Proof of Sequential Continuity

What a Sequential Continuity Proof Must Show

A sequential continuity proof begins with an arbitrary sequence approaching the point under consideration. This is the central quantifier in the definition: the conclusion must hold for every such sequence, not merely for a convenient selection. By contrast, to disprove sequential continuity, one carefully chosen sequence is enough.

The Sequential Characterization of Continuity, established in “Continuity Between Metric Spaces,” identifies sequential continuity at a point with continuity there. The Metric Epsilon-Delta Characterization of Continuity from “Epsilon-Delta Continuity in Metric Spaces” gives a second route: one may prove continuity by controlling output distances whenever input distances are sufficiently small. In this tutorial we focus on how to organize sequence arguments and, especially, how to construct useful sequences when continuity fails.

Let $f:X\to Y$ be a function between metric spaces $(X,d_X)$ and $(Y,d_Y)$, and fix $x\in X$. Sequential continuity at $x$ means that whenever $x_n\to x$, the image sequence satisfies $f(x_n)\to f(x)$. In a proof, this means starting with an arbitrary $(x_n)$ with $d_X(x_n,x)\to0$, then showing that $d_Y(f(x_n),f(x))\to0$.

A useful proof discipline is to keep the input and output distances visible. An estimate of the form

d_Y(f(x_n),f(x))\leq C\,d_X(x_n,x)

with a fixed constant $C$ immediately gives the required convergence. More often, one first bounds some part of the output difference and then uses the triangle inequality to finish. Either way, the proof must account for every sequence approaching $x$.

A Test Using Sequences That Avoid the Limit Point

Sometimes a function has a special definition at a single point, while its behavior elsewhere is easier to describe. In that situation, it can be convenient to test sequences that never take the value of the limit point. The following criterion justifies that approach and includes the case in which there are no such sequences.

Theorem (Punctured-Sequence Test): The function $f$ is sequentially continuous at $x$ if and only if, for every sequence $(y_n)$ in $X$ such that $y_n\to x$ and $y_n\ne x$ for every $n$, one has $f(y_n)\to f(x)$.

Proof. If $f$ is sequentially continuous at $x$, then the stated conclusion holds for every sequence satisfying the test, since each is a sequence converging to $x$.

For the converse, suppose the stated test holds, and take any sequence $(x_n)$ in $X$ with $x_n\to x$. We show that $f(x_n)\to f(x)$. If not, the definition of convergence in the metric space $Y$ gives some $\varepsilon>0$ and a subsequence $(x_{n_k})$ such that

d_Y(f(x_{n_k}),f(x))\geq\varepsilon \quad\text{for every }k.

Every term of this subsequence must differ from $x$, because $x_{n_k}=x$ would imply $f(x_{n_k})=f(x)$, contradicting the displayed inequality. Also, $x_{n_k}\to x$, since it is a subsequence of a convergent sequence. The assumed test therefore gives $f(x_{n_k})\to f(x)$, contradicting the inequality. Hence $f(x_n)\to f(x)$. Since $(x_n)$ was arbitrary, $f$ is sequentially continuous at $x$. $\square$

This criterion does not require $x$ to be a limit point of $X$. If $x$ is isolated, there may be no sequence with every term different from $x$ that converges to $x$. In that case the test is vacuous, but the conclusion remains valid: every sequence converging to an isolated point is eventually equal to that point. The proof above also handles this case, because any subsequence witnessing failure would have to consist entirely of points different from $x$.

Worked Example: Testing the Identity on a Sequence of Isolated Points

Let $X=\{0\}\cup\{1/m:m\in\mathbb{N},\,m\geq1\}$, with the usual metric, and let $f:X\to\mathbb{R}$ be the identity function $f(t)=t$. At $x=0$, consider any sequence $(y_n)$ in $X$ with $y_n\ne0$ and $y_n\to0$. Since $f(y_n)=y_n$ for each $n$, we have

|f(y_n)-f(0)|=|y_n-0|=|y_n|\longrightarrow0.

The Punctured-Sequence Test proves sequential continuity at $0$. At a point $1/m$, the point is isolated in $X$. For $m=1$, the distance to the nearest other point is $1/2$. For $m\geq2$, the distances to the adjacent points are

\frac1m-\frac1{m+1}=\frac1{m(m+1)} \quad\text{and}\quad \frac1{m-1}-\frac1m=\frac1{m(m-1)}.

Both are positive, and all other points of $X$ are at least as far away as one of these adjacent points. Thus some ball around $1/m$ contains no other point of $X$. Any sequence in $X$ converging to $1/m$ must eventually equal $1/m$, so its image under the identity also converges to $f(1/m)$. This example illustrates why isolated points must not be overlooked when using a test involving sequences that avoid the limit point.

Constructing a Sequence That Witnesses Failure

When continuity fails, the epsilon-delta definition gives more than a general warning: it supplies a method for choosing a sequence. The next result makes the choice flexible. The sequence can be required to approach the point at any specified rate, while its function values remain a fixed positive distance from the proposed limit.

Theorem (Arbitrarily Fast Discontinuity Witness): Suppose $f$ is not continuous at $x$. Then there are $\varepsilon_0>0$ and, for every positive sequence $(r_n)$ with $r_n\to0$, points $u_n\in X$ such that $$ d_X(u_n,x)<r_n \quad\text{and}\quad d_Y(f(u_n),f(x))\geq\varepsilon_0 \quad\text{for every }n. $$

Proof. By the Metric Epsilon-Delta Characterization of Continuity, failure of continuity at $x$ means that there is some $\varepsilon_0>0$ such that, for every $\delta>0$, one can find $u\in X$ satisfying

d_X(u,x)<\delta \quad\text{and}\quad d_Y(f(u),f(x))\geq\varepsilon_0.

