Proposition 6

In a compact space $X$, the component of $x$, the set $A_x$, and the intersection of the neighbourhoods of $x$ which are both open and closed, are all three identical.

It is enough to show that $A_x$ is connected: for in any uniform space $X$ the component of $x$ is contained in the intersection of the neighbourhoods of $x$ which are both open and closed, and this intersection is contained in $A_x$ by Proposition 5.

Suppose $A_x$ is not connected. Since $A_x$ is closed, there are two non-empty disjoint closed sets $B$ and $C$ such that $B \cup C = A_x$. By Proposition 4 of § 3 there is an entourage $U$ of $X$ such that $U(B) \cap U(C)$ is empty.

Let $W$ be an open entourage such that $\hat{W} \subset U$, and let $H$ be the closed set which is the complement of $W(B) \cup W(C)$ in $X$. Suppose for example that $x \in B$, and let $y$ be a point of $C$. Then for each symmetric entourage $V \subset W$ one sees immediately, by induction on $i$, that every $V$-chain $(x_i)_{0 \leq i \leq n}$ joining $x$ and $y$ in $X$ must have a point in $H$, by the choice of $W$. Since by hypothesis $x$ and $y$ can be joined by a $V$-chain for each symmetric entourage $V$, we see that $H \cap A_{x, v}$ is not empty if $V \subset W$. On the other hand, if $V' \subset V$ then clearly $A_{x, v'} \subset A_{x, v}$; thus it follows that, as $V$ runs through the set of symmetric entourages of $X$, the sets $H \cap A_{x, v}$ form a filter base of closed sets in the compact space $H$. Hence all these sets have a common point and therefore $H$ meets $A_x$; but this contradicts the definition of $H$.