Proposition 8

Let P be a nonempty Polish topological space and let R be an equivalence relation in P whose graph is a meagre subset of $P\times P$. There exists a continuous injective mapping of the set $\{0,1\}^{\mathbf{N}}$, endowed with the product topology of the discrete topologies, into P whose image meets each equivalence class with respect to R in at most one point.

Let the space P be endowed with a metric $d$ compatible with its topology for which it is complete. Let $(A_n)_{n\in\mathbf{N}}$ be an increasing sequence of closed subsets of $P\times P$, with empty interiors, such that the graph $\Gamma_R$ of the relation R is contained in the union of the $A_n$.

Let $\mathscr{O}$ be the set of nonempty open subsets of P. We shall construct by induction a sequence $(f_n)_{n\in\mathbf{N}}$, where for every $n$, $f_n$ is a mapping of $\{0,1\}^n$ into $\mathscr{O}$, satisfying the following properties :

(i) For every $n\geqslant 1$, every $x\in \{0,1\}^{n-1}$ and every $t\in \{0,1\}$, the closure of the open set $f_n(x, t)$ is contained in $f_{n-1}(x)$;

(ii) For every $x\in \{0,1\}^n$, one has diam($f_n(x)$)$\leqslant 2^{-n}$;

(iii) For every $n\geqslant 1$ and every pair $(x, x')$ of distinct elements of $\{0,1\}^n,f_n(x)\times f_n(x')$ does not meet $A_{n-1}$.

Choose a nonempty open set U of X of diameter $\leqslant 1$ and define $f_0$ as the constant mapping with image $\{U\}$. Suppose that the mappings $f_0, . . . , f_n$ have been constructed.

Let $p:\{0,1\}^{n+1}\rightarrow \{0,1\}^n$ be the mapping defined by $p(x_0, . . . , x_n) =$ $(x_0, . . . , x_{n-1})$. By lemma 3 above, applied to the family of open sets $(f_n(p(x)))$ for $x\in \{0,1\}^{n+1}$ and to the dense open set $\complement A_n$ of $P\times P$, there exists a family $(g(x))_{x\in \{0,1\}^{n+1}}$ of nonempty open sets of P such that $g(x)\subset f_n(p(x))$ for every $x$ and such that $(g(x)\times g(x'))\cap A_n$ be empty for every pair $(x, x')$ of distinct elements of $\{0,1\}^{n+1}$. One then defines the mapping $f_{n+1}$ by choosing, for each element $x\in \{0,1\}^{n+1}$, a nonempty open subset of $g(x)$ whose diameter is $\leqslant 2^{-n-1}$ and whose closure is contained in $g(x)$.

For every element $x= (x_n)_{n\in\mathbf{N}}$ of $\{0,1\}^{\mathbf{N}}$, the sequence of sets $(f_n(x_0, . . . , x_{n-1}))_{n\in\mathbf{N}}$ is a decreasing sequence of open subsets of X each of which contains the closure of the next and whose diameter tends to 0; the intersection of this sequence of sets is therefore reduced to a point (TG, II, p. 15), which we denote by $f(x)$. If two points $x,x'$ of $\{0,1\}^{\mathbf{N}}$ satisfy $x_i=x'_i$ for $i\leqslant n$, one has $d(f(x), f(x'))\leqslant 2^{-n}$. Consequently, the mapping $f:\{0,1\}^{\mathbf{N}}\rightarrow P$ is continuous. Let $x$ and $x'$ be distinct elements of $\{0,1\}^{\mathbf{N}}$. For $n\in \mathbf{N}$ such that $(x_0, . . . , x_n)=\not$ $(x'_0, . . . , x'_n)$, the open set $f_{n+1}(x_0, . . . , x_n)\times f_{n+1}(x'_0, . . . , x'_n)$ is disjoint from $A_n$, by definition of $f_{n+1}$, hence the pair $(f(x), f(x'))$ does not belong to $A_n$. It follows that $f(x)$ and $f(x')$ are not equivalent for the relation R. Consequently, $f$ is injective and its image meets each equivalence class for R in at most one point. The proposition is thus proved.