Sequential Continuity

Continuity Viewed Through Sequences

The previous tutorial proved the composition theorem using nested epsilon-delta choices. There is another way to describe what a function does near a point: follow any sequence of inputs that approaches that point and examine the corresponding sequence of outputs. This viewpoint turns a local question about all sufficiently nearby inputs into a question about how the function acts on convergent sequences.

A sequence in a domain \(E\) is allowed to repeat values, including the point it approaches. It need not consist of distinct points. This matters, for example, when a point is isolated in \(E\): a sequence in \(E\) that converges to that point must eventually equal it. Throughout, convergence of a sequence in \(E\subseteq\mathbb{R}\) refers to its usual convergence as a real sequence, with every term belonging to \(E\).

Definition: Let \(E\subseteq\mathbb{R}\), let \(f:E\to\mathbb{R}\), and let \(a\in E\). The function \(f\) is sequentially continuous at \(a\) if, for every sequence \((x_n)\) in \(E\) that converges to \(a\), the sequence \((f(x_n))\) converges to \(f(a)\).

The definition quantifies over every sequence in the domain that approaches \(a\). It does not ask whether the outputs converge to some possibly different value: the required limit is specifically \(f(a)\). Nor does it require the sequence to approach \(a\) from both sides. The domain may contain only some nearby points, and the input sequences must respect that domain.

Continuity Implies Sequential Continuity

The first connection follows directly from the epsilon-delta definition. Once a sequence of inputs is eventually within the \(\delta\) supplied by continuity, its outputs are eventually within the desired \(\varepsilon\) of \(f(a)\).

Theorem: Let \(E\subseteq\mathbb{R}\), let \(f:E\to\mathbb{R}\), and let \(a\in E\). If \(f\) is continuous at \(a\), then \(f\) is sequentially continuous at \(a\).

Proof. Let \((x_n)\) be any sequence in \(E\) converging to \(a\). We show that \(f(x_n)\to f(a)\). Fix an arbitrary \(\varepsilon>0\). By continuity of \(f\) at \(a\), there is a \(\delta>0\) such that, for every \(x\in E\),

|x-a|<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon.

Since \(x_n\to a\), there is a positive integer \(N\) such that \(n\geq N\) implies \(|x_n-a|<\delta\). Each \(x_n\) belongs to \(E\), so the continuity implication applies to \(x_n\) for every \(n\geq N\). Therefore

n\geq N \quad\Longrightarrow\quad |f(x_n)-f(a)|<\varepsilon.

This is the definition of convergence of \(f(x_n)\) to \(f(a)\). The sequence \((x_n)\) was arbitrary, so \(f\) is sequentially continuous at \(a\). \(\square\)

The proof uses only continuity at the one point \(a\). It does not require continuity of \(f\) throughout \(E\), and it does not require \(E\) to contain an interval around \(a\). The sequence is drawn from \(E\), which is exactly the domain on which the continuity condition is understood.

Worked Example: Following a Sequence Through a Square

Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), let \(a=-2\), and consider any sequence \((x_n)\) converging to \(-2\). We can check directly that the outputs converge to \(f(-2)=4\). Factor the difference:

$$ |f(x_n)-f(-2)| =|x_n^2-4| =|x_n+2||x_n-2|. $$

Because \(x_n\to-2\), there is an \(N_0\) such that \(n\geq N_0\) implies \(|x_n+2|<1\). For such \(n\), the triangle inequality gives

|x_n-2| =|(x_n+2)-4| \leq |x_n+2|+4 <5.

Now let \(\varepsilon>0\). Since \(x_n\to-2\), there is an \(N_1\) such that \(n\geq N_1\) implies \(|x_n+2|<\varepsilon/5\). For every \(n\geq\max\{N_0,N_1\}\), the factorization and bounds yield

|f(x_n)-4| =|x_n+2||x_n-2| <\frac{\varepsilon}{5}\cdot5 =\varepsilon.

Thus \(f(x_n)\to4=f(-2)\). The estimate illustrates the same principle as the theorem: once the inputs are sufficiently close to the point, their outputs are close to its function value.

The Domain Can Change Which Sequences Are Available

Sequential continuity is always relative to the stated domain. At a point that is isolated in \(E\), every sequence in \(E\) converging to that point is eventually constant. Consequently, every function defined on \(E\) is sequentially continuous at that point, regardless of its behavior elsewhere. This does not mean that the function has to be bounded or well behaved throughout the domain.

Worked Example: Sequential Continuity at an Isolated Point

Let

E=\{0\}\cup\{1/n:n\text{ is a positive integer}\},

and define \(f:E\to\mathbb{R}\) by \(f(0)=0\) and \(f(1/n)=(-1)^n n\). Consider \(a=1/4\), which belongs to \(E\). The only points of \(E\) immediately around \(1/4\) are separated from it: the neighboring listed values are \(1/5\) and \(1/3\). In particular, if \(x\in E\) and \(|x-1/4|<1/40\), then \(x=1/4\). Indeed, \(1/5\) is \(1/20\) from \(1/4\), \(1/3\) is \(1/12\) from \(1/4\), and every other point of \(E\) is at least as far away as one of these or is \(0\).

Let \((x_n)\) be any sequence in \(E\) converging to \(1/4\). By convergence, eventually \(|x_n-1/4|<1/40\), so eventually \(x_n=1/4\). Therefore \(f(x_n)=f(1/4)=4\) for all sufficiently large \(n\), and hence \(f(x_n)\to4=f(1/4)\). This proves sequential continuity at \(1/4\).

