Exercise 7, § 7

Let $Y$ be a nonempty compact space, $\mathscr{R}$ a linear subspace of $\mathscr{C}(Y; \mathbf{R})$ that contains the constants, separates the points of $Y$ and is lattice-ordered (in other words, is a Riesz space).

a) Let $\mathscr{N}$ be a maximal isolated subspace of $\mathscr{R}$ (Ch. II, §1, Exer. 4); show that there exists one and only one point $y_0$ such that the positive linear form $\varphi : f \mapsto f(y_0)$ on $\mathscr{R}$ is latticial (Ch. II, §2, Exer. 5 b)) and $\mathscr{N} = \varphi^{-1}(0)$. (To see that there exists at least one point of $Y$ where all of the functions $f \geq 0$ belonging to $\mathscr{N}$ are zero, argue by contradiction, by showing that in the contrary case, taking into account the compactness of Y and the definition of the order relation in $\mathscr{R}$, one would have $\mathscr{N} = \mathscr{R}$. To show that the set $Z(\mathscr{N})$ of points of Y where all of the functions in $\mathscr{N}$ are zero reduces to a single point, make use of the fact that $\mathscr{N}$ is a hyperplane in $\mathscr{R}$ (Ch. II, §2, Exer. 5) and the fact that $\mathscr{R}$ separates the points of Y and contains the constants.)

b) Under the hypotheses of a), show that $y_0 \in \check{S}_{\mathscr{R}}(Y)$. (Setting $S = \check{S}_{\mathscr{R}}(Y)$, note that the subspace $\mathscr{R}'$ of $\mathscr{C}(S; \mathbf{R})$ formed by the restrictions to S of the functions in $\mathscr{R}$ is canonically isomorphic to $\mathscr{R}$ as an ordered vector space (Exer. 6 c)) and that, by means of the isomorphism inverse to $f \mapsto f|S$, the positive linear form $f \mapsto f(y_0)$ yields a latticial positive linear form on $\mathscr{R}'$; then apply a) to $\mathscr{R}'$.)

c) For every isolated subspace $\mathscr{N}_0 \neq \mathscr{R}$, show that there exists a maximal isolated subspace $\mathscr{N} \supset \mathscr{N}_0$. (Use the fact that if an isolated subspace contains a constant function $\neq 0$, then it is equal to $\mathscr{R}$.)

d) Let Z be the union of the sets $Z(\mathscr{N})$ (each reduced to a point) as $\mathscr{N}$ runs over the set of maximal isolated subspaces of $\mathscr{R}$; show that $Z = \check{S}_{\mathscr{R}}(Y)$. (First prove that Z is closed, by noting that if $y \notin Z$ then the linear form $f \mapsto f(y)$ on $\mathscr{R}$ is not latticial, consequently (using Ch. II, §2, Exer. 5 b)) neither is the linear form $f \mapsto f(y')$ for all points $y'$ sufficiently near $y$. Next, use Prop. 7 of No. 3, by showing that every function $f_0 \in \mathscr{R}$ attains its infimum $\alpha$ in Y at at least one point of Z; for this, setting $M = f_0^{-1}(\alpha)$, one considers the subspace $\mathscr{N}_0$ of functions $f \in \mathscr{R}$ of the form $f_1 - f_2$ with $f_1 \geq 0, f_2 \geq 0, f_1$ and $f_2$ zero on M; prove that $\mathscr{N}_0$ is isolated and use c).)