Proposition 7

Let X be a separated space. Every Souslinian subspace (TG, IX, p. 59, def. 2) of X is approachable.

Let S be a Souslinian subspace of X. By definition, there exists a complete metric space P of countable type and a continuous surjective mapping $g$ of P onto S. By TG, IX, p. 64, lemma 3, there exists a sieve $C = (C_n, p_n, \varphi_n)_{n\in\mathbf{N}}$ of the metric space P.

For every integer $n$ and every $c\in C_n$, let us denote by $F_n(c)$ the image of the set $\varphi_n(c)$ under $g$; let us also put $F^*_n(c) = D^*(F_n(c))$ and

For every $c\in C_n,F^*_n(c)$ is approachable (IV, p. 361, lemma 2, c)). Since $C_{n+1}$ is countable, the union of the approachable subsets $F^*_{n+1}(c')$, where $c'$ runs through $\overset{-1}{p_{n}}(c)$, is again an approachable subset of X. It contains the union of the $F_{n+1}(c')$, which is equal to $F_n(c)$. Hence (loc. cit., d)), $G_n(c)$ is a meagre subset of X. The union G of the subsets $G_n(c)$, for $n\in \mathbf{N}$ and $c\in C_n$, is therefore a meagre subset of X.

Let $c_0\in C_0$ and let $x$ be an element of $F^*_0(c_0)\cap \complement G$; let us prove that $x\in F_0(c_0)$. Since $x\notin G_0(c_0)$ and $x\in F^*_0(c_0)$, there exists $c_1\in \overset{-1}{p_{0}}(c_0)$ such that $x\in F^*_1(c_1)$, by relation (1). By recurrence, there exists an element $c= (c_n)_n$ of $\prod_nC_n$ such that, for every integer $n\in \mathbf{N}$, one has $x\in F^*_n(c_n)$ and $p_n(c_{n+1}) =c_n$.

The subsets $\varphi_n(c_n)$ form a basis of a Cauchy filter in P; this filter converges to a point $p$ of P, since P is complete. The image of this filter under $g$ is a filter F on X, having as basis the set of the $F_n(c_n)$ for $n\in \mathbf{N}$, which converges to $g(p)$. Since $F^*_n(c_n)$ is contained in $\overline{F_n(c_n)}$ and $x$ belongs to each of the $F_n(c_n)$, $x$ is an adherent point of F, hence $x=g(p)$ because the space X is separated (TG, I, p. 52, prop. 1). The point $p$ belongs to $\overline{\varphi_1(c_1)}$, hence to $\varphi_0(c_0)$. Consequently, $x\in F_0(c_0)$.

Consequently, the set $F^*_0(c_0)$ - $F_0(c_0)$ is contained in G. It is therefore a meagre subset of X, hence also an approachable subset. Since $F_0(c_0) = F^*_0(c_0)$ - $(F^*_0(c_0)$ - $F_0(c_0))$ and $F^*_0(c_0)$ is approachable, $F_0(c_0)$ is approachable, because the approachable subsets of X form a tribe. Since $C_0$ is countable, the set S which is the union of the $F_0(c_0)$, for $c_0\in C_0$, is again approachable.