The function is nevertheless unbounded on \(E\): for every positive integer \(n\), \(|f(1/n)|=n\). That behavior occurs near \(0\), not near the isolated point \(1/4\). The example shows why one must keep the point and domain in the definition rather than infer local behavior from the function’s overall behavior.

A Sequence Can Expose a Failure

To disprove sequential continuity at \(a\), it is enough to find one sequence in the domain that converges to \(a\) but whose function values do not converge to \(f(a)\). The sequence is a witness to failure. This makes the definition useful not only for establishing continuity, but also for locating the precise obstruction.

Worked Example: A Rational-Irrational Function at Zero

Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=1\) when \(x\) is rational and \(f(x)=0\) when \(x\) is irrational. Since \(0\) is rational, \(f(0)=1\). For each positive integer \(n\), let \(x_n=\sqrt{2}/n\). Each \(x_n\) is irrational: if \(\sqrt{2}/n\) were rational, multiplying by the integer \(n\) would make \(\sqrt{2}\) rational. Also,

|x_n-0|=\frac{\sqrt{2}}{n}\longrightarrow0.

Thus \(x_n\to0\), but every \(x_n\) is irrational, so \(f(x_n)=0\) for every \(n\). The output sequence is constantly \(0\), and therefore

f(x_n)\longrightarrow0\ne1=f(0).

This one sequence proves that \(f\) is not sequentially continuous at \(0\). For comparison, the rational sequence \(y_n=1/n\) also converges to \(0\), but \(f(y_n)=1\) for every \(n\). The two sequences approaching the same point produce incompatible output behavior.

Sequential Continuity and Composition

The composition theorem from the previous tutorial says that ordinary continuity passes through a composition. The sequential formulation has a parallel result. Its proof follows the sequence in two stages: first the inner function sends the input sequence to a sequence approaching \(f(a)\); then sequential continuity of the outer function controls the resulting outputs.

Theorem: Let \(E,F\subseteq\mathbb{R}\), let \(f:E\to F\), and let \(g:F\to\mathbb{R}\). Suppose \(a\in E\), \(f\) is sequentially continuous at \(a\), and \(g\) is sequentially continuous at \(f(a)\). Then \(g\circ f\) is sequentially continuous at \(a\).

Proof. Let \((x_n)\) be any sequence in \(E\) converging to \(a\). Since \(f\) is sequentially continuous at \(a\),

f(x_n)\longrightarrow f(a).

Every term \(f(x_n)\) belongs to \(F\), since \(f:E\to F\). Thus \((f(x_n))\) is a sequence in the domain of \(g\) converging to \(f(a)\). Sequential continuity of \(g\) at \(f(a)\) gives

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

For every \(n\), \(g(f(x_n))=(g\circ f)(x_n)\), and \(g(f(a))=(g\circ f)(a)\). Therefore \((g\circ f)(x_n)\to(g\circ f)(a)\). Since the sequence \((x_n)\) was arbitrary, \(g\circ f\) is sequentially continuous at \(a\). \(\square\)

Worked Example: Two Sequentially Continuous Maps in Succession

Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x+3\), and let \(g:\mathbb{R}\to\mathbb{R}\) be \(g(t)=t^2\). Take any sequence \(x_n\to2\). Then

f(x_n)=x_n+3\longrightarrow5=f(2).

The outputs of \(f\) therefore form a sequence converging to \(5\). Squaring those terms gives a sequence converging to \(25=g(5)\); directly, the estimate can be verified by factoring:

$$ |g(f(x_n))-g(f(2))| =|(x_n+3)^2-25| =|x_n-2||x_n+8|. $$

Because \(x_n\to2\), eventually \(|x_n-2|<1\), which implies \(|x_n+8|=|(x_n-2)+10|<11\). Given \(\varepsilon>0\), eventually \(|x_n-2|<\varepsilon/11\) as well. Beyond both corresponding indices,

|g(f(x_n))-g(f(2))| <\frac{\varepsilon}{11}\cdot11 =\varepsilon.

Hence \(g(f(x_n))\to25\). The composition is \(g(f(x))=(x+3)^2\), and its value at \(2\) is indeed \(25\), as the sequential composition theorem predicts.

What This Direction Establishes

The theorem proved here gives one implication: continuity at a point guarantees sequential continuity there. The converse is an important part of the sequential criterion for continuity, but it requires a separate argument. The next tutorial proves that criterion. Until then, sequential continuity can be used safely as a consequence of continuity, and a single failing sequence can be used to rule it out.

A common mistake is to test only one convenient sequence and conclude that a function is sequentially continuous. The definition requires the condition for every sequence in the domain converging to \(a\). One sequence is enough to disprove the property, but proving it requires control of all such sequences. A second mistake is to use sequences not contained in the domain; their function values may not even be defined.

Check Your Understanding

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

What must happen to \(f(x_n)\) whenever \((x_n)\) is a sequence in \(E\) converging to \(a\) for \(f\) to be sequentially continuous at \(a\)?
In the proof that continuity implies sequential continuity, where is convergence of the input sequence used?
Why is every function sequentially continuous at an isolated point of its domain?
Which sequence demonstrates that the rational-irrational function in the example is not sequentially continuous at zero?
In the sequential composition theorem, why must \(f(x_n)\) belong to \(F\)?
Does checking one sequence that converges to \(a\) establish sequential continuity? Explain the role of “every” in the definition.

What You'll Learn