Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 4F (Metric and Topological Spaces)

This does not define a topology on $\mathbb{Z}$. Every singleton $\{2k\}$ is declared open, but their union over $k\in \mathbb{Z}$ is $2\mathbb{Z}$, an infinite proper subset. It is therefore not declared open, violating closure under arbitrary unions. This is the example that finite subsets do not define a topology on an infinite set. The empty set, whole space and finite-intersection requirements do hold, but they do not compensate for the failed union axiom. This is different from the cofinite topology, which uses finite complements rather than finite sets.

If an open set contains an even integer, invariance under every even translation makes it contain all even integers. Similarly, containing an odd integer makes it contain all odd integers. Conversely each parity class is invariant under those translations. Thus the proposed open sets are exactlyThey form a partition topology. Both the empty and whole sets are included, and any union or finite intersection is again a union of parity classes. Hence this is a topological space. It is not the discrete topology: no singleton is open.