Bourbaki

Exercise 25, § 9

General Topology, I, 148

Bourbaki marks this one as harder than the rest.

a) Let XX be an accessible space, and let Φ\Phi be the set of filters on XX which have a base consisting of closed sets. Show that the set Φ\Phi is inductive, and let X′′X'' be the set of maximal filters in Φ\Phi.

b) For each open subset UU of XX, let U∗U^* be the set of filters F∈X′′\mathfrak{F} \in X'' such that U∈FU \in \mathfrak{F}. If U,VU, V are two open subsets of XX, then

(U∩V)∗=U∗∩V∗and(U∪V)∗=U∗∪V∗ (U \cap V)^* = U^* \cap V^* \quad \text{and} \quad (U \cup V)^* = U^* \cup V^*

(remark that if F∈X′′\mathfrak{F} \in X'' is such that U∉FU \notin \mathfrak{F}, then X=U∈FX = U \in \mathfrak{F}). Show that the sets U∗U^* form a base of a topology on X′′X'' and that in this topology X′′X'' is quasi-compact [verify axiom (C''''), using the remark above] and accessible [if F,G\mathfrak{F}, \mathfrak{G} are two distinct elements of X′′X'', show that there exist two disjoint closed subsets AA and BB in XX such that A∈FA \in \mathfrak{F} and B∈GB \in \mathfrak{G}].

c) For each x∈Xx \in X let φ(x)\varphi(x) be the ultrafilter on XX which has the single set {x}\{ x \} as a base. Show that φ\varphi is a homeomorphism of XX onto a dense subset of X′′X'' and that if XX is quasi-compact (and accessible), then φ(X)=X′′\varphi(X) = X''.

d) Show that for each continuous mapping ff of XX into a compact space YY there is a unique continuous mapping g:X′′→Yg : X'' \to Y such that f=g∘φf = g \circ \varphi.

e) Let X~\tilde{X} be the universal Hausdorff space associated with X′′X'' (§8§ 8, Exercise 27); show that the canonical mapping ψ:X′′→X~\psi : X'' \to \tilde{X} is surjective and that X~\tilde{X} is compact. Thus every continuous mapping of XX into a compact space YY can be written uniquely in the form f=h∘ψ∘φf = h \circ \psi \circ \varphi, where h:X~→Yh : \tilde{X} \to Y is continuous.

Solution

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Transcribed from the printed volume. Not translated.