Let $(A_i)_{i \in I}$ be a family of connected subsets of $X$, all of which contain the same point $x$; we have to show that

is connected. If not, there are two open sets $B$ and $C$ such that $B \cap A$ and $C \cap A$ are non-empty, and $A \subset B \cup C$ and $A \cap B \cap C = \emptyset$. $x$ belongs to one of the sets $B, C$, say $x \in B$; on the other hand one of the sets $A_i$, say $A_x$, meets $C$; we have therefore $A_x \subset B \cup C$, $A_x \cap B \cap C = \emptyset$ and $B \cap A_x$ and $C \cap A_x$ are non-empty. Hence $A_x$ is not connected, which is a contradiction.

By induction on $n$ we see immediately that the set $B_n = \bigcup_{i=0}^n A_i$ is connected for all $n$, by Proposition 2. The sets $B_n$ have a non-empty intersection; hence their union, equal to $\bigcup_{n=0}^\infty A_n$, is connected by Proposition 2.

For otherwise the intersections of $B$ with the interior and exterior of $A$ would be two open subsets of $B$ which form a partition of $B$, and $B$ would not be connected.

Suppose $f(A)$ is not connected. Then there exist two sets $M', N'$ which are open in $f^{-1}(A)$ and which form a partition of $f(A)$; hence $A \cap f^{-1}(M')$ and $A \cap f^{-1}(N')$ are open in $A$ and form a partition of $A$; this contradicts the hypothesis that $A$ is connected.

The inverse image of a connected set under a continuous mapping need not be connected; consider for example a mapping of a discrete space into a space consisting of one point.

From Proposition 4 we derive another characterization of non-connected spaces:

The condition is sufficient by Proposition 4. Conversely, if $X$ is not connected, there exist two non-empty disjoint open subsets $A, B$ whose union is $X$, and the mapping $f$ of $X$ onto a discrete space of two elements $\{a, b\}$, defined by $f(A) = \{a\}$ and $f(B) = \{b\}$, is continuous.

This is an immediate consequence of Proposition 4 of no. 2.

Suppose $X$ is not connected. Then there is a partition of $X$ into two open sets $A, B$. The sets $A, B$ are saturated with respect to $R$; for if $x \in A$ then the equivalence class $M$ of $x$ cannot meet $B$, otherwise the sets $A \cap M, B \cap M$ would form a partition of $M$ into two sets open in $M$, which is impossible since $M$ is connected. The canonical images of $A$ and $B$ are therefore open sets in $X/R$ and form a partition of $X/R$; this contradicts the assumption that $X/R$ is connected.

Let $X = \prod_{i \in I} X_i$ be a product of topological spaces. If the $X_i$ are non-empty, we have $X_i = \operatorname{pr}_i X$ for each $i \in I$; hence if $X$ is connected so are the $X_i$ (no. 2, Proposition 4). Conversely, suppose that each of the $X_i$ is connected and $X$ is not. By Proposition 5 of no. 2, there exists a continuous surjective mapping $f : X \to X'$, where $X'$ is a discrete space which contains more than one point. Let $a = (a_i)$ be any point of $X$, and $x$ any index; the partial mapping $f_x : X_x \to X'$, defined by $f_x(x) = f((y_i))$ where $y_x = x$ and $y_i = a_i$ if $i \neq x$, is continuous on $X_x$; since $X_x$ is connected, $f_x$ must be constant on $X_x$. It follows immediately by induction that $f(x) = f(a)$ for all points $x = (x_i)$ such that $x_i = a_i$ for all but a finite number of indices $i \in I$. But these points $x$ form a dense subset of $X$ (§ 4, no. 3, Proposition 8). Hence $f$ is continuous on $X$ and constant on a dense subset of $X$, and therefore constant on $X$ (§ 8, no. 1, Proposition 2, Corollary 1). But this contradicts the definition of $f$.

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.

This product set is connected (no. 4, Proposition 8). Conversely, if $A$ is a connected subset of $X$ which contains $x$, then $\operatorname{pr}_i(A)$ is a connected set (no. 2, Proposition 4) which contains $x$; since $A \subset \prod_i \operatorname{pr}_i(A)$, it follows that $A$ is contained in the product of the components of the $x_i$.

The condition is sufficient, since the component of $x$ relative to an open neighbourhood of $x$ is then a neighbourhood of $x$ in $X$.

Conversely, let $A$ be an open subset of a locally connected space $X$, let $B$ be a component of $A$, and let $x \in B$. Let $V$ be a connected neighbourhood of $x$ contained in $A$; by the definition of components, $V$ is contained in $B$; hence $B$ is open in $X$ (§ 1, no. 2, Proposition 1).

The components of a locally connected space $X$ therefore form a partition of $X$ into open sets, and hence $X$ is the sum (§ 2, no. 4, Example 3) of its components.

For $V$ is open and closed in $U$, hence a frontier point of $V$ (relative to $X$) cannot belong to $U$, for it would also be a frontier point of $V$ relative to $U$, and there is none.

