Exercise 18, § 9

In a topological space X, one says that a point $a$ is totally adherent to a subset A of X if for every neighbourhood U of $a$, one has $\mathrm{Card}(A) = \mathrm{Card}(A \cap U)$. Now prove that in order that X be quasi-compact, it is necessary and sufficient that for every infinite subset A of X, there exists a point totally adherent to A. (If X is not quasi-compact, show that there exists an initial ordinal $\omega_\alpha$ (E, III, p. 87, exerc. 10) and an open covering $(U_\xi)$ of X, where $\xi$ ranges over the set of ordinals $< \omega_\alpha$, such that for each index $\xi < \omega_\alpha$, the complement $A_\xi$ in X of the union of the $U_n$ of index $\eta < \xi$ has a cardinal equal to $\aleph_\alpha$; deduce that one can define by transfinite induction a family $(x_\xi)$ ($\xi < \omega_\alpha$) of points of X such that $x_\xi \in A_\xi$ for every $\xi$ and $x_\xi \neq x_\eta$ for $\eta < \xi$.)