PID plus p greater than omega-one eliminates S-spaces

Assume for contradiction that some subspace $W\subseteq K$ is not Lindelöf. Regularity, Hausdorffness, and hereditary separability pass to subspaces, so replace $K$ by $W$. Choose an open cover with no countable subcover. Recursively for $\alpha <{\omega }_1$, select a cover member $U_{\alpha }$ and $x_{\alpha }\in U_{\alpha }$ outside ${\bigcup }_{\beta <\alpha }{U}_{\beta }$. Let $X=\{x_{\alpha }:\alpha <{\omega }_1\}$ and relabel $U_{x_{\alpha }}=U_{\alpha }$; then $U_x\cap X$ is countable for every $x\in X$. By regularity choose open $V_x$ with $x\in V_x\subseteq \overline{V_x}\subseteq U_x$. Define $\mathcal{I}=\{A\in [X{]}^{\le \omega }:(\mathrm{\forall }x\in X)\mathrm{\mid }A\cap \overline{V_x}\mathrm{\mid }<\omega \}$. It is an ideal containing all finite sets.

We first derive the needed domination fact from $\mathfrak{p}$. If $\mathcal{F}\subseteq {\omega }^{\omega }$ has size less than $\mathfrak{p}$, consider, on the countable set ${\omega }^{<\omega }$, the sets $A_f=\{s:(\mathrm{\forall }i<\mathrm{\mid }s\mathrm{\mid })f(i)\le s(i)\}$ for $f\in \mathcal{F}$ and $C_n=\{s:\mathrm{\mid }s\mathrm{\mid }\ge n\}$. Every finite intersection is infinite. A pseudointersection $B$ exists by the definition of $\mathfrak{p}$; thin it to distinct $s_n$ with $\mathrm{\mid }{s}_n\mathrm{\mid }>n$, and put $g(n)=s_n(n)$. Since $B{\subseteq }^{\ast }{A}_f$, $g$ eventually dominates every $f\in \mathcal{F}$. Now take $A_n\in \mathcal{I}$, replace them by their increasing finite unions, and enumerate each infinite $A_n$ as $\{a_{n,k}:k<\omega \}$. For each $x$, choose $f_x(n)$ past the finite set $A_n\cap \overline{V_x}$. As $\mathrm{\mid }X\mathrm{\mid }={\omega }_1<\mathfrak{p}$, choose one eventual dominator $g$ for all $f_x$, and set $A={\bigcup }_n\{a_{n,k}:k\ge g(n)\}$, ignoring finite $A_n$. Each $A_n{\subseteq }^{\ast }A$, while for fixed $x$ all sufficiently large rows avoid $\overline{V_x}$ and the finitely many remaining rows meet it finitely. Thus $A\in \mathcal{I}$, proving that $\mathcal{I}$ is a P-ideal.

Apply PID to $\mathcal{I}$. In the first alternative take uncountable $Y\subseteq X$ with $[Y{]}^{\le \omega }\subseteq \mathcal{I}$. For $y\in Y$, the set $Y\cap V_y$ must be finite; otherwise a countably infinite subset of it would belong to $\mathcal{I}$ yet meet $\overline{V_y}$ infinitely. Since the space is Hausdorff and hence $T_1$, delete the finitely many other points of $Y\cap V_y$ to obtain a relative open neighborhood isolating $y$. Thus $Y$ is an uncountable discrete subspace, which is not separable, contradicting hereditary separability.

In PID's second alternative write $X={\bigcup }_n{Y}_n$ with every $Y_n\perp \mathcal{I}$. Some $Y=Y_n$ is uncountable, and hereditary separability gives a countable dense $D\subseteq Y$. The family $\{D\setminus \overline{V_x}:x\in X\}$ has the strong finite intersection property. Indeed, if $D\setminus {\bigcup }_{x\in F}\overline{V_x}$ were finite for some finite $F$, then $Y\subseteq \bar{D}\subseteq {\bigcup }_{x\in F}\overline{V_x}$ together with finitely many points. But every $\overline{V_x}\subseteq U_x$ and each $U_x\cap X$ is countable, forcing $Y$ countable, a contradiction. Since $\mathrm{\mid }X\mathrm{\mid }={\omega }_1<\mathfrak{p}$, F1 gives an infinite pseudointersection $a\subseteq D$. Then $a\cap \overline{V_x}$ is finite for every $x$, so $a\in \mathcal{I}$; but $a\subseteq Y$ contradicts $Y\perp \mathcal{I}$. Thus the second alternative is impossible as well.

Both PID alternatives contradict hereditary separability, so the assumed non-Lindelöf subspace $W$ cannot exist. Hence every subspace of the original $K$ is Lindelöf: $K$ is hereditarily Lindelöf. By the S-space definition in F1, no regular Hausdorff hereditarily separable non-Lindelöf space exists. AC is used exactly as recorded in A1.