Lemma 2

If $(W, U)$ is a distinguished pair then the set of all distinguished pairs $(W', U')$ such that $W'$ meets $W$ is countable.

Since $\mathfrak{B}$ is countable it is enough to show that, given $U' \in \mathfrak{B}$, the set of distinguished pairs $(W', U')$ such that $W'$ meets $W$ is countable. Now these sets $W'$ are open, since $X$ is locally connected (no. 6, Proposition 11) and mutually disjoint since they are components of $\overline{p}^{-1}(U')$; hence the sets $W' \cap W$ are open and mutually disjoint. But $W$ contains a countable subset which is dense in $W$; hence the set of $W'$ such that $W' \cap W$ is not empty is also countable.

To prove Theorem 1, consider the following relation R between two points x, x' of X: "There exists a finite sequence of distinguished pairs (W_i, U_i) (1 \leq i \leq n) such that x \in W_1 and x' \in W_n and that W_i \cap W_{i+1} \neq \emptyset for \leq 1 i \leq n — 1."

Lemma 1 states that R is reflexive, and it is readily verified that R is symmetric and transitive, so that R is an equivalence relation; also, since the W_i are open, every equivalence class mod R is open in X. But X is connected; hence there can be only one equivalence class, i.e. any two points of X are congruent mod R. We shall deduce from this that X is the union of a countable family of first elements of distinguished pairs, and this will prove Theorem 1. For this, let x be any point of X, and define by induction on n a sequence (C_n) of open subsets of X as follows: by Lemma 1 there is a distinguished pair (W_1, U_1) such that x \in W_1, and we take C_1 = W_1; if n > 1 then C_n is to be the union of all first elements W of distinguished pairs (W, U) such that W meets C_{n-1}. By induction on n one shows immediately, by virtue of Lemma 2, that C_n is a countable union of first elements of distinguished pairs. Finally, every x' \in X belongs to some C_n; for there is a finite sequence

of distinguished pairs such that x \in W'_1, x' \in W'_m and W'_i \cap W'_{i+1} \neq \emptyset for 1 \leq i \leq m — 1, and by induction on i we see that W'_i \subset C_{i+1} for all i, so that x' \in C_{m+1}.

Q.E.D.