Exercise 24, § 9

(a) Now prove that in a countable absolutely closed space X, the set of isolated points is everywhere dense. (Reason by contradiction, by arranging the points of X in a sequence, and forming by recurrence a countable open covering $(\overline{U}_n)$ of X such that no finite union of the $\overline{U}_n$ is a covering of X). b) Give an example of a completely separated absolutely closed space which is not a Lindelöf space (in the product $[0, 1] \times [0, 1]$, consider the complements of the sets $A \times \{0\}$, where A runs through the set of subsets of $[0, 1]$, and apply the general method of I, p. 103, exerc. 20 c)). This space satisfies the property (D{II}) of I, p. 90, exerc. 7, but not the property (D_{III}). c) Let $X_0$ be a compact space without isolated point, $\mathfrak{M}$ the set of complements of the countable subsets of $X_0$, the sets of $\mathfrak{M}$ being therefore everywhere dense (cf. a)). Let $\mathfrak{T}$ be the topology generated by the open sets of $X_0$ and the sets of $\mathfrak{M}$. Now prove that the space X obtained by endowing $X_0$ with the topology $\mathfrak{T}$ is absolutely closed, completely separated, and is a Lindelöf space not satisfying the condition (D_{II}) of exerc. 7 of I, p. 90, and in which no point has a countable fundamental system of neighbourhoods and no infinite sequence whose terms are all distinct has a value of adherence. If in $X_0$ every open subspace is a Lindelöf space (cf. I, p. 107, exerc. 16 c)), every open subspace of X is a Lindelöf space and in particular X satisfies (D_{III}) (I, p. 107, exerc. 16 a)). This is so when $X_0 = \{0, 1\}$; but now prove that in this case there are countable closed subsets of X which are not countable intersections of open sets of X (cf. IX, § 5, n° 3).*