Lemma 3

Every metric space X of countable type possesses a sifting. If also X is zero-dimensional, then X has a strict sifting.

Note first that if $Y$ is a metric space of countable type and if $\varepsilon$ is a real number $> 0$, then there is a countable covering of $Y$ by sets of diameter $\leq \varepsilon$ ($§ 2$, no. 8, Proposition 13). If, moreover, $Y$ is zero-dimensional, there is such a covering $(V_n)$ formed of sets which are both open and closed; if $W_n$ is the intersection of $V_n$ and $\bigcap_{k < n} (Y - V_k)$, we see that the $W_n$ are closed, of diameter $\leq \varepsilon$, pairwise disjoint and cover $X$. In any case, the closures of the non-empty sets of the covering form a countable covering of $Y$ whose elements are non-empty closed sets of diameter $\leq \varepsilon$.

Let $X$ be a metric space of countable type. Let $C_0$ be the set of indices of a countable covering of $X$ formed of non-empty closed sets of diameter $\leq 1$, which are pairwise disjoint if $X$ is zero-dimensional; $\varphi_0$ shall be the mapping which associates with each index $c \in C_0$ the corresponding set of the covering. Suppose that we have already defined the $C_i$ and the $\varphi_i$ and the surjective mappings $p_i : C_{i+1} \to C_i$ for $i \leq n$ in such a way that condition b) is satisfied for these indices. If $c \in C_n$, the space $\varphi_n(c)$ has a countable covering by non-empty closed sets of diameter $\leq 1/2^{n+1}$, which are mutually disjoint if $X$ [and therefore $\varphi_n(c)$] is zero-dimensional; if $I(c)$ denotes the index set of this covering, we take $c_{n+1}$ to be the sum of the sets $I(c)$ as $c$ runs through $C_n$; for each $c' \in C_{n+1}$, let $p_n(c')$ denote the element $c \in C_n$ such that $c' \in I(c)$, and let $\varphi_{n+1}(c')$ denote the set with index $c'$ in the covering of $\varphi_n(c)$ under consideration. Clearly we thus define by induction a sifting of $X$, and this sifting is strict if $X$ is zero-dimensional; hence Lemma 3.

Now suppose that $X$ is a complete metric space of countable type, and consider a sifting of $X$ by a sieve $C$ and mappings $\varphi_n$. If $\gamma = (c_n)$ is a point of the space $L(C)$ associated with $C$, the sequence $(\varphi_n(c_n))$ is a decreasing sequence of closed sets in $X$ whose diameters tend to 0; the intersection of this sequence of sets consists therefore of a single point, which we denote by $f(\gamma)$. Thus we have defined a mapping $f : L(C) \to X$. If two points $\gamma, \gamma'$ of $L(C)$ have the same $i$-th coordinates for $i \leq n$, it is clear that the distance between $f(\gamma)$ and $f(\gamma')$ is $\leq 1/2^n$, and therefore $f$ is continuous, by virtue of the definition of the topology of $L(C)$. For each $x \in X$ it follows from the definition of a sifting that we can define by induction on $n$ a sequence $\gamma = (c_n)$ such that $x \in \varphi_n(c_n)$ for each $n \geq 0$, and $c_n = p_n(c_{n+1})$; hence $x = f(\gamma)$ and so $f$ is surjective. Furthermore, if the sifting is strict, the sequence $\gamma = (c_n)$ such that $x = f(\gamma)$ is unique, and so in this case $f$ is bijective. $f$ is said to be the mapping induced by the sifting considered.