Exercise 25, § 9

(a) Let X be a separated space, and let $\Gamma$ be the set of filters on X admitting a basis formed of open sets, and having no adherent point in X. Now prove that the set $\Gamma$ is inductive if it is not empty (in other words, if X is not absolutely closed). (b) Let $\Omega$ be the set of maximal filters in $\Gamma$, and let $X'$ be the sum set of X and $\Omega$. For every $x \in X$, denote by $\mathfrak{B}(x)$ the set of neighbourhoods of $x$ in X; for every ω ∈ Ω, denote by $\mathfrak{B}(\omega)$ the set of parts of the form $\{\omega\} \cup M$, where M runs through the filter ω. There is a topology on $X'$ for which $\mathfrak{B}(x')$ is a fundamental system of neighbourhoods of $x'$ for every $x' \in X'$. For this topology, $X'$ is an absolutely closed space, and possesses the following universal property: for every continuous open mapping $f : X \to Y$ into an absolutely closed space Y, there exists one and only one continuous mapping $g : X' \to Y$ which extends $f$ (consider the images by $f$ of the filters belonging to $\Omega$; observe that in an absolutely closed space, a maximal filter in the set of filters having a basis of open sets is necessarily convergent). In particular, the topology of $X'$ is X-maximal (I, p. 94, exerc. 11). Deduce that there exist absolutely closed spaces whose topology is quasi-maximal (I, p. 91, exerc. 6).