Theorem 1

Suppose that I is directed, that the sets $\mathfrak{S}_\alpha$ satisfy conditions (i) and (ii), and that the inverse system $(\mathrm{E}_\alpha, f_{\alpha\beta})$ has the following properties:

(iii) For each pair of indices $\alpha$, $\beta$ such that $\alpha \leqslant \beta$, and each $x_\alpha \in \mathrm{E}_\alpha$, we have $\overset{-1}{f}_{\alpha\beta}(x_\alpha) \in \mathfrak{S}_\beta$.

(iv) For each pair of indices $\alpha$, $\beta$ such that $\alpha \leqslant \beta$, and each $\mathrm{M}_\beta \in \mathfrak{S}_\beta$, we have $f_{\alpha\beta}(\mathrm{M}_\beta) \in \mathfrak{S}_\alpha$.

Let $\mathrm{E} = \varprojlim \mathrm{E}_\alpha$ and let $f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha$ be the canonical mapping for each $\alpha \in \mathrm{I}$. Then:

(a) For each $\alpha \in \mathrm{I}$ we have

(b) If $\mathrm{E}_\alpha$ is non-empty for each $\alpha \in \mathrm{I}$, then E is non-empty.

Let $\Sigma$ be the set of families $\mathfrak{A} = (\mathrm{A}_\alpha)_{\alpha \in \mathrm{I}}$ which satisfy the following conditions :

If $\mathfrak{A} = (A_\alpha)$ and $\mathfrak{A}' = (A'_\alpha)$ are any two elements of $\Sigma$, let the relation $\mathfrak{A} \leqslant \mathfrak{A}'$ mean that $A_\alpha \supset A'_\alpha$ for all $\alpha$. Clearly $\Sigma$ is ordered by this relation.

(1) Let us first show that the ordered set $\Sigma$ is inductive. Let L be a totally ordered set and let $\lambda \to \mathfrak{A}^\lambda = (A^\lambda_\alpha)_{\alpha \in I}$ be a strictly increasing mapping of L into $\Sigma$. For each $\alpha \in I$, put $B_\alpha = \bigcap_{\lambda \in L} A^\lambda_\alpha$. Then it is immediately seen that the family $\mathfrak{B} = (B_\alpha)_{\alpha \in I}$ satisfies (21); by reason of (i) and (ii), it also satisfies (20), hence belongs to $\Sigma$; and it is clear that $\mathfrak{B}$ is an upper bound of the set of the $\mathfrak{A}^\lambda$.

(2) Let $\mathfrak{A} = (A_\alpha)$ be a maximal element of $\Sigma$. We shall show that $A_\alpha = f_{\alpha\beta}(A_\beta)$ whenever $\alpha \leqslant \beta$. Let $A'_\alpha = \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(A_\beta)$ for all $\alpha \in I$, and let us show that $\mathfrak{A}' = (A'_\alpha)$ belongs to $\Sigma$. Note first that if $\alpha \leqslant \beta \leqslant \gamma$, we have $f_{\alpha\gamma}(A_\gamma) = f_{\alpha\beta}(f_{\beta\gamma}(A_\gamma)) \subset f_{\alpha\beta}(A_\beta)$ by (21); moreover, $f_{\alpha\beta}(A_\beta) \in \mathfrak{S}_\alpha$ by (iv), and $f_{\alpha\beta}(A_\beta) \neq \emptyset$ by (20). Hence conditions (i) and (ii) show that $\mathfrak{A}'$ satisfies (20). Also, if $\alpha \leqslant \beta$, we have

and for each $\delta \geqslant \alpha$ there exists $\gamma \in I$ such that $\gamma \geqslant \delta$ and $\gamma \geqslant \beta$, so that $f_{\alpha\gamma}(A_\gamma) \subset f_{\alpha\delta}(A_\delta)$ and consequently

Hence $\mathfrak{A}'$ satisfies (21) and therefore belongs to $\Sigma$. Since $A'_\alpha \subset A_\alpha$ for all $\alpha$, the maximality of $\mathfrak{A}$ in $\Sigma$ implies $\mathfrak{A}' = \mathfrak{A}$, and our assertion is proved.

