Exercise 4.19

Suppose $f$ is not continuous at $x_0$. Then there is a sequence of real numbers $y_n$ converging to $x_0$ such that $f(y_n)$ doesn’t converge to $f(x_0)$, that is, there is a $\delta >0$ such that $(f(x_0)-\delta ,f(x_0)+\delta )$ contains only a finite number of the $f(y_n)$. Hence there is an infinite subsequence $x_n$ of the $y_n$ such that either all the $f(x_n)$ are greater than $f(x_0)+\delta$ or less than $f(x_0)-\delta$. Suppose the first case is true (the second case can be handled by considering the function $-f$, which also satisifies the hypotheses). Following the hint, let $r$ be a rational number such that $f(x_n)>r>f(x_0)$ for all $n$. Then by the intermediate value property satisfied by $f$, for each $n$ there is a number $t_n$ such that $t_n$ lies between $x_0$ and $x_n$ and $f(t_n)=r$. Since $t_n$ converges to $x_0$, $x_0$ is in the closure of the set of all $x$ such that $f(x)=r$. This set is closed by hypothesis, so we must have $f(x_0)=r$, contradicting the assumption that $f(x_0)<r$.