Proposition 5

Let $(\mathrm{E}_\alpha, f_{\alpha\beta})$ be an inverse system of sets relative to a directed set I which has a countable cofinal subset, and suppose furthermore that the $f_{\alpha\beta}$ are surjective. Then, if $\mathrm{E} = \varprojlim \mathrm{E}_\alpha$, the canonical mapping $f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha$ is surjective for each $\alpha \in \mathrm{I}$ (and, a fortiori, E is not empty provided that none of the $\mathrm{E}_\alpha$ is empty).

Let $(\alpha_n)$ be a sequence of elements of I which form a cofinal subset of I. Since I is directed, we can define a sequence $(\beta_n)$ of elements of I inductively by the conditions $\beta_0 = \alpha_0, \beta_n \geqslant \beta_i$ for $i < n$ and $\beta_n \geqslant \alpha_n$. Clearly the sequence $(\beta_n)$ is increasing and forms a cofinal subset of I. In view of Proposition 1 of no. 1 and the relations $f_\alpha = f_{\alpha\beta_n} \circ f_{\beta_n}$ for $\alpha \leqslant \beta_n$, we need only prove the Proposition for the case $\mathrm{I} = \mathbf{N}$. Moreover, it is clear that it suffices to prove that $f_0$ is surjective. Let $x_0 \in \mathrm{E}_0$. Define $x_n \in \mathrm{E}_n$ (for $n \geqslant 1$) inductively to be an element of the set $\overset{-1}{f}_{n-1,n}(x_{n-1})$, which is possible because the latter set is non-empty by hypothesis. We then show by induction on $n - m$ that $x_m = f_{mn}(x_n)$ for $m \leqslant n$, and it follows that $x = (x_n)$ belongs to E.

The second criterion concerns inverse systems $(\mathrm{E}_\alpha, f_{\alpha\beta})$ relative to an index set I such that for each $\alpha \in \mathrm{I}$ we are given a set $\mathfrak{S}_\alpha$ of subsets of $\mathrm{E}_\alpha$ which satisfy the following conditions :

(i) Every intersection of sets belonging to $\mathfrak{S}_\alpha$ also belongs to $\mathfrak{S}_\alpha$.

It follows in particular (by considering the intersection of the empty family) that $\mathrm{E}_\alpha \in \mathfrak{S}_\alpha$.

(ii) If a set of subsets $\mathfrak{F} \subset \mathfrak{S}_\alpha$ is such that every finite intersection of sets belonging to $\mathfrak{F}$ is non-empty, then $\bigcap\limits_{\mathbf{M} \in \mathfrak{F}} \mathrm{M}$ is non-empty.

In view of (i), it is clear that (ii) is equivalent to the following condition :

(ii)′ If $\mathfrak{G} \subset \mathfrak{S}_\alpha$ is left directed (with respect to inclusion) and does not contain the empty set, then $\bigcap\limits_{\mathbf{M} \in \mathfrak{G}} \mathrm{M}$ is non-empty.