Let $r_{1}, r_{2},...$ be an enumeration of rational in $[0,1].$ For $n=1, 2,..$ let
$$t_{n}(x)=0, (0\leq x < r_{n})$$
$$t_{n}(x)=\frac{1}{n^2}, (r_{n}\leq x \leq 1)$$
and define
$$ f(x)=\sum_{n=1}^{\infty}t_{n}(x)$$
Then $f$ is continuous at all irrational points.
How can i approach this kind of question? how to think? Please help.
If $y$ is irrational and if $\varepsilon>0$, then there is a natural $N$ such that$$\sum_{n=N+1}^\infty\frac1{n^2}<\varepsilon.$$Then$$(\forall x\in[0,1]):\left\lvert f(x)-\sum_{n=1}^Nt_n(x)\right\rvert=\sum_{n=N+1}^\infty t_n(x)<\varepsilon.$$So, we have a series of functions, each of which is continuous at $y$, which converges uniformly to $f$. Therefore, $f$ is continuous at $y$