There is sequence of continuous functions $f_{n}$ on the unit interval $[0,1]$ which converges to a function $f$ such that $f$ is discontinuous at rational points, a dense subset of the interval.

Let $f(x)=\begin{cases} 0& x \;\text{is irrational or } x\in \{0,1\}\\ 1/n & x=m/n,\;\;\;\;(m,n)=1 \end{cases}$

Let $\{r_{0},r_{1},\ldots, r_{n}\ldots, \}$ be the sequence of rational numbers in $[0,1]$.

Let $f_{n}$ be the unique continuous picewise linear function which satisfies $f_{n}(r_{j})=f(r_{j})\;\text{for}\;\;j=0,1,\ldots,n$ and vanishes at end points of the interval.

It is easy to show that $f_{n}$ converges to $f$ and $f$ is discountinuous at rational points

There is sequence of continuous functions $f_{n}$ on the unit interval $[0,1]$ which converges to a function $f$ such that $f$ is discontinuous at rational points of $(0,1)$, a dense subset of the interval.

Let $f(x)=\begin{cases} 0& x \;\text{is irrational or } x\in \{0,1\}\\ 1/n & x=m/n,\;\;\;\;(m,n)=1 \end{cases}$

Let $\{r_{0},r_{1},\ldots, r_{n}\ldots, \}$ be the sequence of rational numbers in $[0,1]$.

Let $f_{n}$ be the unique continuous picewise linear function which satisfies $f_{n}(r_{j})=f(r_{j})\;\text{for}\;\;j=0,1,\ldots,n$ and vanishes at end points of the interval.

It is easy to show that $f_{n}$ converges to $f$ and $f$ is discountinuous at rational points of $(0,1)$.

There is an example on the unit interval $[0,1]$.

Let $f(x)=\begin{cases} 0& x \;\text{is irrational or } x\in \{0,1\}\\ 1/n & x=m/n,\;\;\;\;(m,n)=1 \end{cases}$

Let $\{r_{0},r_{1},\ldots, r_{n}\ldots, \}$ be the sequence of rational numbers in $[0,1]$.

Let $f_{n}$ be the unique continuous picewise linear function which satisfies $f_{n}(r_{j})=f(r_{j})\;\text{for}\;\;j=0,1,\ldots,n$ and vanishes at end points of the interval.

It is easy to show that $f_{n}$ converges to $f$ and $f$ is discountinuous at rational points

There is sequence of continuous functions $f_{n}$ on the unit interval $[0,1]$ which converges to a function $f$ such that $f$ is discontinuous at rational points, a dense subset of the interval.

Let $f(x)=\begin{cases} 0& x \;\text{is irrational or } x\in \{0,1\}\\ 1/n & x=m/n,\;\;\;\;(m,n)=1 \end{cases}$

Let $\{r_{0},r_{1},\ldots, r_{n}\ldots, \}$ be the sequence of rational numbers in $[0,1]$.

Let $f_{n}$ be the unique continuous picewise linear function which satisfies $f_{n}(r_{j})=f(r_{j})\;\text{for}\;\;j=0,1,\ldots,n$ and vanishes at end points of the interval.

It is easy to show that $f_{n}$ converges to $f$ and $f$ is discountinuous at rational points