Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 4E / i / Solution

If $f(n)\to \infty$, then for each $k$ the complement of $f^{-1}(I_k)$ contains only finitely many indices. Thus every nonempty open tail has a cofinite preimage, and the empty set has empty preimage. Hence $f$ is continuous from the cofinite space to the tail-topology space.