Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / E / E3 / b

Let $K_a$ be the union of all connected subsets containing $a$. This collection is nonempty because it includes $\{a\}$. Any two of its members meet at $a$, so (a)(iii) makes their full union connected. It contains $a$ and by its definition contains every connected subset through $a$. Thushas exactly the required largest-set property.

If $K_a\cap K_b\ne \varnothing$, their union is connected by (a)(iii). It contains $a$, so maximality gives $K_a\cup K_b\subseteq K_a$ and hence $K_b\subseteq K_a$. Interchanging the roles gives the reverse inclusion. Thus the two components coincide; otherwise they are disjoint.

The inclusion of $X$ into $\mathbb{R}$ is continuous, so a connected component $K$ of $X$ is an interval in $\mathbb{R}$. For $x\in K$, openness of $X$ gives $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq X$. This interval is connected and meets $K$, so their union is connected and maximality forces the interval into $K$. Thus $K$ is open in $\mathbb{R}$. A nonempty open connected subset of the line is an open interval, allowing an infinite endpoint or the whole line. Hence every component is an open interval.

The components cover $X$ and are pairwise disjoint by (b)(ii); by (iii) they are open intervals. Each nonempty interval contains a rational number. Enumerate the rationals and assign to each interval the first rational it contains. Disjointness makes these assigned rationals distinct, giving an injection from the set of components into $\mathbb{N}$. Thus $\boxed{X\text{ is a countable union of pairwise disjoint open intervals}}$, with the empty union for $X=\varnothing$.