Bourbaki

Exercise 22, § 4

General Topology, X, p. 61

Bourbaki marks this one as harder than the rest.

Let XX be a compact space, AA the normed algebra C(X;C)C(X; \mathbf{C}) over C\mathbf{C}. For every closed ideal aa of AA, prove that the relation f∈af \in a implies fˉ∈a\bar{f} \in a (remark that fˉ\bar{f} is the uniform limit of functions of the form gfgf). Extend then to the algebra AA the results of Exercise 13 of XX, p. 57.

¶ 23 a) Let XX be a compact space, AA a finite rank algebra over the field R\mathbf{R}, admitting a unit element, and endowed with the topology defined in VI, p. 5. Let BB be a subalgebra of the R\mathbf{R}-algebra C(X;A)C(X; A); suppose that for every ε>0\varepsilon > 0, every pair of distinct points x,yx, y of XX and every pair of elements u,vu, v of AA, there exists f∈Bf \in B such that ∥f(x)−u∥≤ε\| f(x) - u \| \leq \varepsilon and ∥f(y)−v∥≤ε(∥z∥\| f(y) - v \| \leq \varepsilon (\| z \| being any norm defining the topology of AA); suppose moreover that if uu and vv belong to R\mathbf{R}, there is a function f∈Bf \in B satisfying the preceding conditions and taking its values in R\mathbf{R}. Under these conditions, prove that BB is dense in C(X;A)C(X; A) for the topology of uniform convergence. (First prove that every function of C(X;R)C(X; \mathbf{R}) can be uniformly approximated by functions of BB, then that the same is true of every constant a∈Aa \in A, using the assumption and a partition of unity).

b) Take A=HA = \mathbf{H}, field of quaternions over R\mathbf{R}. Let BB be a subalgebra of the R\mathbf{R}-algebra C(X;H)C(X; \mathbf{H}) such that: 1° for every function f∈Bf \in B, the conjugate fˉ\bar{f} of ff (defined by fˉ(x)=f(xˉ)\bar{f}(x) = f(\bar{x}) for every x∈Xx \in X) belongs to BB; 2° for every ε>0\varepsilon > 0, every pair of distinct points x,yx, y of XX and every pair of elements u,vu, v of H\mathbf{H}, there exists f∈Bf \in B such that ∥f(x)−u∥≤ε\| f(x) - u \| \leq \varepsilon and ∥f(y)−v∥≤ε\| f(y) - v \| \leq \varepsilon. Prove that BB is everywhere dense in C(X;H)C(X; \mathbf{H}) (use aa).

c) Extend to the algebra C(X;H)C(X; \mathbf{H}) the results of XX, p. 57, Exercise 13, and now prove in particular that every closed ideal of this algebra is two-sided (reason as in Exercise 22 using bb).

¶ 24) a) Let XX be a compact totally disconnected space, AA a topological ring. We equip the ring C(X;A)C(X; A) with the topology of uniform convergence, which is compatible with the ring structure. Let aa be a left ideal of C(X;A)C(X; A), and for every x∈Xx \in X, let F(x)F(x) be the closure in AA of the sets of the f(x)f(x) for f∈af \in a, which is a closed left ideal of AA. Now prove that aˉ\bar{a} is identical with the set of the f∈C(X;A)f \in C(X; A) such that f(x)∈F(x)f(x) \in F(x) for every x∈Xx \in X (observe that for every x∈Xx \in X the neighbourhoods of xx which are both open and closed form a fundamental system of neighbourhoods of xx).

b) Suppose in addition that there exists in AA a fundamental system of neighbourhoods of 0 which are two-sided ideals of AA. Let BB be a subring of C(X;A)C(X; A) containing the constants and separating the points of XX; now prove that BB is dense in C(X;A)C(X; A). (For every closed subset FF of XX, every point x∉Fx \notin F and every neighbourhood UU of 0 in AA, now prove that there exists f∈Bf \in B such that f(x)=1f(x) = 1 and f(y)∈Uf(y) \in U for every y∈Fy \in F).

25) Let XX be a locally compact space; we call strict topology on the vector space Cb(X;R)C^b(X; \mathbf{R}) of continuous and bounded numerical functions the topology defined by the family of distances dφ(f,g)=sup⁡x∈Xφ(x)∣f(x)−g(x)∣d_\varphi(f, g) = \sup_{x \in X} \varphi(x)|f(x) - g(x)|, where φ\varphi is a continuous numerical function, ≥0\geq 0 and tending to 0 at the point at infinity.

a) Now prove that for the uniform structure defined by these distances, Cb(X;R)C^b(X; \mathbf{R}) is complete and that the strict topology is compatible with the algebra structure on R\mathbf{R}.

b) Now prove that if AA is a closed subalgebra of Cb(X;R)C^b(X; \mathbf{R}) for the strict topology, separating the points of XX, and such that, for every x∈Xx \in X, there exists f∈Af \in A such that f(x)≠0f(x) \neq 0, then A=Cb(X;R)A = C^b(X; \mathbf{R}).

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.