Exercise 13, § 4

Let $X$ be a compact space, $A$ the normed algebra $C(X; \mathbf{R})$. For every subset $M$ of $A$, denote by $V(M)$ the set of the $x \in X$ such that $u(x) = 0$ for every $u \in M$. For every subset $Y$ of $X$, denote by $\mathfrak{F}(Y)$ the set of the $u \in A$ such that $u(x) = 0$ in $Y$.

a) $V(M)$ is a closed subspace of $X$, $\mathfrak{F}(Y)$ a closed ideal of $A$. If $a$ is the ideal of $A$ generated by a subset $M$ of $A$, $V(M) = V(a)$; $V$ and $\mathfrak{F}$ are decreasing mappings for the relation of inclusion, and one has $V(\{0\}) = X$, $V(A) = \varnothing$, $V(\{u\}) = \varnothing$ for every invertible element $u$ of $A$, $V\left( \bigcup_{\lambda \in L} M_\lambda \right) = V\left( \sum_{\lambda \in L} M_\lambda \right) = \bigcup_{\lambda \in L} V(M_\lambda)$ for every family $(M_\lambda)_{\lambda \in L}$ of subsets of $A$, $V(M.M') = V(M) \cup V(M')$; $\mathfrak{F}(X) = \{0\}$, $\mathfrak{F}(\varnothing) = A$, $\mathfrak{F}\left( \bigcup_{\lambda \in L} Y_\lambda \right) = \bigcap_{\lambda \in L} \mathfrak{F}(Y_\lambda)$ for every family $(Y_\lambda)_{\lambda \in L}$ of subsets of $X$.

b) Prove that for every ideal $a$ of $A$, one has $\mathfrak{F}(V(a)) = \overline{a}$, and for every subset $Y$ of $X$, $V(\mathfrak{F}(Y)) = \overline{Y}$. (To prove the first equality, observe that $V(\overline{a}) = V(a)$ and that one may therefore suppose that $a$ is closed; remark that if $x, y$ are two distinct points of $X - V(a)$, there exists $u \in a$ such that $u(x) \neq u(y)$ and use prop. 4 of $X$, p. 37. To prove the second equality, use the Urysohn theorem).

c) Deduce from b) that $x \mapsto \mathscr{F}(\{x\})$ is a bijection of $X$ onto the set of maximal ideals of $A$, which are therefore closed. For an ideal of $A$ to be closed, it is necessary and sufficient that it be an intersection of maximal ideals. The ring $A$ is without radical (A, VIII, § 6, No. 3). d) Show that the mapping $e \mapsto V(\{e\})$ is a bijection of the set of idempotents of $A$ onto the set of subsets of $X$ which are both open and closed. For a point $x \in X$ to be isolated, it is necessary and sufficient that the maximal ideal $\mathscr{F}(\{x\})$ be principal.