If $x \in A$, every neighbourhood of $x$ in $A$ is of the form $V \cap A$, where $V$ is a neighbourhood of $x$ in $X$. Since $V \cap B = (V \cap A) \cap B$, it follows that $x$ lies in the closure of $B$ with respect to $A$ if and only if $x$ lies in the closure of $B$ with respect to $X$.

It follows that if $A, B, C$ are three subsets of $X$ such that $A \supset B \supset C$, and if $B$ is dense in $A$, and $C$ is dense in $B$, then $C$ is dense in $A$ (transitivity of density), for we have $\overline{A} = \overline{B} = \overline{C}$ in $X$.

For $V$ contains $U \cap A$, where $U$ is an open subset of $X$ which contains $x$, hence $\overline{V}$ contains $U \cap \overline{A} = U$ (§ 1, no. 6, Proposition 5).

Under these conditions, a subset $B$ of $X$ is open (resp. closed) in $X$ if and only if each of the sets $B \cap A_i$ is open (resp. closed) in $A_i$.

Clearly if $B$ is open (resp. closed) in $X$, then $B \cap A_i$ is open (resp. closed) in $A_i$. Conversely, suppose first that condition a) is satisfied; since $(\complement B) \cap A_i = A_i - (B \cap A_i)$, it is enough, by duality, to consider the case in which each of the $B \cap A_i$ is open with respect to $A_i$. In this case $B \cap \dot{A}_i$ is open in $\dot{A}_i$ for each $i \in I$, and therefore open in $X$; and since $B = \bigcup_i (B \cap \dot{A}_i)$ by hypothesis, it follows that $B$ is open in $X$.

Now suppose that b) is satisfied; by duality again, we need only consider the case in which each of the $B \cap A_i$ is closed in $A_i$, and therefore closed in $X$. Since the family $(B \cap A_i)$ is locally finite and $B = \bigcup_i (B \cap A_i)$, it follows from § 1, no. 5, Proposition 4 that $B$ is closed in $X$.

For if $F'$ is a closed subset of $X'$ and if $F = \overline{f}(F')$, then $F \cap A_i$ is closed in $A_i$ for each $i \in I$ ($§ 2$, no. 1, Theorem 1) and therefore $F$ is closed in $X$ by Proposition 3 of no. 1; the result now follows from Theorem 1 of $§ 2$, no. 1.

a) $L$ is locally closed in $X$.

b) $L$ is the intersection of an open subset and a closed subset of $X$.

c) $L$ is open in its closure $\overline{L}$ in $X$.

If $L$ is locally closed, then, for each $x \in L$, there is an open neighbourhood $V_x$ of $x$ in $X$ such that $L \cap V_x$ is closed in $V_x$; $U = \bigcup_{x \in L} V_x$ is open in $X$, and Proposition 3 of no. 1 shows that $L$ is closed in $U$; therefore a) implies b). If $L = U \cap F$, where $U$ is open and $F$ closed in $X$, we have $L \subset F$; hence $L \subset U \cap L \subset U \cap F = L$, which shows that $L = U \cap L$ is open in $L$, so that b) implies c). Finally, if $L = U \cap L$, where $U$ is open in $X$, $L$ is closed in $U$, hence locally closed, and thus c) implies a).

This follows immediately from Proposition 5 above and Theorem 1 of § 2, no. 1.

This is a particular case of § 2, no. 4, Proposition 6; it expresses the fact that the quotient topology is the final topology for the mapping $\varphi$.

Proposition 6 shows that there is a one-to-one correspondence between the continuous mappings of $X/R$ into $Y$ and the continuous maps of $X$ into $Y$ which are constant on each equivalence class mod $R$.

This is a particular case of a general property of quotient structures (Set Theory, Chapter IV, § 2, no. 6, criterion CST 20).

This is a particular case of the transitivity of final topologies (§ 2, no. 4, Proposition 7. Cf. Set Theory, Chapter IV, § 2, no. 3, criterion CST 21).

For the condition b) [resp. c)] expresses that the image under $g$ of every open (resp. closed) set in $X/R$ is an open (resp. closed) set in $f(X)$.

For if $\varphi : X \to X/R$ is the canonical mapping, then $g$ and $\varphi \circ s$ are bijective, continuous and inverse to each other; likewise $s$ and the restriction of $f$ to $s(Y)$ are bijective, continuous and inverse to each other.

If $R$ is an equivalence relation on a topological space $X$ and

A continuous section of a topological space with respect to an equivalence relation need not exist (Exercise 12).

a) $h$ is a homeomorphism.

b) Every open subset of $A$ which is saturated with respect to $R_A$ is the intersection with $A$ of an open subset of $X$ which is saturated with respect to $R$.

c) Every closed subset of $A$ which is saturated with respect to $R_A$ is the intersection with $A$ of a closed subset of $X$ which is saturated with respect to $R$.

The first part of the proposition is immediate (no. 5). The second part follows from Proposition 8 of no. 5: if $B$ is an open (resp. closed) subset of $A$ which is saturated with respect to $R_A$, and $g(B) = f(B)$ is the intersection with $f(A)$ of an open (resp. closed) subset $C$ of $X/R$, then $B$ is the intersection with $A$ of the open (resp. closed) subset $f^{-1}(C)$

(*) This expression means that $f \circ j = \psi \circ h \circ \varphi$.

of X, which is saturated with respect to R; and conversely, if B is the intersection with A of an open (resp. closed) subset D which is saturated with respect to R, then $f(B)$ is the intersection of $f(A)$ and $f(D)$, which is open (resp. closed) in $X/R$.

For if A is open (resp. closed) in X and saturated with respect to R, and if $B \subset A$ is open (resp. closed) in A and saturated with respect to $R_A$, then B is open (resp. closed) in X and saturated with respect to R.

Since each equivalence class mod R meets A, the canonical image of $A/R_A$ in $X/R$ is the whole of $X/R$; on the other hand, if U is open in A and is saturated with respect to $R_A$, it follows from the hypothesis that $\bar{u}^1(U)$ is the set obtained by saturating U with respect to R; since $u$ is continuous, $\bar{u}^1(U)$ is open in X (§ 2, no. 1, Theorem 1). The corollary follows from this fact by virtue of Proposition 10.