Proposition 9

The component of any point in a topological space $X$ is a closed set. The relation "y belongs to the component of x" is an equivalence relation $R \{ x, y \}$ on $X$, and the equivalence classes are the components of $X$. The quotient space $X/R$ is totally disconnected.

The first part of the proposition is an immediate consequence of Definition 3 and the fact that the closure of a connected set is connected (no. 1, Proposition 1). Since the union of connected sets which have a point in common is connected (no. 1, Proposition 2), the relation $R$ is transitive, hence is an equivalence relation (since it is obviously reflexive and symmetric) and the equivalence class of $x$ with respect to $R$ is the component of $x$. It remains to show that $X/R$ is totally disconnected. Let $f : X \to X/R$ be the canonical mapping, and let $F$ be a closed set in $X/R$ containing at least two distinct points; the inverse image $f^{-1}(F)$ of $F$ is closed in $X$, saturated with respect to $R$, and contains at least two distinct components of $X$ and hence is not connected. Hence there exist two non-empty closed sets $B, C$ in $X$ such that $B \cap C = \varnothing$ and $B \cup C = f^{-1}(F)$. The component of any point $x$ of $f^{-1}(F)$ in $f^{-1}(F)$ is the same as the component of $x$ in $X$ (by the definition of $R$) and therefore $B$ and $C$, which are both open and closed in $f^{-1}(F)$, are saturated with respect to $R$. Hence $f(B)$ and $f(C)$ are closed in $X/R$, and $f(B) \cup f(C) = F$ and $f(B) \cap f(C) = \varnothing$; this shows that $F$ is not connected and consequently that $X/R$ is totally disconnected.