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

A subset $A$ is connected if, with its subspace topology, it cannot be written as $A=U\cup V$ with $U,V$ disjoint nonempty relatively open sets. Equivalently its only subsets both open and closed relative to $A$ are $\varnothing ,A$. This definition includes the empty set and singletons as connected sets.

The connected subsets of the real line are exactly the intervals, including singletons, unbounded intervals and the empty set. Equivalently, whenever $x<y$ belong to the set, every $z$ with $x<z<y$ also belongs to it. Endpoints may or may not be included.

Use the separation definition in (i). If $A={\bigcup }_i{A}_i$ had a separation $A=U\cup V$, the intersections with each connected $A_i$ show that $A_i$ lies wholly in one side. Choose one nonempty $A_{i_0}$; every other $A_i$ intersects it and must lie on that same side. Thus the other side is empty, a contradiction. If the union is empty it is connected by definition. Therefore $\boxed{\bigcup _i{A}_i\text{ is connected}}$.

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 $X=\varnothing$.