta$ may depend on $\varepsilon$, the point $x$, the function, and the metrics. In this pointwise definition, it need not work at other points.

The condition concerns every $u\in X$ within distance $\delta$ of $x$, including $u=x$. At that point the output distance is $d_Y(f(x),f(x))=0<\varepsilon$, as required. The inequalities are strict on both sides, in agreement with the use of open balls.

Equivalence with Open-Set Continuity

The epsilon-delta formulation is not a different notion of continuity. It is exactly the open-set definition stated using balls and distances. This equivalence allows us to choose whichever formulation makes a particular proof simpler.

Theorem (Metric Epsilon-Delta Characterization of Continuity): Let $f:X\to Y$ be a function between metric spaces and let $x\in X$. Then $f$ is continuous at $x$ in the open-set sense if and only if it is epsilon-delta continuous at $x$.

Proof. Suppose first that $f$ is epsilon-delta continuous at $x$. Let $V\subseteq Y$ be open with $f(x)\in V$. Since $V$ is open, there is an $\varepsilon>0$ such that $B_\varepsilon(f(x))\subseteq V$. By epsilon-delta continuity, choose $\delta>0$ such that

d_X(u,x)<\delta \quad\Longrightarrow\quad d_Y\bigl(f(u),f(x)\bigr)<\varepsilon.

The implication says that $f(u)\in B_\varepsilon(f(x))\subseteq V$ for every $u\in B_\delta(x)$. Thus $U=B_\delta(x)$ is an open set containing $x$ and satisfying $f(U)\subseteq V$. This is continuity at $x$ in the open-set sense.

Conversely, suppose $f$ is continuous at $x$ in the open-set sense. Fix any $\varepsilon>0$. The ball $B_\varepsilon(f(x))$ is open in $Y$ and contains $f(x)$. By continuity at $x$, there is an open set $U\subseteq X$ such that $x\in U$ and $f(U)\subseteq B_\varepsilon(f(x))$. Because $U$ is open and contains $x$, there is a $\delta>0$ such that $B_\delta(x)\subseteq U$. For any $u\in X$ with $d_X(u,x)<\delta$, we have $u\in U$, so $f(u)\in B_\varepsilon(f(x))$. Therefore

d_Y\bigl(f(u),f(x)\bigr)<\varepsilon.

This proves epsilon-delta continuity at $x$. Both implications hold for each $x$, so continuity everywhere in the open-set sense is equivalent to epsilon-delta continuity at every point. $\square$

The proof uses only the ball properties of metric spaces and the open-set definition from “Continuity Between Metric Spaces.” In one direction, a ball around the output supplies the target accuracy. In the other, the open neighborhood of the input contains a ball that supplies the required $\delta$.

Worked Examples

Worked Example: The Square Function at an Arbitrary Point

Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=x^2$, with the usual metric, and fix $a\in\mathbb{R}$. Given $\varepsilon>0$, choose

\delta=\min\left\{1,\frac{\varepsilon}{2|a|+1}\right\}.

This is positive because both entries in the minimum are positive. If $|x-a|<\delta$, then $|x-a|<1$. The factorization of the difference of squares gives

|f(x)-f(a)| =|x^2-a^2| =|x-a|\,|x+a|.

Since $x+a=(x-a)+2a$, the triangle inequality yields

|x+a|\leq |x-a|+2|a|<1+2|a|.

Consequently,

|f(x)-f(a)| \leq|x-a|(2|a|+1) <\delta(2|a|+1) \leq\varepsilon.

The last inequality follows from $\delta\leq\varepsilon/(2|a|+1)$. This verifies the epsilon-delta condition at every $a$. Notice that the chosen $\delta$ can become smaller as $|a|$ grows; the argument does not require one tolerance to work at all points.

Worked Example: Distance from a Fixed Point

Let $(X,d)$ be a metric space, fix $p\in X$, and define $h:X\to\mathbb{R}$ by $h(x)=d(x,p)$. We show directly that $h$ is epsilon-delta continuous at any $x\in X$. Given $\varepsilon>0$, take $\delta=\varepsilon$.

For any $u\in X$, the Reverse Triangle Inequality for a Metric gives

\bigl|d(u,p)-d(x,p)\bigr|\leq d(u,x).

Thus, whenever $d(u,x)<\delta$, we obtain

|h(u)-h(x)| =\bigl|d(u,p)-d(x,p)\bigr| \leq d(u,x) <\delta =\varepsilon.

The output distance here is the usual absolute-value distance on $\mathbb{R}$. The estimate works in an arbitrary metric space $X$, not just on the real line, because it relies only on the metric inequality.

Worked Example: Any Map from a Discrete Metric Space

Let $X$ have the discrete metric $\delta_X$, and let $Y$ be any metric space. Consider any function $f:X\to Y$, with no restrictions on its values. Fix $x\in X$ and $\varepsilon>0$. Choose the input tolerance $\delta=1$. If $\delta_X(u,x)<1$, then $u=x$: distinct points in a discrete metric space have distance $1$.

It follows that

d_Y\bigl(f(u),f(x)\bigr) =d_Y\bigl(f(x),f(x)\bigr) =0 <\varepsilon.

Thus every function from a discrete metric space into any metric space is epsilon-delta continuous at every point. The choice $\delta=1$ is sufficient for every positive $\varepsilon$. This conclusion depends on the input metric: nearby points at distances strictly less than $1$ can only be the point itself.

