Limits and continuity · Definition

Sequential continuity

real analysisfunctionscontinuitysequences

Before you start you need to know

It translates continuity into the language of sequences, where known convergence results (monotonicity, subsequences, algebra of limits) can be used directly.

Definition

A function \(f\) is sequentially continuous at \(c\) if, for every sequence \((x_n)\) in the domain converging to \(c\), the sequence of images \(f(x_n)\) converges to \(f(c)\).

\[x_n\to c\ \Longrightarrow\ f(x_n)\to f(c)\]

The central result is the sequential criterion: for a real function at a point of its domain, being continuous at \(c\) and being sequentially continuous at \(c\) are the same. If \(f\) is continuous and \(x_n\to c\), given \(\varepsilon\) the \(\delta\) of the definition works for \(|x_n-c|<\delta\) from some \(n\) on, so \(f(x_n)\to f(c)\). The converse is proved by contrapositive: if \(f\) is not continuous, there is \(\varepsilon\) such that for each \(\delta=1/n\) there is \(x_n\) with \(|x_n-c|<1/n\) and \(|f(x_n)-f(c)|\ge\varepsilon\).

In practice the criterion is mostly used to refute. One sequence \(x_n\to c\) whose images do not converge to \(f(c)\) is enough: with \(x_n=-1/n\) the step function turns out discontinuous at \(0\), and with \(x_n=\sqrt2/n\) so does the Dirichlet function. Two sequences converging to \(c\) with images of different limits also suffice, because at most one of them can converge to \(f(c)\).

It also simplifies proofs of general properties. Since \(f(x_n)\to f(c)\) and \(g(x_n)\to g(c)\), the algebra of limits of sequences immediately gives that \(f+g\), \(fg\) and, with \(g(c)\ne0\), \(f/g\) are continuous at \(c\). Composition is handled the same way, applying the condition first to \(g\) and then to \(f\).

The criterion demands every sequence: checking it on a few concrete sequences, even if they give \(f(c)\), does not prove continuity, and a sequence that does not converge to \(c\) says nothing. In more general spaces the two notions can differ, but in \(\mathbb R\) they coincide, and this equivalence will be used with the intermediate value theorem and with theorems on closed intervals.

Think about it

Prove that \(f(x)=x^2\) is continuous at \(c\) using sequences.

If \(x_n\to c\), by the algebra of limits \(x_n^2=x_n\cdot x_n\to c\cdot c=c^2=f(c)\).

Is checking \(f(x_n)\to f(c)\) for \(x_n=c+1/n\) enough to conclude that \(f\) is continuous at \(c\)?

No. The condition must hold for every sequence converging to \(c\); for instance, a function with a jump to the left of \(c\) satisfies it with \(c+1/n\) and is not continuous.

Exercises 10

  1. Let \(f(x)=x^2\) and \(x_n=1+1/n\). What does \(f(x_n)\) converge to?
  2. Let \(f(x)=1\) if \(x\ge0\) and \(f(x)=-1\) if \(x<0\). Which sequence converging to \(0\) shows that \(f\) is not continuous at \(0\)?
  3. Let \(f\) be the Dirichlet function (\(1\) at rationals, \(0\) at irrationals) and \(x_n=\sqrt2/n\). What is \(\lim f(x_n)\)?
  4. If \(f\) is continuous at \(c\), what holds?
  5. Let \(f(x)=\dfrac{x^2-1}{x-1}\) if \(x\ne1\) and \(f(1)=3\). With \(x_n=1+1/n\), what does \(f(x_n)\) converge to?
  6. To prove with the sequential criterion that \(f\) is not continuous at \(c\), it suffices to exhibit:
  7. Let \(f(x)=\sin(1/x)\) for \(x\ne0\) and \(f(0)=0\). With \(x_n=\dfrac{1}{\pi/2+2\pi n}\), what does \(f(x_n)\) converge to?
  8. The sequential criterion says that, for a real function \(f\) at a point \(c\) of its domain:
  9. Let \(f(x)=x\) if \(x\) is rational and \(f(x)=0\) if irrational. At which points is it continuous?
  10. Let \(f\) and \(g\) be continuous at \(c\), with \(f(c)=3\) and \(g(c)=-2\). If \(x_n\to c\), what does \(f(x_n)\,g(x_n)\) converge to?

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §4.3, Teorema 4.3.2
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 4, Teorema 4.2 (límites por sucesiones)
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §5.1, Teorema 5.1.3