Sequential Continuity in Metric Spaces

Continuity Tested by Sequences

The epsilon-delta definition measures continuity by comparing distances near a point. A complementary approach asks what a function does to sequences approaching that point. The Sequential Characterization of Continuity, established in “Continuity Between Metric Spaces,” says these approaches are equivalent at each point. Here we use that connection to organize the idea of sequential continuity and to study how it behaves under common constructions.

Throughout, let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and let \(f:X\to Y\). A sequence \((x_n)\) in \(X\) converges to \(x\in X\) when \(d_X(x_n,x)\to0\). Sequential continuity asks whether convergence of the inputs to \(x\) forces convergence of the outputs to the particular point \(f(x)\).

Definition: The function \(f:X\to Y\) is sequentially continuous at \(x\in X\) if, for every sequence \((x_n)\) in \(X\) with \(x_n\to x\), the sequence \(\bigl(f(x_n)\bigr)\) converges to \(f(x)\). The function is sequentially continuous if it is sequentially continuous at every point of \(X\).

The target of the output sequence is not arbitrary: it must be \(f(x)\). This requirement distinguishes continuity from the weaker assertion that the outputs happen to converge somewhere. For example, a sequence of inputs approaching \(x\) could have outputs converging to a point different from \(f(x)\); that would violate sequential continuity at \(x\).

By the Sequential Characterization of Continuity, a function between metric spaces is continuous at \(x\) if and only if it is sequentially continuous at \(x\). Thus sequential continuity does not define a new notion of continuity for metric spaces. Its value is methodological: sometimes it is easier to track the terms of a convergent sequence than to choose an epsilon-delta tolerance directly. Conversely, a carefully chosen sequence can expose a failure of continuity.

Working Directly with Convergent Sequences

To prove sequential continuity at \(x\), begin with an arbitrary sequence \(x_n\to x\) in the domain and show that \(d_Y(f(x_n),f(x))\to0\). The word “arbitrary” matters: checking one convenient sequence does not establish the property. To disprove sequential continuity, however, it suffices to find one sequence approaching \(x\) whose images do not converge to \(f(x)\).

Worked Example: A Bounded Rational Function on the Real Line

Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(t)=1/(1+t^2)\). Fix \(a\in\mathbb{R}\), and let \((x_n)\) be any sequence with \(x_n\to a\). Since \(x_n\to a\), there is an index \(N\) such that \(n\geq N\) implies \(|x_n-a|<1\). For these indices,

|x_n|\leq |x_n-a|+|a|<|a|+1.

The denominators in the difference of the function values are at least \(1\), and factoring the difference of squares gives

\left|f(x_n)-f(a)\right| =\frac{|x_n^2-a^2|}{(1+x_n^2)(1+a^2)} \leq |x_n-a|\,|x_n+a|.

For \(n\geq N\), the triangle inequality and the bound above imply

|x_n+a|\leq |x_n|+|a|<2|a|+1.

Consequently, for every \(n\geq N\),

|f(x_n)-f(a)| \leq (2|a|+1)|x_n-a|.

The constant \(2|a|+1\) is fixed, while \(|x_n-a|\to0\). Hence \(|f(x_n)-f(a)|\to0\), so \(f(x_n)\to f(a)\). The sequence was arbitrary, and \(a\) was arbitrary; therefore \(f\) is sequentially continuous on \(\mathbb{R}\).

Worked Example: Coordinate Evaluation on Bounded Sequences

Let \(\ell^\infty\) be the set of bounded real sequences, equipped with the metric induced by the supremum norm. Define \(E_3:\ell^\infty\to\mathbb{R}\) by \(E_3(x)=x_3\), the third coordinate of \(x\). Suppose \(x^{(n)}\to x\) in this metric. By the definition of convergence in the supremum norm,

\|x^{(n)}-x\|_\infty\longrightarrow 0.

For every \(n\), the absolute difference in the third coordinate is bounded by the supremum of all coordinate differences:

|E_3(x^{(n)})-E_3(x)| =|x^{(n)}_3-x_3| \leq \|x^{(n)}-x\|_\infty.

