Exercise 20, § 4

Let $X$ be a compact connected space and let $L$ be the set of all points of $X$ which have a fundamental system of connected neighbourhoods. The points of $S = CL$ are called the singular points of $X$. The points of the interior $\dot{L}$ of $L$, and the components of $\bar{S}$ are called the prime constituents of $X$.

a) Let $A \neq X$ be a non-empty open subset of $X$, let $K$ be a component of $\overline{A}$ and let $F$ be the closure of $\overline{A} \cap CK$. Show that if $A \cap K \cap F$ is not empty then all its points are singular. If $x \in A \cap F \cap K$, the component $Q$ of $x$ in $F$ meets $CA$ [consider a point of $CK$ contained in $V(x)$ and its component in $\overline{A}$, and show by using Exercise 17 that the set of points of $F$ which can be joined to $x$ by a $V$-chain in $F$ meets $CA$]. Deduce that the prime constituent $P$ which contains $x$ meets $CA$ (remark that $P$ contains the component of $x$ in $Q \cap A$, and use Exercise 17 applied to the compact connected space $Q$ and the open subset $Q \cap A$ of $Q$).

b) Deduce from a) that if $P$ is a prime constituent of $X$ and $U$ is a neighbourhood of $P$, then $P$ has a connected neighbourhood contained in $U$ (by reductio ad absurdum).

c) Let $V$ be a symmetric entourage of $X$, and let $V'$ denote the set of pairs $(x, y)$ of points of $X$ which can be joined by a $V$-chain, all of whose points are singular, except perhaps for $x$ and $y$. As $V$ runs through the symmetric entourages of X, the corresponding sets V' form a fundamental system of entourages of a uniformity (in general not Hausdorff) on X. Let X' be the associated Hausdorff space, and show that the inverse images of the points of X' under the canonical mapping X → X' are the prime constituents of X. The (compact connected) space X' is called the space of prime constituents of X. Show that X' is locally connected [use b) and Chapter I, § 10, no. 4, Proposition 8].

d) Let (rn) be the sequence of all the rational numbers contained in ]0,1[, arranged in some order. For each irrational x ∈ [0,1], let

Let X be the closure in $\mathbf{R}^2$ of the graph of f. Show that X is connected and irreducible between its points with abscissas 0 and 1, but that X has only one prime constituent.*