Let $X$ be a locally connected space, $R$ an equivalence relation on $X$, $\varphi : X \to X/R$ the canonical mapping. Let $A$ be an open subset of

X/R and C a component of A. Then $\overline{\varphi}^1(C)$ is a union of components of $\overline{\varphi}^1(A)$; for if $x \in \overline{\varphi}^1(C)$ and if K is the component of x in $\overline{\varphi}^1(A)$ then $\varphi(K)$ is connected (no. 2, Proposition 4), is contained in A, and contains $\varphi(x)$; hence $\varphi(K) \subset C$ by the definition of C, and therefore $K \subset \overline{\varphi}^1(C)$. Since X is locally connected and $\overline{\varphi}^1(A)$ is open in X, it follows from Proposition 11 that $\overline{\varphi}^1(C)$ is open in X; consequently C is open in X/R and hence, by Proposition 11 again, X/R is locally connected.

b) Conversely, if the product of a family $(X_i)$ of non-empty topological spaces is locally connected, then each $X_i$ is locally connected, and $X_i$ is connected for all but a finite number of indices.

a) Let J be the finite subset of I such that $X_i$ is not connected if and only if $i \in J$. Let

be an elementary set containing a point $x = (x_i)$ of X and let K be the finite subset of I such that $U_i \neq X_i$ if and only if $i \in K$. Let $V_i$ be $X_i$ for $i \notin J \cup K$, and let $V_i$ be a connected neighbourhood of $x_i$ contained in $U_i$ for $i \in J \cup K$; then is connected (by Proposition 8 of no. 4) and is a neighbourhood of $x$ contained in U. Hence X is locally connected.

b) Let $a = (a_i)$ be a point of X and let V be a connected neighbourhood of $a$ in X. Since we have $\mathrm{pr}_i V = X_i$ except for a finite number of indices (§ 4, no. 1) it follows from no. 2, Proposition 4 that the $X_i$ are connected, for all but a finite number of indices. On the other hand, for each $x \in I$, each $a_x \in X_x$ and each neighbourhood $V_x$ of $a_x$ in $X_x$, there is a point $x$ of X such that $\mathrm{pr}_x x = a_x$, and

is a neighbourhood of $x$ in X; V therefore contains a connected neighbourhood W of $x$, whose projection $\mathrm{pr}_x W$ is a connected neighbourhood of $a_x$ contained in $V_x$ (no. 2, Proposition 4 and § 4, no. 2, Proposition 5). Hence each $X_x$ is locally connected.

Then the space $X$ is the union of a countable family of open sets each of which is contained in a set of $\mathfrak{B}$.

Let $\mathfrak{B}$ be a countable base of the topology of $Y$. We shall say that a pair $(W, U)$ is distinguished if (i) $U \in \mathfrak{B}$ and (ii) $W$ is a component of $\overline{p}^{-1}(U)$ contained in a set of $\mathfrak{B}$.

The inverse image $\overline{p}^{-1}(p(x))$ is discrete and therefore there is a neighbourhood of $x$ in $X$ all of whose points $x'$ other than $x$ have an image $p(x') \neq p(x)$; since $X$ satisfies $(O_{III})$, there is a closed neighbourhood $V$ of $x$ with this property, and we may assume also that $V$ is contained in a set of $\mathfrak{B}$. Let $F$ be the frontier of $V$ in $X$. By condition (i) of the theorem, $p(F)$ is closed in $Y$; and since $p(F)$ does not contain $p(x)$, there is a set $U \in \mathfrak{B}$ which contains $p(x)$ and does not meet $p(F)$. Let $W$ be the component of $x$ in $\overline{p}^{-1}(U)$; then it is enough to show that $W \subset A$. If this were not so, then $W$ would meet $F$ (no. 1, Proposition 3) and therefore $p(F)$ would meet $U$, contrary to the definition of $U$.

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.

First, the hypotheses imply that X is regular (§ 8, no. 4, Proposition 13). Let us show that the conditions of Theorem 1 are satisfied if we take \mathfrak{B} to be the set of all closed subsets V of X such that the restriction of p to V is a homeomorphism of V onto a closed subspace of Y. By hypothesis, the interiors of the sets of \mathfrak{B} cover Y and, by virtue of the assumption on Y, each V \in \mathfrak{B} has a countable base and therefore contains a countable dense subset (§ 1, no. 6, Proposition 6). Furthermore, if x \in p^{-1}(y), there is a neighbourhood V \in \mathfrak{B} of x in X such that V con-

() \ It can be shown that these conditions imply that the topology of Y is metrizable (Chapter IX, § 4, Exercise 22). \* tains no point of $\overline{p}^{-1}(y)$ other than $x$, and hence $\overline{p}^{-1}(y)$ is a discrete space. We may therefore apply Theorem 1, which shows that $X$ is the union of a countable family $(T_n)_{n \geq 0}$ of open sets, such that each subspace $T_n$ has a countable base $(U_{mn})_{m \geq 0}$. Then the $U_{mn}$ ($m \geq 0, n \geq 0$) form a base of the topology of $X$ ($§ 3$, no. 1, Remark).

For each $x \in X$, let $V_x$ be a compact neighbourhood of $x$ in $X$ which has a countable base. It follows from $§ 9$, no. 4, Theorem 2, Corollary 2, that the set $\mathfrak{B}$ of the $V_x$ satisfies the conditions of Theorem 1, and we complete the proof as in Corollary 1.

Notice that, in this corollary, it can happen that the restriction of $p$ to an arbitrarily small neighbourhood $V$ of a point of $X$ is not a homeomorphism of $V$ onto $p(V)$.

Clearly $X$ is locally connected. Also each $x \in X$ has an open neighbourhood $U$ in $X$ such that the restriction of $p$ to $U$ maps $U$ homeomorphically onto an open subspace $p(U)$ of $Y$. Since $p(U)$ is a locally compact subspace of $Y$ ($§ 9$, no. 7, Proposition 13), there is a compact neighbourhood $W$ of $p(x)$ contained in $p(U)$, whence $U \cap \overline{p}^{-1}(W)$ is a compact neighbourhood of $x$ contained in $U$; thus $X$ is locally compact, since by hypothesis $X$ is Hausdorff. $U \cap \overline{p}^{-1}(W)$, being compact, is closed in $X$ ($§ 9$, no. 3, Proposition 4) and the conditions of Corollary 1 are therefore satisfied; hence the topology of $X$ has a countable base.