Continuity Through Convergent Sequences
Continuity is often defined by controlling how close inputs must be to keep outputs close. In metric spaces, there is an equivalent test using convergent sequences: a map is continuous when it sends every sequence converging to a point to a sequence converging to the value at that point. The word every is essential. A single well-behaved sequence provides only one test, not a proof of continuity.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, and let \(f:X\to Y\). Continuity at \(x\in X\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that, for every \(z\in X\), \(\ d_X(z,x)<\delta\ \) implies \(\ d_Y(f(z),f(x))<\varepsilon\). A sequence \((x_n)\) in \(X\) converges to \(x\) when, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(n\ge N\) implies \(d_X(x_n,x)<\varepsilon\). Convergence in \(Y\) is defined in the same way using \(d_Y\).
Proof. Suppose first that \(f\) is continuous at \(x\), and let \((x_n)\) be any sequence in \(X\) with \(x_n\to x\). Fix \(\varepsilon>0\). By continuity at \(x\), there is a \(\delta>0\) such that \(d_X(z,x)<\delta\) implies \(d_Y(f(z),f(x))<\varepsilon\). Since \(x_n\to x\), there is an \(N\) such that \(n\ge N\) implies \(d_X(x_n,x)<\delta\). For every such \(n\), applying the continuity implication with \(z=x_n\) gives \(d_Y(f(x_n),f(x))<\varepsilon\). Thus \(f(x_n)\to f(x)\).
Conversely, suppose every sequence in \(X\) converging to \(x\) has its image converge to \(f(x)\). We prove continuity at \(x\) by contradiction. If \(f\) were not continuous at \(x\), the negation of the epsilon-delta definition would give an \(\varepsilon_0>0\) such that, for every \(\delta>0\), there is a \(z\in X\) satisfying \(d_X(z,x)<\delta\) and \(d_Y(f(z),f(x))\ge\varepsilon_0\). For each \(n\in\mathbb{N}\), use \(\delta=1/n\) and choose such a point \(x_n\). Then \(d_X(x_n,x)<1/n\), so \(x_n\to x\). But every term satisfies \(d_Y(f(x_n),f(x))\ge\varepsilon_0\), which prevents \(f(x_n)\) from converging to \(f(x)\): the convergence condition for \(\varepsilon_0\) would require these distances to be less than \(\varepsilon_0\) eventually. This contradicts the assumed sequence property. Therefore \(f\) is continuous at \(x\). \(\square\)
The construction in the reverse direction explains why metrics matter: if continuity fails, the distances \(1/n\) let us choose inputs arbitrarily close to \(x\), one for each \(n\), while keeping their outputs at least a fixed positive distance from \(f(x)\). This produces a sequence that directly detects the failure.
Proof. If \(f\) is continuous on \(X\), it is continuous at every \(x\in X\), so the theorem gives the stated sequence implication at each point. Conversely, if the implication holds for every \(x\in X\), the theorem shows that \(f\) is continuous at each such \(x\), which is exactly continuity on \(X\). If \(X\) is empty, both conditions hold vacuously: there are no points at which continuity must be checked. \(\square\)
Using the Criterion at a Point
Worked Example: The Square Function at 2
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), with the standard metric on both copies of \(\mathbb{R}\). We show by the sequential criterion that \(f\) is continuous at \(2\). Take any sequence \((x_n)\) with \(x_n\to2\). Since it converges, there is an \(N\) such that \(n\ge N\) implies \(|x_n-2|<1\). For those \(n\), the triangle inequality gives \(|x_n+2|\le |x_n-2|+4<5\). Therefore \[ |f(x_n)-f(2)|=|x_n^2-4|=|x_n-2||x_n+2|<5|x_n-2|. \] Because \(|x_n-2|\to0\), the right-hand side tends to \(0\), so \(f(x_n)\to f(2)=4\). The sequence was arbitrary, and the criterion proves continuity at \(2\).
Check with a particular sequence: take \(x_n=2+1/n\), which converges to \(2\) because \(|x_n-2|=1/n\to0\). Substitution gives \(f(x_n)=(2+1/n)^2=4+4/n+1/n^2\), so \(f(x_n)-f(2)=4/n+1/n^2\to0\). At \(n=2\), \(x_2=2+1/2=5/2\), \(f(x_2)=25/4\), and \(f(x_2)-4=25/4-16/4=9/4\). The formula gives \(4/2+1/2^2=2+1/4=9/4\), as required. This particular check agrees with the general argument; it does not replace it.
A useful way to prove continuity at a point is therefore to start with an arbitrary sequence converging to that point and estimate the output distance. The sequence is arbitrary throughout the argument: a bound that works only for a specially chosen sequence cannot establish the universal condition in the theorem.
