Past exam of the options II course of the University of Oxford / 2010 / E / E3

A subset $A$ is connected if its subspace topology admits no separation into two disjoint nonempty relatively open subsets. Equivalently, every continuous map $A\to \{0,1\}$, with the target discrete, is constant: a nonconstant map gives the separation by its two fibers, and a separation defines such a map.

The connected subsets of the usual real line are the empty set and all intervals, including singletons, bounded intervals of any endpoint type, rays, and the whole line. Equivalently, a nonempty subset is connected exactly when it contains every point between any two of its points.

Use the characterization by continuous maps to the discrete two-point space. Let $A={\bigcup }_i{A}_i$ and let $f:A\to \{0,1\}$ be continuous. Its restriction to each connected $A_i$ is constant. Since $A_i\cap A_j\ne \varnothing$, those two constants agree for every pair $i,j$. Hence $f$ is constant on their union. Thus the union is connected. If the indexing set is empty, the union is empty and is connected directly.

LetThis family is nonempty because it contains $\{a\}$. All its members intersect at $a$, so part (a)(iii) proves that $K_a$ is connected. It contains $a$ and, by its construction, contains every connected subset containing $a$. Thus it has exactly the required largest-subset property. This is the connected component containing $a$.

Suppose two components $K$ and $L$ intersect. Their union is connected by part (a)(iii). It contains the point used to define $K$, so maximality forces $K\cup L\subseteq K$, giving $L\subseteq K$. The same reasoning with $L$ gives $K\subseteq L$. Hence $K=L$. Thus distinct connected components are disjoint.

Let $C$ be a component of the open set $X\subseteq \mathbb{R}$. Being a connected subset of the real line, $C$ is an interval. For any $x\in C$, openness of $X$ gives $ϵ>0$ with $(x-ϵ,x+ϵ)\subseteq X$. This interval is connected and meets $C$, so their union is connected. Maximality of $C$ makes the entire interval lie in $C$. Thus every point of $C$ is interior, so each component is an open interval, allowing unbounded endpoints.

Every point belongs to its component, so the components cover $X$; by part (ii) they are pairwise disjoint, and by part (iii) they are nonempty open intervals. Each contains a rational number. Fix an enumeration of $\mathbb{Q}$ and choose the first rational belonging to each component. Distinct components receive distinct rationals, giving an injection from the component family into a countable set. ThereforeThe empty open set is represented by the empty union.