Pointwise Epsilon-Delta Continuity

Continuity at One Point

Continuity at a point is a local condition: inputs sufficiently close to a chosen point must have outputs close to the output at that point. The epsilon-delta definition makes both distances explicit. It applies even when the metric on the domain and the metric on the codomain are different.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f:X\to Y$. The topology on each space is the one induced by its stated metric. Fix $x_0\in X$. In the definition, $\varepsilon$ measures an allowed output distance, while $\delta$ is chosen to control the input distance.

Definition: The map $f$ is continuous at $x_0$ if for every $\varepsilon>0$ there exists $\delta>0$ such that, for every $x\in X$, $$ d_X(x,x_0)<\delta \quad\Longrightarrow\quad d_Y\bigl(f(x),f(x_0)\bigr)<\varepsilon. $$ The same $\delta$ must work for all $x$ satisfying the input condition for that $\varepsilon$. The point $x=x_0$ is included; its output distance is $d_Y(f(x_0),f(x_0))=0$.

The order of the quantifiers matters: first an arbitrary positive $\varepsilon$ is given, then a positive $\delta$ may be chosen, and then the implication must hold for every $x\in X$. The choice of $\delta$ may depend on $\varepsilon$, the map, and the point $x_0$. It cannot depend on which particular $x$ satisfying $d_X(x,x_0)<\delta$ is tested.

Identity and Constant Maps

Proposition: For every metric space $(X,d)$ and every $x_0\in X$, the identity map $\operatorname{id}_X:X\to X$ is continuous at $x_0$. Every constant map from a metric space to a metric space is continuous at each point of its domain.

Proof. For the identity map, let $\varepsilon>0$ and choose $\delta=\varepsilon$, which is positive. If $d(x,x_0)<\delta$, then $d(\operatorname{id}_X(x),\operatorname{id}_X(x_0))=d(x,x_0)<\delta=\varepsilon$. This is precisely the required implication. For a constant map $f(x)=c$, where $c\in Y$, fix any $x_0\in X$ and any $\varepsilon>0$. Choose $\delta=1$. For every $x$ with $d_X(x,x_0)<1$, $d_Y(f(x),f(x_0))=d_Y(c,c)=0<\varepsilon$. Thus the definition holds at every point. $\square$

Worked Example: The Identity on the Real Line

Give $\mathbb{R}$ its standard metric $d(s,t)=|s-t|$. Verify directly that $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x$, is continuous at $x_0=3$ for the requested output tolerance $\varepsilon=0.2$.

Choose $\delta=0.2$. If $|x-3|<0.2$, then \[ |f(x)-f(3)|=|x-3|<0.2=\varepsilon. \] For instance, $x=3.1$ satisfies $|3.1-3|=0.1<0.2$, and its output distance is $|f(3.1)-f(3)|=|3.1-3|=0.1<0.2$.

Check: The general implication above follows because the output distance equals the input distance for every $x$, not just for the sample $x=3.1$. The choice $\delta=\varepsilon=0.2$ therefore works for all eligible inputs.

Worked Example: A Constant Map

Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=7$, and check continuity at $x_0=-2$ for $\varepsilon=0.01$. Choose $\delta=1$. For every real $x$ satisfying $|x-(-2)|=|x+2|<1$, the output distance is \[ |f(x)-f(-2)|=|7-7|=0<0.01. \]

Check: The chosen $\delta=1$ is positive, and the conclusion does not require a more precise estimate on $x$: both outputs equal $7$. Since the argument works for any positive $\varepsilon$, the map is continuous at $-2$.

Distance to a Fixed Point

For a metric space $(X,d)$ and a fixed point $a\in X$, the distance-to-$a$ function is $g_a:X\to\mathbb{R}$, defined by $g_a(x)=d(x,a)$. Its codomain has the standard metric on $\mathbb{R}$, so the required output distance is $|g_a(x)-g_a(x_0)|$. The key estimate is the reverse triangle inequality.

Lemma (Reverse triangle inequality): For any points $u,v,a$ in a metric space, $$ \bigl|d(u,a)-d(v,a)\bigr|\le d(u,v). $$

Proof. The triangle inequality applied to $u,v,a$ gives $d(u,a)\le d(u,v)+d(v,a)$, hence $d(u,a)-d(v,a)\le d(u,v)$. Applying it with $u$ and $v$ interchanged gives $d(v,a)-d(u,a)\le d(u,v)$. A real number bounded above by $d(u,v)$ and its negative also bounded above by $d(u,v)$ has absolute value at most $d(u,v)$. Therefore $\bigl|d(u,a)-d(v,a)\bigr|\le d(u,v)$. $\square$

Theorem: For every metric space $(X,d)$, every $a\in X$, and every $x_0\in X$, the map $g_a(x)=d(x,a)$ is continuous at $x_0$.

