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 ${\omega }_1\to ({\omega }_1,\alpha {)}^2$ 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 $⟨X,\tau ⟩$ is a regular topological space which is not hereditarily Lindelöf, then there exists a sequence $⟨(x_{\gamma },U_{\gamma })\mid \gamma <{\omega }_1⟩$ such that for all $\gamma <{\omega }_1$:

• $x_{\gamma }\in U_{\gamma }\in \tau$;
• $\{x_{\beta }\mid \gamma <\beta <{\omega }_1\}\cap \overline{U_{\gamma }}=\mathrm{\varnothing }$.

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 $⟨(x_{\gamma },{\alpha }_{\gamma })\mid \gamma <{\omega }_1⟩$ 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 $x_{\gamma }\notin F_{\gamma }$. 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, $\overline{U_{\gamma }}\cap \{x_{\beta }\mid \gamma <\beta <{\omega }_1\}=\mathrm{\varnothing }$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Theorem (Todorcevic, 1983). PID + $\mathfrak{p}>{\omega }_1$ implies that there are no $S$-spaces.
Proof. Suppose that $\mathbb{X}=⟨X,\tau ⟩$ is a $T_1$ regular space which is not a hereditarily Lindelöf. We shall prove that the space is neither hereditarily separable.
Let $⟨(x_{\beta },U_{\beta })\mid \beta <{\omega }_1⟩$ 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  $$\mathcal{I}:=\{Y\in [{\omega }_1{]}^{\le {\mathrm{\aleph }}_0}\mid |\{\beta <{\omega }_1\mid Y\cap I_{\beta }\text{ is infinite}\}|\le {\mathrm{\aleph }}_0\}.$$
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:

$▸$ There exists an uncountable $A\subseteq {\omega }_1$ such that $[A{]}^{{\mathrm{\aleph }}_0}\subseteq \mathcal{I}$.
Then, by Lemma 1 of the earlier post, there exists an uncountable $B\subseteq A$ such that $\alpha \notin I_{\beta }$ (and hence $\alpha \notin U_{\beta }$) 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.

$▸$ There exists a sequence $⟨Z_n\mid n<\omega ⟩$ whose union is ${\omega }_1$, and $[Z_n{]}^{{\mathrm{\aleph }}_0}\cap \mathcal{I}=\mathrm{\varnothing }$ 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 $Y\in [Z{]}^{{\mathrm{\aleph }}_0}$.
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 $\mathfrak{p}>{\mathrm{\aleph }}_1$, we may now pick an infinite pseduointersection $P$ of the family $\{Y\setminus I_{\beta }\mid \beta <{\omega }_1\}$. As $P\in [Z{]}^{{\mathrm{\aleph }}_0}$, and the latter is disjoint from $\mathcal{I}$, we get that $P\notin \mathcal{I}$, 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 $P{\subseteq }^{\ast }(Y\setminus I_{\beta })$. This is a contradiction. So, $Z$ is indeed a nonseparable subspace of $\mathbb{X}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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).