Lemma 5

Let X be a Hausdorff space and let A, A' be two disjoint Souslin subspaces of X. Then there is a Borel set B of X such that $B \supset A$ and $B \cap A' = \emptyset$.

By Proposition 13 of no. 5 there exist two sieves C, C' and continuous mappings f of L(C) onto A, $f'$ of L(C') onto A', constructed by the method described in no. 5. For each $n \geq 0$ and each $c \in C_n$, let $q_n(c)$ denote the subspace of L(C) formed of all sequences $(c_k)_{k \geq 0}$ such that $c_n = c$; $q_n(c)$ is a closed subspace of L(C). For each $\gamma = (c_n) \in L(C)$ the sequence of closed sets $q_n(c_n)$ is decreasing and forms a filter base with $\gamma$ as limit. Moreover, for each $c \in C_n$, the sets $q_{n+1}(d)$, where d runs through the set $p_n^{-1}(c)$ in $C_{n+1}$, form a partition of $q_n(c)$. Make the analogous definitions of notation for the sieve $c'$.

We shall argue by contradiction and assume that every Borel set which contains A meets A'. In the first place, it follows from Lemma 4 and the definition of a sifting that there exist $c_0 \in C_0$ and $c'_0 \in C'_0$ such that every Borel set containing $f(q_0(c_0))$ meets $f'(q'_0(c'_0))$. We can then define, by induction on n,

as follows: suppose that $c_i$ and $c'_i$ have already been defined for $i < n$, in such a way that for each index $i < n$ every Borel set containing $f(q_i(c_i))$ meets $f'(q'_i(c'_i))$; applying Lemma 4 and the definition of a sifting to the sets $f(q_{n-1}(c_{n-1}))$ and $f'(q'_{n-1}(c'_{n-1}))$, we see that there exist $c_n \in C_n$ and $c'_n \in C'_n$ such that $p_{n-1}(c_n) = c_{n-1}$ and $p_{n-1}(c'_n) = c'_{n-1}$ and such that every Borel set containing $f(q_n(c_n))$ meets $f'(q'_n(c'_n))$. Now the sequence $f(q_n(c_n))$ converges to a point $a = f(\gamma) \in A$, and the sequence $f'(q'_n(c'_n))$ converges to a point $a' = f'(\gamma') \in A'$. Since $A \cap A' = \emptyset$ and $X$ is Hausdorff, there is a closed neighbourhood $V$ of $a$ which does not contain $a'$, and thus for $n$ sufficiently large, $V$ contains $f(q_n(c_n))$ and does not meet $f'(q'_n(c'_n))$. This is a contradiction, since $V$ is a Borel set.

To prove Theorem 2, let $Y_n$ denote the union of the sets $X_i$ such that $i \neq n$; then $Y_n$ is a Souslin subspace of $X$ (no. 2, Proposition 8). For each index $n$ there exists a Borel set $B'_n$ which contains $X_n$ and does not meet $Y_n$, by Lemma 5. Let $B_n$ be the intersection of $B'_n$ and $\bigcap_{i < n} (X - B'_i)$. Then the $B_n$ are Borel sets, are mutually disjoint, and are such that $B_n \supset X_n$ for each $n$.