function continuous at only one point

Let us show that the function $f:ℝ\to ℝ$,

is continuous at $x=0$, but discontinuous for all $x\in ℝ\setminus \{0\}$[1].

We shall use the following characterization of continuity for $f$:$f$ is continuous at $a\in ℝ$ if and only if ${lim}_{k\to \mathrm{\infty }}f(x_k)=f(a)$for all sequences $(x_k)\subset ℝ$ such that ${lim}_{k\to \mathrm{\infty }}{x}_k=a$.

It is not difficult to see that $f$ is continuous at $x=0$. Indeed, if $x_k$ is asequence converging to $0$. Then

$\underset{k\to \mathrm{\infty }}{lim}|f(x_k)|$$=$$\underset{k\to \mathrm{\infty }}{lim}|f(x_k)|$

$=$$\underset{k\to \mathrm{\infty }}{lim}|x_k|$

$=$$0.$

Suppose $a\ne 0$. Then there exists a sequence of irrational numbers $x_1,x_2,\mathrm{\dots }$converging to $a$. For instance, if $a$ is irrational, we can take $x_k=a+1/k$, andif $a$ is rational, we can take $x_k=a+\sqrt{2}/k$. For this sequence we have

$\underset{k\to \mathrm{\infty }}{lim}f(x_k)$$=$$-\underset{k\to \mathrm{\infty }}{lim}{x}_k$

$=$$-a.$

On the other hand, we can also construct a sequence of rational numbers $y_1,y_2,\mathrm{\dots }$converging to $a$. For example, if $a$ is irrational, this follows as the rational numbersare dense in $ℝ$, and if $a$ is rational, we can set $y_k=x_k+1/k$.For this sequence we have

$\underset{k\to \mathrm{\infty }}{lim}f(y_k)$$=$$\underset{k\to \mathrm{\infty }}{lim}{y}_k$

$=$$a.$

In conclusion $f$ is not continuous at $a$.