Using Sequences to Detect Discontinuity
To prove continuity at a point, one must control the function on every sufficiently nearby input. To prove discontinuity, a single well-chosen sequence can be enough: its inputs approach the point, but its outputs fail to approach the function value there. This turns a local failure of an epsilon-delta condition into a concrete calculation.
Throughout, let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, let \(D\subseteq X\) have the subspace metric, and let \(f:D\to Y\). Fix \(c\in D\). A sequence \((x_n)\) in \(D\) converges to \(c\) if for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(n\ge N\) implies \(d_X(x_n,c)<\varepsilon\). The definition of \(f\) being continuous at \(c\) is: for every \(\varepsilon>0\), there is \(\delta>0\) such that, for all \(x\in D\), \(d_X(x,c)<\delta\) implies \(d_Y(f(x),f(c))<\varepsilon\).
Proof. Suppose first that \(f\) is discontinuous at \(c\). Negating the definition of continuity gives an \(\varepsilon_0>0\) such that, for every \(\delta>0\), there is an \(x\in D\) with \(d_X(x,c)<\delta\) and \(d_Y(f(x),f(c))\ge\varepsilon_0\). For each \(n\in\mathbb{N}\), apply this statement with \(\delta=1/n\), and choose such a point \(x_n\). Then \(d_X(x_n,c)<1/n\), so \(x_n\to c\). But \(d_Y(f(x_n),f(c))\ge\varepsilon_0\) for every \(n\), so the image sequence does not converge to \(f(c)\).
Conversely, suppose \(x_n\to c\) and \(f(x_n)\not\to f(c)\). If \(f\) were continuous at \(c\), then for any \(\varepsilon>0\) its continuity condition would provide \(\delta>0\) such that \(d_X(x,c)<\delta\) implies \(d_Y(f(x),f(c))<\varepsilon\). Since \(x_n\to c\), eventually \(d_X(x_n,c)<\delta\). The corresponding image values would then eventually satisfy \(d_Y(f(x_n),f(c))<\varepsilon\). This holds for every \(\varepsilon>0\), so \(f(x_n)\to f(c)\), a contradiction. Therefore \(f\) is discontinuous at \(c\). \(\square\)
The first half of the proof gives a systematic construction when no obvious sequence comes to mind: failure of continuity supplies a fixed positive output error \(\varepsilon_0\), and inputs can be chosen within \(1/n\) of \(c\) while preserving that error. In examples, however, explicit sequences are usually simpler and show exactly how the function behaves.
What Counts as a Successful Witness?
A sequence that approaches \(c\) is not enough by itself. Its images must fail to converge to \(f(c)\). There are two useful ways this can happen: the images may fail to converge at all, or they may converge to a value different from \(f(c)\). In either case the sequential criterion proves discontinuity.
Proof. If \(f\) were continuous at \(c\), the sequential criterion for metric continuity would give \(f(x_n)\to f(c)\) and \(f(y_n)\to f(c)\). Limits in a metric space are unique: if a sequence converges to both \(a\) and \(b\), then for every \(n\), \(d_Y(a,b)\le d_Y(a,z_n)+d_Y(z_n,b)\); taking the two distances on the right sufficiently small forces \(d_Y(a,b)=0\), hence \(a=b\). Thus \(L_1=f(c)=L_2\), contrary to \(L_1\ne L_2\). In the second case, continuity would likewise force \(f(x_n)\to f(c)\); uniqueness of limits would give \(L=f(c)\), again a contradiction. \(\square\)
Worked Discontinuity Tests
Worked Example: An Oscillating Function
Define \(f:\mathbb{R}\to\mathbb{R}\), with both copies of \(\mathbb{R}\) carrying the standard metric, by \(f(0)=0\) and \(f(x)=\sin(1/x)\) for \(x\ne0\). Test continuity at \(0\).
For \(n\in\mathbb{N}\), set \(x_n=1/(\pi/2+2\pi n)\). The denominator is positive, so \(x_n>0\), and \(\pi/2+2\pi n>2\pi n\). Therefore \(0<x_n<1/(2\pi n)\), which tends to \(0\); hence \(x_n\to0\). Since \(x_n\ne0\), the formula for nonzero inputs applies, and
Thus \(f(x_n)=1\) for every \(n\), whereas \(f(0)=0\). The image sequence is constantly \(1\), so its distance from \(f(0)\) is \(1\) at every term; it does not converge to \(0\). The sequential criterion proves that \(f\) is discontinuous at \(0\).
Check: the inputs approach \(0\), while the output discrepancy is exactly \(|f(x_n)-f(0)|=|1-0|=1\) for every \(n\). In particular, taking the required output tolerance to be \(1/2\) shows that this sequence never meets it.
Worked Example: A Sign Function at Its Jump
Define \(g:\mathbb{R}\to\mathbb{R}\), again with the standard metric on each copy, by \(g(x)=1\) for \(x>0\), \(g(x)=-1\) for \(x<0\), and \(g(0)=0\). Use two sequences to test continuity at \(0\).
