Exercise 17, § 9

In a topological space $X$, a point $a$ is said to be totally adherent to a subset $A$ of $X$ if for each neighbourhood $U$ of $a$ we have $\operatorname{card}(A) = \operatorname{card}(A \cap U)$. Show that $X$ is quasi-compact if and only if for each infinite subset $A$ of $X$ there is a point of $X$ which is totally adherent to $A$. (If $X$ is not quasi-compact, show that there is an initial ordinal $\omega_\alpha$ (Set Theory, Chapter III, § 6, Exercise 10) and an open covering $(U_\xi)$ of $X$, where $\xi$ runs through 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 $U_\eta$ with indices $\eta < \xi$ has a cardinal equal to $\aleph_\alpha$; deduce that it is possible to define a family $(x_\xi)$ ($\xi < \omega_\alpha$) of points of $X$ by transfinite induction such that $x_\xi \in A_\xi$ for each $\xi$ and $x_\xi \neq x_\eta$ whenever $< \eta \xi$).