The S-space problem, and the cardinal invariant $\mathfrak p$

Recall that an $S$-space is a regular hereditarily separable topological space which is not hereditarily Lindelöf. Do they exist? Consistently, yes. However, Szentmiklóssy proved that compact $S$-spaces do not exist, assuming Martin’s Axiom. Pushing this further, Todorcevic later proved that there are no $S$-spaces under PFA.

In the above-mentioned paper of Todorcevic, it is also proved that PFA entails $\omega_1\rightarrow(\omega_1,\alpha)^2$ for all $\alpha<\omega_1$. As you might remember, a proof of the latter was given in an earlier post. In this post, we shall show how the same ideas of the preceding yields the result about $S$-spaces.

Lemma. If $\langle X,\tau\rangle$ is a regular topological space which is not hereditarily Lindelöf, then there exists a sequence $\langle (x_\gamma,U_\gamma)\mid \gamma<\omega_1\rangle$ such that for all $\gamma<\omega_1$:

• $x_\gamma\in U_\gamma\in\tau$;
• $\{x_\beta\mid \gamma<\beta<\omega_1\}\cap \overline{U_\gamma}=\emptyset$.

Proof. Let $\{V_\alpha\mid\alpha<\kappa\}$ be an uncountable cover of some subspace $Y$ of $X$, that has no countable subcover. Then, it is easy to recursively construct $\langle (x_\gamma,\alpha_\gamma)\mid \gamma<\omega_1\rangle$ such that:

• $x_0$ is an arbitrary element of $Y$, and $\alpha_0$ is some ordinal in $\kappa$ satisfying $x_0\in V_{\alpha_0}$;
• $x_\gamma\in Y\setminus\bigcup_{\beta<\gamma}V_{\alpha_\beta}$, and $\alpha_\gamma$ is some ordinal in $\kappa$ satisfying $x_\gamma\in V_{\alpha_\gamma}$ for all nonzero $\gamma<\omega_1$.

Denote $F_\gamma:=X\setminus\bigcup_{\beta\le \gamma}V_{\alpha_\beta}$. Then $F_\gamma$ is a closed subset of $X$, with $x_\gamma\not\in F_\gamma$. Thus, by regularity of $X$, we can find an open set $U_\gamma$ such that $x_\gamma\in U_\gamma$, while $\overline{U_\gamma}\cap F_\gamma$ is empty. So $\overline{U_\gamma}\subseteq \bigcup_{\beta\le \gamma}V_{\alpha_\beta}$, and in particular, $\overline{U_\gamma}\cap\{x_\beta\mid\gamma<\beta<\omega_1\}=\emptyset$. $\square$

Theorem (Todorcevic, 1983). PID + $\mathfrak p>\omega_1$ implies that there are no $S$-spaces.
Proof. Suppose that $\mathbb X=\langle X,\tau\rangle$ is a $T_1$ regular space which is not a hereditarily Lindelöf. We shall prove that the space is neither hereditarily separable.
Let $\langle (x_\beta,U_\beta)\mid \beta<\omega_1\rangle$ be given by the previous lemma. For notational simplicity, we shall identify $x_\beta$ with $\beta$, for all $\beta<\omega_1$. Write $I_\beta:=\overline{U_\beta}\setminus\{\beta\}$. Then, after the notational simplification, we get that $\omega_1$ is the underlying set of some uncountable subspace of $\mathbb X$, and $I_\beta\cap\omega_1\subseteq\beta$. Put  $$\mathcal I:=\{ Y\in[\omega_1]^{\le\aleph_0}\mid |\{\beta<\omega_1\mid Y\cap I_\beta\text{ is infinite}\}|\le\aleph_0\}.$$
By Lemma 2 from the earlier post, $\mathcal I$ is a P-ideal. Hence, the P-Ideal dichotomy entails that one of the following must hold:

$\blacktriangleright$ There exists an uncountable $A\subseteq\omega_1$ such that $[A]^{\aleph_0}\subseteq\mathcal I$.
Then, by Lemma 1 of the earlier post, there exists an uncountable $B\subseteq A$ such that $\alpha\not\in I_\beta$ (and hence $\alpha\not\in U_\beta$) for all $\alpha<\beta$ in $B$. So $B$ (or $\{ x_\beta\mid \beta\in B\}$, if you like) is an uncountable discrete subspace of $\mathbb X$, and hence the latter is not hereditarily separable.

$\blacktriangleright$ There exists a sequence $\langle Z_n\mid n<\omega\rangle$ whose union is $\omega_1$, and $[Z_n]^{\aleph_0}\cap\mathcal I=\emptyset$ for all $n<\omega$. Pick $n<\omega$ such that $Z_n$ is uncountable, and simply denote $Z:=Z_n$.
We claim that $Z$ is not separable. Suppose not, and pick a dense subset $Y\in[Z]^{\aleph_0}$.
Notice that for every $\mathcal F\in[\omega_1]^{<\omega}$, the set $Y_{\mathcal F}:=Y\setminus\bigcup_{\beta\in\mathcal F}I_\beta$ is infinite (because otherwise, the set $Y$ would have been covered by the relatively-closed and countable set $\bigcup\{(I_\beta\cup\{\beta\})\cap Z\mid \beta\in\mathcal F\cup Y_{\mathcal F}\}$, contradicting the choice of $Y$ as a dense subset of the uncountable space $Z$). As $\mathfrak p>\aleph_1$, we may now pick an infinite pseduointersection $P$ of the family $\{ Y\setminus I_\beta\mid \beta<\omega_1\}$. As $P\in[Z]^{\aleph_0}$, and the latter is disjoint from $\mathcal I$, we get that $P\not\in\mathcal I$, and there must exist some (in fact, uncountably many) $\beta<\omega_1$ such that $P\cap I_\beta$ is infinite. Fix such a $\beta$. Then , $P\cap I_\beta$ is infinite, while $P\subseteq^* (Y\setminus I_\beta)$. This is a contradiction. So, $Z$ is indeed a nonseparable subspace of $\mathbb X$. $\square$

Open problem. In his paper, Todorcevic introduced the so-called models of the form PFA(T)[T]. Here PFA(T) stands for a certain weakening of the usual PFA, in which a Souslin tree T exists. Then, roughly speaking, a model of the form PFA(T)[T] is the outcome of forcing with the Souslin tree $T$ over the ground model of PFA(T).
Todorecevic proved that any such model of PFA(T)[T] satisfies PID. However, it is open whether PFA(T)[T] refutes the existence of $S$-spaces, since after forcing with the Souslin tree $T$, we have $\mathfrak p=\omega_1$ (see Farah’s lemma).