Exercise 4.18

(Matt “frito” Lundy)
Because every point $x\in R$ is a limit point of $R$, we can show that for $r\in Q$, we have $\underset{x\to r}{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x)=0\ne f(r)$ and for $y\in R,y\notin Q$, we have $\underset{x\to y}{\lim <!--\; FUNCTION\; APPLICATION\; -->}f(x)=0=f(y)$.

Fix $x\in R$. Let $\{x_n\}$ be a sequence in $R$ such that $x_n\ne x$ and $x_n\to x$ as $n\to \infty$.

If $\{x_n\}$ has only finitely many $x_n\in Q$, then there exists some $N\in N$ such that $n\ge N$ implies that $f(x_n)=0<𝜀$, because each $x_n$ is irrational after some finite number of terms. If $\{x_n\}$ has infinitely many $x_n\in Q$, let $\{x_{n_l}\}$ be a subsequence of $\{x_n\}$ consisting of all the rational $x_n$. $x_{n_l}$ must converge to $x$ because $x_n$ converges to $x$.

Each $x_{n_l}=\frac{p_{n_l}}{q_{n_l}}$ where $p_{n_l}\in Z$, $q_{n_l}\in N$ and $\gcd <!--\; FUNCTION\; APPLICATION\; -->(p_{n_l},q_{n_l})=1$. We claim that $q_{n_l}\to +\infty$, as $n_l\to +\infty$, so that

Suppose that $\underset{n_l\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{q}_{n_l}\ne +\infty$, then there exists some $M\in N$ such that $q_{n_l}<M$ for all $n_l$. Then we have:

If $\{p_{n_l}\}$ is unbounded, then, $\{q_{n_l}\}$ bounded means that (3) is unbounded. If $\{p_{n_l}\}$ is bounded, then there are finitely many values of $p_{n_l}$ and $q_{n_l}$. $x\ne x_n$ means that $q_{n_l}x-p_{n_l}\ne 0$ for any $n_l$, so let $\eta =\min <!--\; FUNCTION\; APPLICATION\; -->\left(\left|q_{n_l}x-p_{n_l}\right|\right)>0$ then:

So,

leads to

not converging to

, a contradiction. So

, and for any

,

, and thus

is continuous at every irrational

, and discontinuous at every rational

.