Proof of the Sequential Criterion

From Sequential Continuity Back to Continuity

In the previous tutorial, continuity at a point was shown to imply sequential continuity there: whenever inputs in the domain converge to the point, their function values converge to the value at that point. The converse is also true. The key idea is to turn a failure of continuity into a sequence that approaches the point while keeping every output a fixed positive distance from the function value.

This construction explains why the word every matters in the definition of sequential continuity. If even one sequence approaching \(a\) fails to have outputs converging to \(f(a)\), then continuity at \(a\) is impossible. Conversely, if no such sequence fails, the epsilon-delta condition must hold.

Proof. If \(f\) is continuous at \(a\), the previous tutorial’s theorem, “Continuity Implies Sequential Continuity,” gives the required conclusion. It remains to prove the converse.

Suppose \(f\) is sequentially continuous at \(a\). We prove that \(f\) is continuous at \(a\) by contradiction. If \(f\) were not continuous at \(a\), the negation of the epsilon-delta definition would give some \(\varepsilon_0>0\) such that, for every \(\delta>0\), there exists an \(x\in E\) satisfying

For each positive integer \(n\), apply this statement with \(\delta=1/n\). Choose \(x_n\in E\) such that

The first inequality shows that \(x_n\to a\). Indeed, given any \(\eta>0\), choose a positive integer \(N\) with \(1/N<\eta\). Whenever \(n\geq N\),

But the second inequality holds for every \(n\), so \(f(x_n)\) cannot converge to \(f(a)\). For example, convergence to \(f(a)\) would require the distance from \(f(x_n)\) to \(f(a)\) eventually to be less than \(\varepsilon_0/2\), whereas that distance is always at least \(\varepsilon_0\). This contradicts sequential continuity at \(a\). Therefore \(f\) is continuous at \(a\). \(\square\)

The proof uses no assumption that \(E\) contains an interval around \(a\). Every selected \(x_n\) belongs to \(E\), exactly as required by continuity on that domain. Also, the sequence need not have distinct terms. The inequalities themselves guarantee the needed approach to \(a\) and the persistent failure of the output values to approach \(f(a)\).

How the Witness Sequence Is Built

The construction follows a general pattern. A failed epsilon-delta condition says there is one fixed output error \(\varepsilon_0\) that cannot be eliminated, no matter how small an input neighborhood is chosen. Choosing successively smaller neighborhoods, such as radius \(1/n\), produces a sequence that gets arbitrarily close to \(a\) while retaining that same output error.

The output bound must use a positive number that is fixed independently of \(n\). Merely knowing that each output differs from \(f(a)\) is not enough: those differences might still tend to zero. The fixed lower bound \(\varepsilon_0\) is what makes the sequence a decisive witness.

A Sequence Characterizes Discontinuity

The criterion gives an equivalent way to describe failure. Rather than checking all possible epsilon-delta choices directly, one can look for a single sequence in the domain whose inputs approach \(a\) but whose outputs do not approach \(f(a)\).

Proof. If \(f\) is not continuous at \(a\), the construction in the proof of the Sequential Criterion gives a sequence \((x_n)\) in \(E\) with \(x_n\to a\) and \(|f(x_n)-f(a)|\geq\varepsilon_0\) for every \(n\), for some \(\varepsilon_0>0\). Therefore \(f(x_n)\) does not converge to \(f(a)\).

Conversely, suppose such a sequence exists. If \(f\) were continuous at \(a\), the previously proved implication “Continuity Implies Sequential Continuity” would force \(f(x_n)\to f(a)\). That contradicts the choice of the sequence. Hence \(f\) is not continuous at \(a\). \(\square\)

This theorem is especially useful when the sequence is easy to identify from the function’s definition. To prove discontinuity, one needs only one valid witness. To prove continuity by sequences, however, one must consider every sequence in the domain converging to the point.

Using the Criterion on Different Domains

Sequences in the criterion must lie in the domain. This matters when the domain has gaps or isolated points. An isolated point \(a\in E\) has a neighborhood containing no other points of \(E\). Any sequence in \(E\) converging to \(a\) must then eventually equal \(a\), so its function values eventually equal \(f(a)\). The criterion consequently gives continuity at \(a\), regardless of how the function behaves at other points.

What the Criterion Does—and Does Not—Ask

The Sequential Criterion for Continuity is a complete replacement for the pointwise epsilon-delta definition when working in subsets of \(\mathbb{R}\): the two conditions are equivalent. It gives a flexible proof method because sequences can be chosen to reflect particular features of a domain or function. The next tutorial compares the epsilon-delta and sequential formulations directly.

A common error is to check one sequence and conclude that a function is continuous. One successful sequence says nothing about all the other sequences that may approach the same point. A single failing sequence proves discontinuity, but a continuity proof through sequences must begin with an arbitrary sequence in \(E\) converging to \(a\) and establish convergence of its function values.

Another important detail is the target of convergence. The required conclusion is always \(f(x_n)\to f(a)\). If \(f(x_n)\) converges to a different value, that sequence still shows failure of sequential continuity. The jump-function example illustrates this directly.