Let us show that the function ,

We shall use the following characterization of continuity for $f$ : $f$ is continuous at if and only if $\lim_{k\to \infty} f(x_k)=f(a)$ for all sequences such that $\lim_{k\to \infty} x_k=a$ .

It is not difficult to see that $f$ is continuous at $x=0$ . Indeed, if $x_k$ is a sequence converging to $0$ . Then \begin{eqnarray*} \lim_{k\to \infty} |f(x_k)| &=& \lim_{k\to \infty} |f(x_k)| \\ &=& \lim_{k\to \infty} |x_k| \\ &=& 0. \end{eqnarray*} Suppose $a\neq 0$ . Then there exists a sequence of irrational numbers $x_1, x_2, \ldots$ converging to $a$ . For instance, if $a$ is irrational, we can take $x_k = a+1/k$ , and if $a$ is rational, we can take $x_k = a+\sqrt{2}/k$ . For this sequence we have \begin{eqnarray*} \lim_{k\to \infty} f(x_k) &=& -\lim_{k\to \infty} x_k \\ &=& -a. \end{eqnarray*} On the other hand, we can also construct a sequence of rational numbers $y_1,y_2,\ldots$ converging to $a$ . For example, if $a$ is irrational, this follows as the rational numbers are dense in , and if $a$ is rational, we can set $y_k = x_k+ 1/k$ . For this sequence we have \begin{eqnarray*} \lim_{k\to \infty} f(y_k) &=& \lim_{k\to \infty} y_k \\ &=& a. \end{eqnarray*}In conclusion $f$ is not continuous at $a$ .

Bibliography

1

Homepage of Thomas Vogel, A function which is continuous at only one point.