Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2007 / 4

The topology on $A$ is the subspace topology inherited from the complex plane. A subset $B$ is open in $A$ if $B=A\cap U$ for some open subset $U\subseteq \mathbb{C}$. Equivalently, every $b\in B$ has a radius $ϵ>0$ such that $A\cap \{z:|z-b|<ϵ\}\subseteq B$.

A subset $B$ is closed in $A$ if $A\setminus B$ is open in $A$, equivalently if $B=A\cap F$ for a closed subset $F$ of the complex plane. These are relative properties: the same subset need not be open or closed in the ambient plane.

The set $A$ is disconnected space if it has a separation of a topological space: two nonempty disjoint subsets $U,V$ open in $A$ with $A=U\cup V$. Each is then also closed in $A$, being the other's complement.

The set $A$ is connected if no such separation of a topological space exists. Equivalently its only subsets that are both open and closed are $\varnothing$ and $A$. A separation of a topological space uses openness in $A$ rather than necessarily in the entire complex plane.

Because the image is integer-valued, the open interval $(n-\frac{1}{2},n+\frac{1}{2})$ contains exactly one possible image value, namely $n$. ThusThe right side is open in $A$ by continuity, soThis includes the possibility of an empty fiber. The source notation $f^{-1}(n)$ is understood as the preimage of the singleton $\{n\}$.

Since $A$ is nonempty, choose $a_0\in A$ and put $n_0=f(a_0)$. The fiber $U=f^{-1}(\{n_0\})$ is nonempty and open by part (c). Its complement is the union of the other integer fibers and is therefore also open. If the image contained a second value, that complement would be nonempty, and these two disjoint open subsets would separate $A$. connectedness excludes this. HenceEquivalently a continuous function from a connected space to the discrete space of integers is constant. The proof uses relative openness and does not require the original subset $A$ to be open in the plane.