Remark

Suppose that condition (iii) in Theorem 1 is replaced by the following weaker condition :

(iii)$'$ For each $\alpha \in \mathrm{I}$ and each non-empty set $\mathrm{M}_\alpha \in \mathfrak{S}_\alpha$ there exists $x_\alpha \in \mathrm{M}_\alpha$ such that $\overset{-1}{f}_{\alpha\beta}(x_\alpha) \in \mathfrak{S}_\beta$ for each $\beta \geqslant \alpha$.

Then the conclusion (b) of Theorem 1 remains valid; the proofs of parts (1) and (2) of Theorem 1 remain unchanged and the proof of part (3) remains valid provided that we are careful to take $x_\alpha \in \mathrm{A}_\alpha$ such that $\overset{-1}{f}_{\alpha\beta}(x_\alpha)$ belongs to $\mathfrak{S}_\beta$ for all $\beta \geqslant \alpha$. Finally, the proof of (4) shows that if

and if we choose an $x_\alpha$ in this set such that $\overset{-1}{f}_{\alpha\beta}(x_\alpha) \in \mathfrak{S}_\beta$ whenever $\beta \geqslant \alpha$, there exists $y \in E$ such that $f_\alpha(y) = x_\alpha$, which proves our assertion.

Examples