Exercise 18.6

Proof. As it is easier to do so, we show this using the $𝜖$-$\delta$ definition of continuity, which we know implies the open set definition by Exercise 18.1. First we note that $f(0)=0$ since $0$ is rational. Now consider any $𝜖>0$ and let $\delta =𝜖$. Suppose real $y$ where $\left|y-0\right|=\left|y\right|<\delta$. If $y$ is rational then $y=0$ so that $\left|f(y)-f(0)\right|=\left|0-0\right|=\left|0\right|=0<𝜖$. If $y$ is irrational then $\left|f(y)-f(0)\right|=\left|y-0\right|=\left|y\right|<\delta =𝜖$ again. Since $𝜖$ was arbitrary this shows that $f$ is continuous at $x=0$.

Now consider any $x\ne 0$. Let $𝜖=\left|x\right|∕2$, noting that $𝜖>0$ since $x\ne 0$ so that $\left|x\right|>0$. Now consider any $\delta >0$.

Case: $x\in \mathbb{Q}$. Then $f(x)=0$ but there is clearly an irrational $y$ close enough to $x$ so that $\left|y-x\right|<\min <!--\; FUNCTION\; APPLICATION\; -->(𝜖,\delta )$, and hence both $\left|y-x\right|<𝜖$ and $\left|y-x\right|<\delta$. We also have that $f(y)=y$. We then have that

so that

Case: $x\notin \mathbb{Q}$. Then $f(x)=x$, and there is clearly a rational $y$ close enough to $x$ that $\left|y-x\right|<\delta$. We then also have $f(y)=0$ so that

since $𝜖>0$.

Hence in either case there is a $y$ such that $\left|y-x\right|<\delta$ but $\left|f(y)-f(x)\right|\ge 𝜖$. This suffices to show that $f$ is not continuous at $x$. □