Proposition 6

*Let $\mathbf{E}$ be a prehilbertian space and let $\Phi$ be a non-empty, directed decreasing set of non-empty Hausdorff and complete convex subsets of $\mathbf{E}$. For every $x \in \mathbf{E}$ and every subset $\mathbf{H}$ of $\mathbf{E}$, put $d(x, \mathbf{H}) = \inf_{z \in \mathbf{H}} \| x - z \|$. In order that the intersection M of the sets H belonging to $\Phi$ be non-empty, it is necessary and sufficient that there exists $x_0$ in E such that $\sup_{H \in \Phi} d(x_0, H)$ is finite. For every $x \in E$ we then have

(limit with respect to the directed set $\Phi$).

If M is non-empty, $d(x, H) \leq d(x, M)$ for all $H \in \Phi$ and all $x \in E$.

Conversely, suppose that there exists a point $x_0$ in E and a real number $C \geq 0$ such that $d(x_0, H) \leq C$ for all $H \in \Phi$. Let $x \in E$; then

hence the number $d = \sup_{H \in \Phi} d(x, H)$ is finite. Let B be the set of all $z \in E$ such that $\|x - z\| \leq d$. Since B is convex and closed in E, the sets $H \cap B$, for H ranging over $\Phi$, are convex, Hausdorff and complete. Let $\varepsilon > 0$; there exists a set $H \in \Phi$ such that $d(x, H) \geq d - \varepsilon$, and if $\varepsilon < d/2$, the diameter of $H \cap B$ is bounded by $\sqrt{12 \varepsilon (d - \varepsilon)}$ by prop. 5 (V, p. 9). In other words, for all $H_0 \in \Phi$, the closed sets $H \cap B$, for $H \in \Phi$ and $H \subset H_0$, form a base of the Cauchy filter on the Hausdorff and complete space $H_0$. Hence the intersection of the sets $H \cap B$ (for $H \in \Phi$) reduces to a point y. We get $y \in M$ and $\|x - y\| = d = d(x, M)$. Since M is closed in $H_0$, it is a Hausdorff, convex and complete set in E, and so $y = p_M(x)$. For every $H \in \Phi$, we have $p_H(x) \in H \cap B$, from which we get that $p_M(x) = \lim_{H \in \Phi} p_H(x)$.