Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 10E / e / ii / Solution

Using the preceding $f$, define $g(0)=f(0)+1$ and $g(x)=f(x)$ for $x>0$. Changing the value only at zero makes zero discontinuous while preserving continuity at every $1/n$ and discontinuity everywhere else. Thus the continuity set is precisely $\{1/n:n\ge 1\}$.