The right-hand side tends to zero, so \(E_3(x^{(n)})\to E_3(x)\). This proves sequential continuity of the coordinate-evaluation map. The same argument works for any fixed coordinate: convergence in the supremum norm controls every individual coordinate difference.

Worked Example: A Sequence Detects Discontinuity

Define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(0)=1\) and \(g(t)=0\) whenever \(t\ne0\). Consider \(x_n=1/n\). Each \(x_n\) is nonzero, so \(g(x_n)=0\) for every \(n\). Meanwhile \(x_n\to0\), but \(g(0)=1\). Thus

g(x_n)=0\longrightarrow 0\ne 1=g(0).

The sequence violates the defining condition for sequential continuity at \(0\). Therefore \(g\) is not sequentially continuous at \(0\), and, by the Sequential Characterization of Continuity, it is not continuous there. One sequence is enough for this negative conclusion; testing only this sequence, or any finite collection of sequences, would not be enough to prove continuity.

Composition Preserves Sequential Continuity

Composition is a natural setting for the sequence-based definition. If inputs approaching \(x\) produce intermediate values approaching \(f(x)\), and the second map sends those intermediate values toward the appropriate output, then the composed map preserves the original limit as well.

Theorem (Composition of Sequentially Continuous Maps): Let \(f:X\to Y\) and \(g:Y\to Z\) be functions between metric spaces. If \(f\) is sequentially continuous at \(x\in X\) and \(g\) is sequentially continuous at \(f(x)\), then \(g\circ f\) is sequentially continuous at \(x\).

Proof. Let \((x_n)\) be any sequence in \(X\) such that \(x_n\to x\). Sequential continuity of \(f\) at \(x\) gives

f(x_n)\longrightarrow f(x)

in \(Y\). We can therefore apply sequential continuity of \(g\) at \(f(x)\) to the sequence \((f(x_n))\). It follows that

g(f(x_n))\longrightarrow g(f(x)).

Since \(g(f(x_n))=(g\circ f)(x_n)\) and \(g(f(x))=(g\circ f)(x)\), this is precisely sequential continuity of \(g\circ f\) at \(x\). \(\square\)

If \(f\) and \(g\) are sequentially continuous everywhere, the theorem applies at every \(x\in X\), so their composition is sequentially continuous everywhere. This result also follows from the previously established composition theorem for continuous maps and the equivalence between continuity and sequential continuity. The direct proof shows exactly how the intermediate sequence carries the limit through the two maps.

Maps into Finite Product Spaces

For a map into a product space, convergence can be tested coordinate by coordinate. The Sum and Maximum Metrics on a Product, established earlier in “Examples of Metrics,” make this precise. We use the maximum metric on \(Y_1\times Y_2\), defined by

d_{\max}\bigl((y_1,y_2),(z_1,z_2)\bigr) =\max\{d_1(y_1,z_1),d_2(y_2,z_2)\}.

Theorem (Sequential Continuity into a Product): Let \(X,Y_1,Y_2\) be metric spaces, and give \(Y_1\times Y_2\) the maximum metric. For \(F:X\to Y_1\times Y_2\), write \(F(x)=(f_1(x),f_2(x))\). Then \(F\) is sequentially continuous at \(x\in X\) if and only if both \(f_1\) and \(f_2\) are sequentially continuous at \(x\).

Proof. First suppose \(F\) is sequentially continuous at \(x\), and let \(x_n\to x\). Then \(F(x_n)\to F(x)\) in the maximum metric. For each \(i\in\{1,2\}\),

d_i\bigl(f_i(x_n),f_i(x)\bigr) \leq d_{\max}\bigl(F(x_n),F(x)\bigr).

The right-hand side tends to zero, so each coordinate sequence \(f_i(x_n)\) converges to \(f_i(x)\). As this holds for every sequence \(x_n\to x\), both coordinate maps are sequentially continuous at \(x\).

Conversely, suppose both coordinate maps are sequentially continuous at \(x\), and take any sequence \(x_n\to x\). Then

d_1\bigl(f_1(x_n),f_1(x)\bigr)\to0 \quad\text{and}\quad d_2\bigl(f_2(x_n),f_2(x)\bigr)\to0.

