Bourbaki

Exercise 23, § 4

General Topology, IX, p. 106

Bourbaki marks this one as harder than the rest.

(a) In order that every finite open covering of a completely regular space be divisible (IX, p. 105, exerc. 20), it is necessary that X be normal; conversely, if X is normal, every finite open covering of X is normal (IX, p. 105, exerc. 20); when R\mathscr{R} runs through the set of finite open coverings of X, the sets VRV_{\mathscr{R}} form a fundamental system of entourages of the uniform structure Uω∗\mathscr{U}_\omega^* induced on X by the uniform structure of βX\beta X (Use IX, p. 48, cor. 1 and axiom (Ov) to define a finite family of continuous distances on X of the form (x,y)↦∣f(x)−f(y)∣(x, y) \mapsto |f(x) - f(y)|, in such a way that an entourage of the uniform structure defined by this family is contained in VRV_{\mathscr{R}}). (b) Let X be a normal space; for every closed subset Y of X, let YβY^\beta be the closure of Y in βX\beta X. Now prove that the continuous mapping βY→Yβ\beta Y \to Y^\beta which extends the canonical injection Y→YβY \to Y^\beta is a homeomorphism (use a), or directly the universal property of βY\beta Y and th. 2 of IX, p. 44). If Y, Z are two closed subsets of X, prove that

(Y∩Z)β=Yβ∩Zβ (Y \cap Z)^\beta = Y^\beta \cap Z^\beta
in βX\beta X (if x0∉(Y∩Z)βx_0 \notin (Y \cap Z)^\beta and x0∈Yβx_0 \in Y^\beta, consider a closed neighbourhood V of x0x_0 in βX\beta X, such that V∩YV \cap Y and Z are disjoint).

Solution

No solution has been written yet.

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