Worked Example: The Reciprocal on the Positive Real Line

Give $X=(0,\infty)$ the restricted usual metric, give $\mathbb{R}$ its usual metric, and define $q(u)=1/u$. Fix $a>0$. For a prescribed $\varepsilon>0$, set

\delta=\min\left\{\frac{a}{2},\frac{\varepsilon a^2}{2}\right\}.

If $u\in(0,\infty)$ and $|u-a|<\delta$, then $|u-a|<a/2$, so $u>a/2$. Therefore $au>a^2/2$, and

\left|\frac{1}{u}-\frac{1}{a}\right| =\frac{|u-a|}{au} \leq\frac{2|u-a|}{a^2} <\frac{2\delta}{a^2} \leq\varepsilon.

The restriction $a>0$ ensures the displayed $\delta$ is positive and that nearby inputs can be kept away from zero. The estimate proves continuity at each point of the domain $(0,\infty)$; it makes no claim about defining the reciprocal at zero.

Quantitative Estimates and Composition

The epsilon-delta definition becomes especially useful when a map satisfies an explicit distance bound. A Lipschitz estimate provides such a bound uniformly at every point.

Definition: A function $f:X\to Y$ between metric spaces is Lipschitz with constant $L\geq 0$ if $$ d_Y\bigl(f(u),f(v)\bigr)\leq Ld_X(u,v) $$ for all $u,v\in X$.

Theorem (Lipschitz Maps Are Epsilon-Delta Continuous): Every Lipschitz map between metric spaces is epsilon-delta continuous at every point.

Proof. Fix $x\in X$ and $\varepsilon>0$. If $L>0$, choose $\delta=\varepsilon/L$. Whenever $d_X(u,x)<\delta$, the Lipschitz estimate gives

d_Y\bigl(f(u),f(x)\bigr) \leq Ld_X(u,x) <L\delta =\varepsilon.

If $L=0$, then $d_Y(f(u),f(x))\leq 0$ for every $u\in X$. Since distances are nonnegative, this distance is zero, and hence is less than every $\varepsilon>0$. Any $\delta>0$, for example $\delta=1$, works. This covers both possible cases for $L$ and proves the theorem. $\square$

Compositions also admit a direct tolerance calculation. The proof of the Composition of Continuous Maps Theorem in “Continuity Between Metric Spaces” uses inverse images of open sets. Here the same conclusion follows by passing the required accuracy backward through the two maps.

Theorem (Epsilon-Delta Continuity of Compositions): Let $f:X\to Y$ be epsilon-delta continuous at $x\in X$, and let $g:Y\to Z$ be epsilon-delta continuous at $f(x)$. Then $g\circ f$ is epsilon-delta continuous at $x$.

Proof. Let $\varepsilon>0$ be an output tolerance in $Z$. By epsilon-delta continuity of $g$ at $f(x)$, there is an $\eta>0$ such that, for every $y\in Y$,

d_Y(y,f(x))<\eta \quad\Longrightarrow\quad d_Z\bigl(g(y),g(f(x))\bigr)<\varepsilon.

By epsilon-delta continuity of $f$ at $x$, applied to the tolerance $\eta$, there is a $\delta>0$ such that

d_X(u,x)<\delta \quad\Longrightarrow\quad d_Y\bigl(f(u),f(x)\bigr)<\eta.

Combining these implications, every $u\in X$ with $d_X(u,x)<\delta$ satisfies

d_Z\bigl((g\circ f)(u),(g\circ f)(x)\bigr) =d_Z\bigl(g(f(u)),g(f(x))\bigr) <\varepsilon.

This is precisely epsilon-delta continuity of $g\circ f$ at $x$. $\square$

Choosing the Tolerance Correctly

The definition is often used incorrectly by choosing $\delta$ before considering the desired $\varepsilon$, or by giving a $\delta$ that depends on the input $u$. The required order is fixed: for each $\varepsilon>0$, one must give a positive $\delta$ that works for all $u$ satisfying $d_X(u,x)<\delta$. The examples show several ways to do this: factor an algebraic difference, apply a metric inequality, or choose a radius that forces the input to equal the center.

A valid choice of $\delta$ need not be the largest possible one. In the square-function example, the minimum of two positive bounds combines a condition that controls $|x+a|$ with one that controls the final error. In the reciprocal example, one bound keeps $u$ away from zero and the other makes the output error sufficiently small. Checking both purposes is what makes the estimate complete.

Takeaway: Continuity at $x$ means that every positive output tolerance admits a positive input tolerance. This metric formulation is equivalent to open-set continuity, yields explicit estimates for maps such as Lipschitz functions, and passes through compositions by choosing tolerances in reverse order.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

In the epsilon-delta condition, which tolerance is specified first, and what may the chosen input tolerance depend on?
How does an open ball around $f(x)$ help prove that epsilon-delta continuity implies open-set continuity?
For the square function at $a$, why is it useful to require $|x-a|<1$ as well as a bound involving $\varepsilon$?
What value of $\delta$ proves continuity of a map whose domain has the discrete metric?
How should the tolerance for $f$ be selected when proving continuity of $g\circ f$?
Why must the reciprocal example be restricted to positive inputs when using its stated estimate?