Bourbaki

Exercise 24, § 9

General Topology, I, 147

Bourbaki marks this one as harder than the rest.

a) Let XX be a Hausdorff space, and let Γ\Gamma be the set of filters on XX which have a base consisting of open sets and which have no cluster point in XX. Show that the set Γ\Gamma is inductive if it is not empty (i.e., if XX is not absolutely closed).

b) Let Ω\Omega be the set of maximal filters in Γ\Gamma, and let X′X' be the set which is the sum of XX and Ω\Omega. For each x∈Xx \in X let B(x)\mathfrak{B}(x) denote the set of neighbourhoods of xx in XX, and for each ω∈Ω\omega \in \Omega let B(ω)\mathfrak{B}(\omega) denote the set of subsets of X′X' of the form {ω}∪M\{ \omega \} \cup M, where MM belongs to the filter ω\omega. Then there is a topology on X′X' in which B(x′)\mathfrak{B}(x') is a fundamental system of neighbourhoods of x′x' for each x′∈X′x' \in X'. Show that, with this topology, X′X' is absolutely closed and has the following universal property : given any open continuous mapping f:X→Yf : X \to Y of XX into an absolutely closed space YY, there is a unique continuous mapping g:X′→Yg : X' \to Y which extends ff (consider the images under ff of the filters which belong to Ω\Omega, and observe that in an absolutely closed space, a filter which is maximal in the set of filters which have a base consisting of open sets must be convergent). In particular, the topology of X′X' is XX-maximal (§3§ 3, Exercise 11). Deduce that there are absolutely closed spaces whose topology is quasi-maximal (§2§ 2, Exercise 6).

Solution

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