Exercise 19, § 9

It is said that a separated space X is absolutely closed if it possesses the following property: for every homeomorphism $f$ of X onto a subspace of a separated space $X'$, $f(X)$ is closed in $X'$. Every compact space is absolutely closed.

a) Now prove that the following conditions are equivalent:

(AF) The space X is absolutely closed.

(AF') Every filter basis on X formed by open sets has at least one point of adherence.

(AF'') Every family of closed sets in X, whose intersection is empty, contains a subfamily whose intersection of the interiors is empty.

(AF''') Every open covering of X contains a finite family whose closures form a covering of X.

b) Let X be a separated space, U an everywhere dense open set in X. Suppose that every filter basis on U formed by open sets has at least one point of adherence in X. Now prove that X is then absolutely closed. In particular, if A is an open subset of an absolutely closed space X, the subspace $\overline{A}$ of X is absolutely closed.

c) Let X be an absolutely closed space, $f$ a continuous mapping of X into a separated space $X'$. Now prove that $f(X)$ is an absolutely closed subspace of $X'$.

d) Now prove that the product of two absolutely closed spaces is absolutely closed (reason as in prop. 17 of I, p. 70).