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

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.