Proof. Let $\varepsilon>0$, and choose $\delta=\varepsilon$. If $d(x,x_0)<\delta$, the reverse triangle inequality, with $u=x$ and $v=x_0$, gives \[ \bigl|g_a(x)-g_a(x_0)\bigr| =\bigl|d(x,a)-d(x_0,a)\bigr| \le d(x,x_0)<\delta=\varepsilon. \] This is the epsilon-delta condition with the standard metric on the codomain $\mathbb{R}$. $\square$

Worked Example: Distance to a Point on the Real Line

Use the standard metric on $\mathbb{R}$, fix $a=2$, and consider $g_2(x)=|x-2|$. Verify continuity at $x_0=5$ for $\varepsilon=0.3$. First, $g_2(5)=|5-2|=3$. Choose $\delta=0.3$. For any $x$ with $|x-5|<0.3$, the reverse triangle inequality gives \[ \bigl||x-2|-3\bigr| =\bigl||x-2|-|5-2|\bigr| \le |x-5|<0.3. \]

As a numerical check, take $x=5.1$. Its input distance is $|5.1-5|=0.1<0.3$. Its output is $g_2(5.1)=|5.1-2|=3.1$, so its output distance from $g_2(5)=3$ is $|3.1-3|=0.1<0.3$.

Check: Substituting the general output values gives $g_2(x)=|x-2|$ and $g_2(5)=|5-2|=3$, exactly the expression bounded above by $|x-5|$. The selected $\delta$ thus works for every eligible $x$, not merely for the numerical check.

The Metric as a Two-Variable Distance Function

There is also a distance function with two inputs: the metric itself, regarded as a map $D:X\times X\to\mathbb{R}$ with $D(u,v)=d(u,v)$. To state continuity at a pair $(x_0,y_0)$, give $X\times X$ the metric \[ \rho\bigl((u,v),(x,y)\bigr)=d(u,x)+d(v,y). \] This is a metric: it is nonnegative, symmetric, and equals zero exactly when $u=x$ and $v=y$; its triangle inequality follows by adding the triangle inequalities for the two coordinates. The codomain $\mathbb{R}$ has its standard metric.

Proposition: With this metric $\rho$ on $X\times X$, the map $D(u,v)=d(u,v)$ is continuous at every $(x_0,y_0)\in X\times X$.

Proof. For any $u,v,x,y\in X$, two applications of the triangle inequality give $d(u,v)\le d(u,x)+d(x,y)+d(y,v)$, so $d(u,v)-d(x,y)\le d(u,x)+d(v,y)$. Interchanging the pairs $(u,v)$ and $(x,y)$ gives $d(x,y)-d(u,v)\le d(u,x)+d(v,y)$. Consequently, \[ |d(u,v)-d(x,y)|\le d(u,x)+d(v,y) =\rho\bigl((u,v),(x,y)\bigr). \] Given $\varepsilon>0$, choose $\delta=\varepsilon$. If $\rho((u,v),(x_0,y_0))<\delta$, this estimate implies $|D(u,v)-D(x_0,y_0)|<\varepsilon$. That is the required pointwise condition. $\square$

A Common Pitfall: Equality Is Not Guaranteed

For the distance-to-$a$ function, the reverse triangle inequality gives a bound, not generally an equality. Replacing it with an equality can invalidate a proof or lead to a false formula. For example, on $\mathbb{R}$ take $a=1$, $x_0=0$, and $x=2$. Then \[ \bigl||2-1|-|0-1|\bigr|=|1-1|=0, \qquad |2-0|=2. \] Thus the output distance is $0$, whereas the input distance is $2$; they are not equal. The correct estimate is $0\le 2$, as the reverse triangle inequality states.

Another point to check when using the definition is that $\delta$ must be positive and must establish a strict output bound. In the examples, $\delta=\varepsilon$ works because the output distance is bounded by the input distance, which is assumed to be strictly less than $\delta$. The chain of inequalities is therefore $\text{output distance}\le\text{input distance}<\delta=\varepsilon$, giving the required strict inequality.

Check Your Understanding

For each question, start from the pointwise epsilon-delta definition and make sure the chosen $\delta$ is positive.

For the identity map on any metric space, what choice of $\delta$ works for a given $\varepsilon>0$, and why?
Let $f:X\to Y$ be constant. Explain why the same positive $\delta$ can be used for every $\varepsilon>0$.
In a metric space $(X,d)$, use the reverse triangle inequality to verify continuity at $x_0$ of $x\mapsto d(x,a)$, where $a\in X$.
For $D(u,v)=d(u,v)$, write an explicit metric on $X\times X$ and state a $\delta$ that works for a given $\varepsilon>0$.
Give real numbers $a,x_0,x$ for which $\bigl||x-a|-|x_0-a|\bigr|\ne |x-x_0|$, and check both sides numerically.