Bourbaki

Exercise 13, § 4

General Topology, X

Bourbaki marks this one as harder than the rest.

Let XX be a compact space and let AA denote the normed algebra C(X;R)\mathscr{C}(X; \mathbf{R}). For each subset MM of AA, let V(M)V(M) denote the set of all x∈Xx \in X such that u(x)=0u(x) = 0 for all u∈Mu \in M. For each subset YY of XX, let J(Y)\mathfrak{J}(Y) denote the set of all u∈Au \in A such that u(x)=0u(x) = 0 for all x∈Yx \in Y.

a) V(M)V(M) is a closed subspace of XX and J(Y)\mathfrak{J}(Y) is a closed ideal of AA. If aa is the ideal of AA generated by a subset MM of AA, then V(M)=V(a)V(M) = V(a); VV and II are order-reversing mappings (with respect to the order relation of inclusion), and we have V{0}=X; V(A)=∅; V({u})=∅V\{0\} = X; \ V(A) = \varnothing; \ V(\{u\}) = \varnothing for every unit uu of AA;

V(⋃λ∈LMλ)=V(∑λ∈LMλ)=⋂λ∈LV(Mλ) V\left( \bigcup_{\lambda \in L} M_\lambda \right) = V\left( \sum_{\lambda \in L} M_\lambda \right) = \bigcap_{\lambda \in L} V(M_\lambda)

for every family (Mλ)λ∈L(M_\lambda)_{\lambda \in L} of subsets of AA; V(M.M′)=V(M)∪V(M′)V(M.M') = V(M) \cup V(M'); J(X)={0}; J(C)=A; J(⋃λ∈LYλ)=⋂λ∈LJ(Yλ)\mathfrak{J}(X) = \{0\}; \ \mathfrak{J}(\mathscr{C}) = A; \ \mathfrak{J}\left( \bigcup_{\lambda \in L} Y_\lambda \right) = \bigcap_{\lambda \in L} \mathfrak{J}(Y_\lambda) for every family (Yλ)λ∈L(Y_\lambda)_{\lambda \in L} of subsets of XX.

b) Show that if aa is any ideal of AA, then J(V(a))=a‾\mathfrak{J}(V(a)) = \overline{a}, and that if YY is any subset of XX, then V(J)(Y)=Y‾V(\mathfrak{J})(Y) = \overline{Y}. (For the first of these relations, observe that V(a‾)=V(a)V(\overline{a}) = V(a) and that we may therefore suppose aa closed; note that if x,yx, y are two distinct points of X−V(a)X - V(a), there exists u∈au \in a such that u(x)≠u(y)u(x) \neq u(y), and use Proposition 4 of no. 2. To prove the second relation, use Urysohn's theorem.)

c) Deduce from b) that x→J({x})x \to \mathfrak{J}(\{x\}) is a bijection of XX onto the set of maximal ideals of AA, which are therefore closed. An ideal of AA is closed if and only if it is an intersection of maximal ideals. The ring AA has no radical.

d) Show that the mapping e→V({e})e \to V(\{e\}) is a bijection of the set of idempotents of AA onto the set of subsets of XX which are both open and closed.

A point x∈Xx \in X is isolated if and only if the maximal ideal J({x})\mathfrak{J}(\{x\}) is principal.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.