Now let $(r_n)$ be any positive sequence tending to zero. For each $n$, apply the failure condition with $\delta=r_n$, and choose a corresponding point $u_n$. These choices satisfy both required inequalities. Since $d_X(u_n,x)<r_n$ and $r_n\to0$, we have $u_n\to x$. Yet $f(u_n)$ cannot converge to $f(x)$, because its distance from $f(x)$ is at least $\varepsilon_0$ for every $n$. Thus the chosen sequence witnesses the failure of sequential continuity. $\square$

The fixed lower bound $\varepsilon_0$ is essential: merely finding outputs that differ from $f(x)$ is not enough to show failure, because those differences might still tend to zero. The prescribed radii $r_n$, on the other hand, can be chosen as small as desired. This freedom is useful when a proof requires the inputs to approach $x$ faster than some other quantity changes.

Worked Example: A Sequence Reveals Oscillation at the Origin

Define $h:\mathbb{R}\to\mathbb{R}$ by $h(0)=0$ and $h(t)=\sin(1/t)$ for $t\ne0$. For each positive integer $n$, set

x_n=\frac{1}{\pi/2+2\pi n}.

The denominators tend to infinity, so $x_n\to0$, and each $x_n\ne0$. Substitution into the definition gives

h(x_n) =\sin\left(\frac{1}{x_n}\right) =\sin(\pi/2+2\pi n) =1 \quad\text{for every }n.

But $h(0)=0$. Thus the image sequence remains at distance $1$ from $h(0)$, so it does not converge to $h(0)$. This explicit sequence proves that $h$ is not sequentially continuous at $0$. It is an instance of the discontinuity-witness method: the inputs approach the point while the output error stays uniformly bounded below.

Turning a Metric Estimate into a Sequence Proof

A common route to sequential continuity is to first establish a direct estimate between output and input distances. The estimate can then be applied to each term of an arbitrary convergent sequence. Distance to a nonempty set gives a useful example.

Theorem (Distance to a Set Is Lipschitz): Let $A$ be a nonempty subset of a metric space $(X,d)$. Define $D(x)=d(x,A)=\inf\{d(x,a):a\in A\}$. Then $$ |D(x)-D(y)|\leq d(x,y) \quad\text{for all }x,y\in X. $$

Proof. Fix $x,y\in X$. For every $a\in A$, the triangle inequality gives

d(x,a)\leq d(x,y)+d(y,a).

Taking the infimum over $a\in A$ yields $D(x)\leq d(x,y)+D(y)$. Interchanging $x$ and $y$ gives $D(y)\leq d(x,y)+D(x)$. Combining these two inequalities proves

-d(x,y)\leq D(x)-D(y)\leq d(x,y),

which is equivalent to $|D(x)-D(y)|\leq d(x,y)$. $\square$

Worked Example: Distance from a Point to Two Fixed Locations

In $\mathbb{R}$, let $A=\{-1,3\}$, and define $D(t)=d(t,A)$. By the definition of distance to this finite set,

D(t)=\min\{|t+1|,|t-3|\}.

The Distance to a Set Is Lipschitz Theorem gives $|D(s)-D(t)|\leq|s-t|$ for all real $s,t$. Now let $(t_n)$ be any sequence with $t_n\to t$. Applying this inequality term by term gives

|D(t_n)-D(t)|\leq|t_n-t|\longrightarrow0.

Therefore $D(t_n)\to D(t)$, and $D$ is sequentially continuous at every real number. This argument avoids splitting the real line into regions where one or the other absolute value is smaller; a single metric estimate handles all points, including those where the two distances agree.

Proof Strategy and a Common Pitfall

A direct proof of sequential continuity and a sequence-based disproof have opposite logical shapes. To prove the property at $x$, fix an arbitrary convergent sequence and derive convergence of its images. To disprove it, select one sequence whose images fail to converge to $f(x)$. Confusing these two tasks leads to a frequent error: checking that one or several familiar sequences behave well does not establish continuity.

For a proof, begin arbitrarily.
State that $(x_n)$ is any sequence in $X$ with $x_n\to x$. Do not impose extra conditions unless you later justify them.

Estimate the output error.
Use the definitions and metric inequalities to bound $d_Y(f(x_n),f(x))$ by quantities that tend to zero.

For a disproof, keep a fixed error.
Choose inputs approaching $x$ while ensuring their images stay at least some fixed positive distance from $f(x)$.

The Punctured-Sequence Test can shorten a proof when the function is described differently at $x$ and away from $x$, but the isolated-point case remains part of the argument. The Arbitrarily Fast Discontinuity Witness provides the reverse tool: if continuity fails, the problematic inputs can be selected within any prescribed shrinking radii. Together, these methods clarify what sequence-based continuity proofs must establish and how a failed proof can be converted into a concrete counterexample.

Check Your Understanding

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

In a proof of sequential continuity at $x$, why must the convergent input sequence be arbitrary?
Why does the Punctured-Sequence Test remain valid when $x$ is isolated?
In the proof of the Punctured-Sequence Test, what property of a subsequence allows the contradiction to be applied?
What does the fixed lower bound $\varepsilon_0$ ensure in an Arbitrarily Fast Discontinuity Witness?
Which two applications of the triangle inequality yield the Lipschitz estimate for distance to a set?

What You'll Learn