Theorem 4

Let X be a Polish space and let R be an equivalence relation on X, such that the equivalence classes mod R are closed in X and the saturation (with respect to R) of each closed set in X is a Borel set. Then there is a Borel set in X which meets each equivalence class in exactly one point.

Consider a metric on $X$ compatible with its topology, and with respect to which $X$ is complete. By Lemma 3 of no. 5, there exists a sifting of $X$, defined by a sieve $C = (C_n, p_n)$ and a sequence of mappings $(\varphi_n)$. For each $c \in C_n$, let $g_n(c)$ be the saturation of the closed set $\varphi_n(c)$ with respect to $R$; by hypothesis, $g_n(c)$ is a Borel set in $X$.

Since each set $C_n$ is countable, we can linearly order each $C_n$ in such a way that the set of elements smaller than a given element is finite. For each $c \in C_n$ we define a set $h_n(c)$ by induction on $n$, as follows. In the first place, for $c \in C_0$, $h_0(c)$ is the intersection of $\varphi_0(c)$ and the sets $X - g_0(c')$, where $c' \in C_0$ and $c' < c$. For $c \in C_{n+1}$, $h_{n+1}(c)$ is the intersection of $\varphi_{n+1}(c), h_n(P_n(c))$ and the sets $X - g_{n+1}(c')$ for $c' \in C_{n+1}$, $p_n(c') = p_n(c)$ and $c' < c$. The $h_n(c)$ are clearly Borel sets.

We shall prove the following assertion: for each integer $n \geq 0$ and each equivalence class $H$ mod $R$, there is a unique element $c \in C_n$ such that $h_n(c)$ meets $H$, and we have

which is therefore a closed set. For $n = 0$, consider the smallest element $c \in C_0$ such that $\varphi_0(c)$ meets $H$; then $\varphi_0(c) \cap H$ does not meet any set $g_0(c')$ for which $c' \in C_0$ and $c' < c$; hence it is contained in $h_0(c) \cap H$ and consequently is equal to this set; moreover, we have $H \subset g_0(c)$ and therefore $h_0(c') \cap H$ is empty for $c' \in C_0$ and $c' > c$; thus the assertion is proved for $n = 0$. We continue by induction on $n$: if there exists $c \in C_{n+1}$ such that $h_{n+1}(c)$ meets $H$, then it follows from the relation $h_{n+1}(c) \subset h_n(p_n(c))$ and the inductive hypothesis that $p_n(c)$ is the unique element $d \in C_n$ such that $h_n(d)$ meets $H$. Observe that $h_n(d)$, which is contained in $\varphi_n(d)$, is contained in the union of the sets $\varphi_n(c)$ for which $c \in p_n^{-1}(d)$, by the definition of a sifting; there is therefore a smallest element $c \in p_n^{-1}(d)$ such that $\varphi_{n+1}(c)$ meets $H$. We have therefore by the inductive hypothesis. Hence and since by definition $\varphi_{n+1}(c) \cap H$ meets none of the sets $g_{n+1}(c')$ for which $c' \in p_n^{-1}(d)$ and $c' < c$, it follows from the definition of $h_{n+1}(c)$ that $\varphi_{n+1}(c) \cap H = h_{n+1}(c) \cap H$. Moreover, we have $H \subset g_{n+1}(c)$ and therefore, if $c' \in p_n^{-1}(d)$ is such that $c' > c$, $h_{n+1}(c') \cap H$ is empty. Hence the assertion is proved for all $n$.

For each integer $n$, let $S_n$ be the union of the sets $h_n(c)$, where $c$ runs through $C_n$. The set $S_n$ is a Borel set, and we have $S_{n+1} \subset S_n$.

Let S be the intersection of the sets S_n, which is also a Borel set in X; we shall show that S meets each equivalence class H mod R in exactly one point. For each n, let c_n(H) be the unique element c \in C_n such that h_{c_n(H)} meets H; then S_n \cap H = \varphi_n(c_n(H)) \cap H, and S \cap H is the intersection of the sets \varphi_n(c_n(H)) \cap H. Since the sequence (c_n(H)) belongs to L.C., the decreasing sequence of closed sets \varphi_n(c_n(H)), whose diameter tends to 0, converges to a point x \in X, since X is complete. The intersection of the closed sets \varphi_n(c_n(H)) \cap H therefore consists of the point x alone, and the proof of Theorem 4 is complete.