For \(n\in\mathbb{N}\), let \(x_n=1/n\) and \(y_n=-1/n\). Their distances from \(0\) are both \(1/n\), which tends to \(0\), so \(x_n\to0\) and \(y_n\to0\). Every \(x_n\) is positive and every \(y_n\) is negative. Hence \(g(x_n)=1\) for every \(n\), and \(g(y_n)=-1\) for every \(n\). The two image limits are \(1\) and \(-1\), which are unequal. The two-sequence test proves that \(g\) is discontinuous at \(0\).
Check: the value required by continuity is \(g(0)=0\). The image distances are \(|g(x_n)-g(0)|=|1-0|=1\) and \(|g(y_n)-g(0)|=|-1-0|=1\) for every \(n\), so neither sequence of images approaches the required value.
Worked Example: The Rational-Irrational Indicator
Define \(h:\mathbb{R}\to\mathbb{R}\) by \(h(x)=1\) if \(x\in\mathbb{Q}\), and \(h(x)=0\) if \(x\notin\mathbb{Q}\). Test continuity at \(0\), where the domain and codomain have their standard metrics.
The number \(0\) is rational, so \(h(0)=1\). Choose \(x_n=\sqrt{2}/n\) for \(n\in\mathbb{N}\). First, \(\sqrt{2}\) is irrational: if \(\sqrt{2}=p/q\) for relatively prime positive integers \(p,q\), then \(p^2=2q^2\), so \(p\) is even; writing \(p=2k\) gives \(q^2=2k^2\), so \(q\) is even as well, contradicting relative primeness. If \(\sqrt{2}/n\) were rational, multiplying it by the integer \(n\) would make \(\sqrt{2}\) rational. Thus every \(x_n\) is irrational and \(h(x_n)=0\).
Also, \(|x_n-0|=\sqrt{2}/n\to0\), so \(x_n\to0\). But \(h(x_n)=0\) for every \(n\), and this constant image sequence does not converge to \(h(0)=1\): its distance from \(1\) is \(|0-1|=1\) at every term. Therefore \(h\) is discontinuous at \(0\).
Check: the input distance is \(\sqrt{2}/n\), which becomes smaller than any prescribed positive tolerance for sufficiently large \(n\); the output distance from the proposed value remains exactly \(1\). This is precisely the persistent discrepancy required by the discontinuity test.
A Common Mistake and a Reliable Procedure
Common mistake: choosing an arbitrary sequence approaching the point and treating its behavior as a test of discontinuity. For the sign function \(g\) above, the constant sequence \(x_n=0\) converges to \(0\), and \(g(x_n)=0\) for every \(n\). This image sequence does converge to \(g(0)=0\). It does not show that \(g\) is continuous: the criterion requires the image of every sequence approaching \(0\) to converge to \(g(0)\). The correction is to choose inputs that expose the change in the function. The sequence \(1/n\), for example, approaches \(0\) while its images are constantly \(1\), not \(0\).
For a practical proof, first write down the point \(c\) and the value \(f(c)\). Then choose inputs \(x_n\) in the domain and verify \(x_n\to c\) using the domain metric. Compute \(f(x_n)\) from the correct branch or definition, and compare it with \(f(c)\). If the image limit differs from \(f(c)\), or the images fail to converge to \(f(c)\), the criterion applies. A useful check is to identify a fixed \(\varepsilon_0>0\) for which \(d_Y(f(x_n),f(c))\ge\varepsilon_0\) for every \(n\), as in the oscillating and sign examples.
The metric-space hypothesis matters: the proof uses distances and the ability to choose inputs within \(1/n\) of the point. The result is stated here for metric spaces, not for arbitrary topological spaces. Also, the sequence must lie in the actual domain \(D\); if the domain is restricted, a proposed sequence outside it cannot test continuity of \(f:D\to Y\).
Check Your Understanding
For each question, specify the input sequence and compare its image behavior with the function value at the point.
- Suppose \(f:\mathbb{R}\to\mathbb{R}\) has \(f(0)=2\) and \(f(1/n)=5\) for every \(n\in\mathbb{N}\). What does the sequence \(1/n\) show about continuity at \(0\)?
- For the sign function \(g\) in the worked example, compute the image sequences for \(x_n=1/n\) and \(y_n=-1/n\). Why do they prove discontinuity at \(0\)?
- If \(x_n\to c\) and \(f(x_n)\to f(c)\), does that alone prove continuity at \(c\)? Explain using the quantifier in the sequential criterion.
- Suppose \(x_n\to c\) and \(d_Y(f(x_n),f(c))\ge 1/3\) for every \(n\). What conclusion follows?
- Why must a sequence used to test \(f:D\to Y\) consist entirely of points in \(D\)?