(3) We shall establish next that if $\mathfrak{A} = (A_\alpha)$ is a maximal element of $\Sigma$, then each of the $A_\alpha$ consists of a single element. Let $x_\alpha \in A_\alpha$. For each $\beta \geqslant \alpha$, put $B_\beta = A_\beta \cap \overset{-1}{f}_{\alpha\beta}(x_\alpha)$; if $\beta$ is not $\geqslant \alpha$, put $B_\beta = A_\beta$; then we shall see that $\mathfrak{B} = (B_\beta)$ belongs to $\Sigma$. If $\beta$ is not $\geqslant \alpha$, the relation $\beta \leqslant \gamma$ implies $f_{\beta\gamma}(B_\gamma) \subset f_{\beta\gamma}(A_\gamma) \subset A_\beta = B_\beta$. If, on the other hand, $\alpha \leqslant \beta \leqslant \gamma$, then since

we have

and since $f_{\beta\gamma}(A_\gamma) \subset A_\beta$, we again have $f_{\beta\gamma}(B_\gamma) \subset B_\beta$, so that the family $\mathfrak{B}$ satisfies (21). Since $A_\alpha = f_{\alpha\beta}(A_\beta)$ whenever $\alpha \leqslant \beta$ by part (2) of the proof, it is clear that $\mathrm{B}_\beta \neq \emptyset$ for all $\beta \in \mathrm{I}$. Finally, by virtue of (i) and (iii), we have $\mathrm{B}_\beta \in \mathfrak{S}_\beta$ for all $\beta \in \mathrm{I}$, and hence $\mathfrak{B} \in \Sigma$. Since $\mathrm{B}_\beta \subset \mathrm{A}_\beta$ for all $\beta \in \mathrm{I}$, the maximality of $\mathfrak{A}$ implies that $\mathrm{B}_\beta = \mathrm{A}_\beta$ for all $\beta$, and in particular $\mathrm{A}_\alpha = \{x_\alpha\}$.

(4) We are now in a position to prove Theorem 1. Let us start with (a). We have

Conversely, let

and put

if $\beta \geqslant \alpha$, and $\mathrm{B}_\beta = \mathrm{E}_\beta$ otherwise. By the definition of $x_\alpha$, the $\mathrm{B}_\beta$ are non-empty, and we have $\mathrm{B}_\beta \in \mathfrak{S}_\beta$ for all $\beta \in \mathrm{I}$ by virtue of (iii) and (i); moreover, it is evident that $f_{\beta\gamma}(\mathrm{B}_\gamma) \subset \mathrm{B}_\beta$ whenever $\beta \leqslant \gamma$. Hence $\mathfrak{B} = (\mathrm{B}_\beta) \in \Sigma$. Let $\mathfrak{A} = (\mathrm{A}_\alpha)$ be a maximal element of $\Sigma$ such that $\mathfrak{A} \geqslant \mathfrak{B}$ (the existence of $\mathfrak{A}$ follows from (1) and §2, no. 4, Theorem 2, Corollary 1). Since, by (3), $\mathrm{A}_\beta$ is of the form $\{y_\beta\}$ for all $\beta \in \mathrm{I}$, it follows that $y = (y_\beta)$ belongs to E, and $f_\alpha(y) = y_\alpha = x_\alpha$ by definition.

Finally, (a) implies (b). We may as well assume that I is not empty (otherwise there is nothing to prove). The hypothesis that the $\mathrm{E}_\alpha$ are non-empty implies that $f_{\alpha\beta}(\mathrm{E}_\beta) \neq \emptyset$ for all $\beta \geqslant \alpha$. Since the $f_{\alpha\beta}(\mathrm{E}_\beta)$, for fixed $\alpha$ and variable $\beta \geqslant \alpha$, form a left directed set of subsets of $\mathrm{E}_\alpha$ belonging to $\mathfrak{S}_\alpha$, condition (ii)$'$ proves that

Hence $f_\alpha(\mathrm{E}) \neq \emptyset$ by (a), and a fortiori $\mathrm{E} \neq \emptyset$. Q.E.D.