Worked Example: The Reciprocal Function at 2
Let \(X=\mathbb{R}\setminus\{0\}\) with the metric inherited from \(\mathbb{R}\), let \(Y=\mathbb{R}\) with its standard metric, and define \(f:X\to Y\) by \(f(x)=1/x\). To test continuity at \(2\), let \((x_n)\) be any sequence in \(X\) with \(x_n\to2\). Eventually \(|x_n-2|<1\); for these terms, \(x_n>1\), so \(|x_n|>1\). Hence \[ \left|f(x_n)-f(2)\right| =\left|\frac{1}{x_n}-\frac{1}{2}\right| =\frac{|2-x_n|}{2|x_n|} <\frac{|x_n-2|}{2}. \] The right-hand side tends to \(0\), and so \(f(x_n)\to f(2)=1/2\). The sequential criterion proves continuity at \(2\). Notice that the domain excludes zero, but the argument uses only terms sufficiently close to \(2\), where the displayed denominator is bounded away from zero.
Check with a particular sequence: set \(x_n=2+1/n=(2n+1)/n\). Each term is positive and therefore belongs to \(X\), and \(|x_n-2|=1/n\to0\). Direct substitution gives \(f(x_n)=n/(2n+1)\), and \[ \left|f(x_n)-\frac12\right| =\left|\frac{n}{2n+1}-\frac12\right| =\frac{1}{2(2n+1)}. \] At \(n=1\), \(x_1=3\), \(f(x_1)=1/3\), and the difference from \(1/2\) is \(1/6\); the formula gives \(1/[2(2\cdot1+1)]=1/6\). As \(n\) increases, the displayed difference tends to zero, confirming the result for this sequence.
Using a Sequence to Refute Continuity
The theorem also gives a direct way to prove discontinuity at \(x\). It is enough to find one sequence \(x_n\to x\) for which \(f(x_n)\) does not converge to \(f(x)\). The sequence must be in the domain, and the output limit must be compared with the actual value \(f(x)\), not just with some other limit.
Worked Example: A Rational-Irrational Function at 0
Give \(\mathbb{R}\) its standard metric and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=1\) if \(x\in\mathbb{Q}\), and \(f(x)=0\) if \(x\notin\mathbb{Q}\). Since \(0\in\mathbb{Q}\), we have \(f(0)=1\).
First consider \(x_n=1/n\). Each \(1/n\) is rational, and \(|x_n-0|=1/n\to0\), so \(x_n\to0\). But \(f(x_n)=1\) for every \(n\), which does converge to \(f(0)=1\). This sequence does not show discontinuity.
Now use \(y_n=\sqrt{2}/n\). The number \(\sqrt{2}\) is irrational: if \(\sqrt{2}=a/b\) for relatively prime positive integers \(a,b\), then \(a^2=2b^2\), so \(a\) is even; writing \(a=2c\) gives \(b^2=2c^2\), so \(b\) is even as well, contradicting that \(a,b\) are relatively prime. If \(\sqrt{2}/n\) were rational, multiplying by the integer \(n\) would make \(\sqrt{2}\) rational. Thus every \(y_n\) is irrational and \(f(y_n)=0\). Also \(|y_n-0|=\sqrt{2}/n\to0\), so \(y_n\to0\). The output sequence is constantly \(0\), and hence converges to \(0\ne f(0)=1\). By the sequential criterion, \(f\) is not continuous at \(0\).
Check at \(n=2\): \(y_2=\sqrt{2}/2\) is irrational, so \(f(y_2)=0\); its distance from \(0\) is \(\sqrt{2}/2\), exactly as the formula \(|y_n|=\sqrt{2}/n\) gives at \(n=2\). The constant output sequence has distance \(|0-f(0)|=|0-1|=1\) from the required limit for every \(n\), so it cannot converge to \(f(0)\).
Common mistake. The calculation \(x_n=1/n\to0\) and \(f(x_n)=1\to f(0)\) is correct, but concluding from it that \(f\) is continuous at \(0\) is not. The criterion requires the implication to hold for every sequence converging to \(0\). The sequence \(y_n=\sqrt{2}/n\) supplies the missing test and shows that the implication fails. In short: one failing sequence disproves continuity; one successful sequence does not prove it.
What the Criterion Adds
The sequential criterion is especially useful when limits of expressions are easier to handle than neighborhoods. To prove continuity at a point, take an arbitrary convergent sequence and show its images converge to the correct value. To disprove continuity, construct a sequence that approaches the point while its images stay away from that value, or converge to a different one.
In either use, keep the quantifiers in view. A continuity proof must cover every sequence in the domain converging to the chosen point. A discontinuity proof needs only one counterexample sequence, but it must genuinely converge to that point and its image sequence must fail to converge to the function value there. The metric hypothesis is what makes this test equivalent to continuity: the reverse direction in the proof built a sequence from points within distance \(1/n\) of the point of failure.
Check Your Understanding
For each question, state the relevant domain and compare the image limit with the function value where appropriate.
- In the proof of the sequential criterion, why do the output distances being at least \(\varepsilon_0\) for every \(n\) prevent convergence to \(f(x)\)?
- For \(f(x)=x^2\) on \(\mathbb{R}\), use the sequential criterion to test continuity at \(0\) by starting with an arbitrary sequence \(x_n\to0\).
- For \(f(x)=1/x\) on \(\mathbb{R}\setminus\{0\}\), explain why a sequence converging to \(2\) is eventually bounded away from \(0\).
- For the rational-irrational function in the worked example, what does the sequence \(1/n\) establish, and what does it fail to establish?
- State the difference between the sequence needed to prove continuity at a point and the sequence needed to disprove it.