Given \(\varepsilon>0\), there are indices \(N_1,N_2\) such that the first distance is less than \(\varepsilon\) for \(n\geq N_1\), and the second is less than \(\varepsilon\) for \(n\geq N_2\). For \(n\geq\max\{N_1,N_2\}\), both distances are less than \(\varepsilon\). Therefore their maximum is less than \(\varepsilon\), which proves \(F(x_n)\to F(x)\) in the maximum metric. This proves sequential continuity of \(F\) at \(x\). \(\square\)

Worked Example: A Map with Two Real-Valued Coordinates

Define \(F:\mathbb{R}^2\to\mathbb{R}^2\) by \(F(s,t)=(s+t,st)\), using the Euclidean metric on the domain and the maximum metric on the target. Let \((s_n,t_n)\to(s,t)\). Euclidean convergence implies \(s_n\to s\) and \(t_n\to t\), since

|s_n-s|\leq\sqrt{(s_n-s)^2+(t_n-t)^2}, \qquad |t_n-t|\leq\sqrt{(s_n-s)^2+(t_n-t)^2}.

It follows that \(s_n+t_n\to s+t\), because

|(s_n+t_n)-(s+t)|\leq |s_n-s|+|t_n-t|\longrightarrow0.

Also, every convergent real sequence is bounded, so there are \(M>0\) and \(N\) such that \(|s_n|\leq M\) for \(n\geq N\). For those indices,

|s_nt_n-st| \leq |s_n|\,|t_n-t|+|t|\,|s_n-s| \leq M|t_n-t|+|t|\,|s_n-s|\longrightarrow0.

Thus \(s_nt_n\to st\) as well. Both coordinates of \(F(s_n,t_n)\) converge to the corresponding coordinates of \(F(s,t)\); by the product theorem, \(F\) is sequentially continuous. The boundedness step is useful when handling products of convergent sequences: it controls the factor multiplying the error.

Restriction to a Subspace and a Useful Caution

Sequences in a subspace use the restricted metric. The Convergence in a Subspace Theorem from “Convergence in Subspaces” says that a sequence in a subset \(A\subseteq X\) converges to \(a\in A\) in the subspace exactly when it converges to \(a\) in \(X\). This lets us transfer sequential continuity to restrictions without changing the convergence calculation.

Proposition (Restriction Preserves Sequential Continuity): Suppose \(f:X\to Y\) is sequentially continuous, and let \(A\subseteq X\) be nonempty with the restricted metric. Then the restriction \(f|_A:A\to Y\) is sequentially continuous.

Proof. Fix \(a\in A\), and let \((a_n)\) be any sequence in \(A\) converging to \(a\) in the restricted metric. By the Convergence in a Subspace Theorem, \(a_n\to a\) in \(X\). Sequential continuity of \(f\) gives \(f(a_n)\to f(a)\) in \(Y\). Since \(f|_A(a_n)=f(a_n)\) and \(f|_A(a)=f(a)\), this is sequential continuity of the restriction at \(a\). The argument holds for every \(a\in A\), proving the proposition. \(\square\)

A common pitfall is to treat one successful sequence test as evidence that a map is sequentially continuous. The definition quantifies over every sequence approaching every point. A single counterexample sequence disproves the property, as in the function \(g\) above; establishing the property requires an argument that covers arbitrary convergent sequences. In metric spaces, the Sequential Characterization of Continuity makes this universal sequence test equivalent to the usual continuity condition.

Takeaway: Sequential continuity means that every sequence converging to a point has images converging to the image of that point. It is equivalent to continuity for maps between metric spaces, is preserved by composition and restriction, and can be checked coordinate by coordinate for maps into finite products.

Check Your Understanding

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

In the definition of sequential continuity at \(x\), what point must the image sequence converge to?
Why does one sequence approaching a point suffice to disprove sequential continuity there?
In the proof for compositions, which convergence allows the sequential continuity of the second map to be applied?
Why does convergence in the maximum metric imply convergence of each coordinate sequence?
What fact about convergent real sequences is used to handle the product \(s_nt_n\) in the worked example?
Why does a sequence in a subspace that converges to a point of that subspace also converge in the ambient space?