Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 4E (Further Analysis)

For the cofinite topology ${\tau }_1$, the empty set and $\mathbb{N}$ are included. A nonempty union of cofinite sets is cofinite because its complement is contained in the finite complement of any one member. A finite intersection is cofinite because its complement is a finite union of finite sets; intersections involving the empty set are empty. This proves all topology axioms.

For the tail topology on the natural numbers ${\tau }_2$, $\mathbb{N}=I_1$. Any nonempty collection of tails has a smallest starting index, so its union is that tail; a finite intersection of tails is the tail with largest starting index. Empty sets and empty-family conventions supply the remaining cases. Thus ${\tau }_2$ is also a topology.

We classify continuous maps between cofinite and tail topologies. Continuity is equivalent toIf $f$ has unbounded image, each $E_k$ is nonempty, hence cofinite. For every threshold $k$, eventually $f(n)\ge k$, which is exactly $f(n)\to \infty$. If the image is bounded, it is a nonempty finite subset of $\mathbb{N}$ and has a maximum $N$. Then $E_N$ is nonempty and cofinite, but $f(n)\le N$ always, so $f(n)=N$ except at finitely many indices. These exhaust the two possibilities and prove necessity. Their sufficiency is checked separately below.

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.

Suppose $f(n)\le N$ for all $n$ and $f(n)=N$ eventually. If $k>N$, then $f^{-1}(I_k)=\varnothing$. If $k\le N$, the preimage contains all sufficiently large indices and is cofinite. Every open set therefore has open preimage, proving $f$ is continuous. Together with the introductory argument this proves the claimed if-and-only-if classification.