The S-space problem, and the cardinal invariant $\mathfrak{b}$

Recall that an S-space is a regular hereditarily separable topological space which is not hereditarily Lindelöf. In a previous post, we showed that such a space exists after adding a Cohen real. Here, we shall construct one from an arithmetic assumption.

Theorem (Todorcevic, 1989). If $\mathfrak{b}={\omega }_1$, then there exists an $S$-space.
Proof. Let $\overrightarrow{f}=⟨f_{\alpha }\mid \alpha <{\omega }_1⟩$ witness that $\mathfrak{b}={\omega }_1$, that is:

1. $f_{\alpha }\in {}^{\omega }\omega$ is an increasing function for all $\alpha <{\omega }_1$;
2. $\{n<\omega \mid f_{\alpha }(n)\ge f_{\beta }(n)\}$ is finite, whenever $\alpha <\beta <{\omega }_1$;
3. for every $g\in {}^{\omega }\omega$, there exists some $\alpha <{\omega }_1$ such that $\{n<\omega \mid g(n)<f_{\beta }(n)\}$ is infinite, whenever $\alpha <\beta <{\omega }_1$.

Write $X:=\{f_{\alpha }\mid \alpha <{\omega }_1\}$. For all $f\in X$ and $n<\omega$, let $$U(f,n):=\{g\in X\mid \underset{m<n}{\bigwedge }g(m)=f(m)\wedge \underset{m\ge n}{\bigwedge }g(m)\le f(m)\}.$$We then let $\mathcal{B}:=\{U(f,n)\mid f\in X,n<\omega \}$ be a basis for a topology $\tau$ on $X$.
To see that $\mathbb{X}:=⟨X,\tau ⟩$ is Hausdorff, simply notice that if $f,g$ are distinct elements of $X$ and $n<\omega$ is such that $f(n)\ne g(n)$, then $f\in U(f,n+1),g\in U(g,n+1)$, while $U(f,n+1)\cap U(g,n+1)=\mathrm{\varnothing }$.
To see that $\mathbb{X}$ is zero-dimensional (and hence regular), we fix $f\in X$ and $n<\omega$ and show that $X\setminus U(f,n)$ is open. Given $g\in X\setminus U(f,n)$, simply find a large enough $n^{\prime }<\omega$ such that $g\upharpoonright n^{\prime }⊈h$ for all $h\in U(f,n)$. Then $U(g,n^{\prime })\cap U(f,n)=\mathrm{\varnothing }$.

Claim. $\mathbb{X}$ is not Lindelöf.
Proof. Let $\mathcal{U}:=\{U(f,0)\mid f\in X\}$. Suppose towards a contradiction that $\mathcal{U}$ admits a countable subcover. Then there exists some ordinal $\beta <{\omega }_1$ such that $$X=\bigcup \{U(f_{\alpha },0)\mid \alpha <\beta \}.$$ In particular $f_{\beta }$ belongs to the latter, contradicting clause (2) above. End of Claim

Claim. $\mathbb{X}$ is hereditarily separable.
Proof. Suppose not, as witnessed by an uncountable $Y\subseteq X$. Then, as in the proof of the main theorem from here, there exists an increasing function $\psi :{\omega }_1\to Y$ such that $Z:=\psi [Y]$ forms an uncountable discrete subspace of $\mathbb{X}$. Fix $\phi :Z\to \omega$ such that $\{U(f_{\alpha },\phi (\alpha ))\mid \alpha \in Z\}$ witnesses that $Z$ is discrete. Find $n^{\ast }<\omega$ such that ${\phi }^{-1}\{n^{\ast }\}$ is uncountable, and then pick an uncountable $B\subseteq {\phi }^{-1}\{n^{\ast }\}$ for which $\{f_{\beta }\upharpoonright n^{\ast }\mid \beta \in B\}$ is a singleton.
It follows that for all $\alpha <\beta$ in $B$, there exists some $n\ge n^{\ast }$ such that $f_{\alpha }(n)>f_{\beta }(n)$. In particular, for all $\alpha <\beta$ in $B$, we have $\mathrm{\Delta }(\alpha ,\beta )>0$, where $$\mathrm{\Delta }(\alpha ,\beta ):=min\{m<\omega \mid m\le n<\omega \to (f_{\alpha }(n)\le f_{\beta }(n))\}.$$

Thus, to meet the desired contradiction, it suffices to establish the following.

Subclaim. There exists $\alpha <\beta$ in $B$ such that $\mathrm{\Delta }(\alpha ,\beta )=0$.
Proof. Fix a countable elementary submodel $\mathcal{N}⪯(H({\omega }_2),\in )$ with $B,\overrightarrow{f}\in \mathcal{N}$. Let $\delta :=min(B\setminus \mathcal{N})$. Write $B_n:=\{\beta \in B\setminus (\delta +1)\mid \mathrm{\Delta }(\delta ,\beta )=n\}$.
As $B\setminus (\delta +1)=\bigcup _{n<\omega }{B}_n$, let us fix some $n<\omega$ such that $B_n$ is uncountable. By item (3) above, we get that the set $$M:=\{m<\omega \mid sup\{f_{\beta }(m)\mid \beta \in B_n\}=\omega \}$$ is non-empty (in fact, infinite), so consider its minimal element, $m:=min(M)$.
For $t\in {}^m\omega$, denote $B_n^t:=\{\beta \in B_n\mid t\subseteq f_{\beta }\}$. By minimality of $m$, the set $\{t\in {}^m\omega \mid B_n^t\ne \mathrm{\varnothing }\}$ is finite, so  we can easily find some $t\in {}^m\omega$ such that $$sup\{f_{\beta }(m)\mid \beta \in B_n^t\}=\omega .$$ Since $\{\beta \in B\mid t\subseteq f_{\beta }\}$ is a non-empty set that lies in $\mathcal{N}$, let us fix some $\alpha \in \mathcal{N}\cap B$ such that $t\subseteq f_{\alpha }$. Put $k:=\mathrm{\Delta }(\alpha ,\delta )$, and then pick $\beta \in B_n^t$ such that $f_{\beta }(m)>f_{\alpha }(k+n)$. Of course, $\alpha <\delta <\beta$. We claim that $\mathrm{\Delta }(\alpha ,\beta )=0$.

This is best seen by dividing into three cases:

• if $i<m$, then $f_{\alpha }(i)=t(i)=f_{\beta }(i)$;
• if $m\le i\le k+n$, then $f_{\alpha }(i)\le f_{\alpha }(k+n)<f_{\beta }(m)\le f_{\beta }(i)$ (recall that the elements of $\overrightarrow{f}$ are increasing functions!);
•  if $k+n<i<\omega$, then $\mathrm{\Delta }(\alpha ,\delta )=k<i$ and $f_{\alpha }(i)\le f_{\delta }(i)$, as well as $\mathrm{\Delta }(\delta ,\beta )=n<i$ and $f_{\delta }(i)\le f_{\beta }(i)$. Altogether, $f_{\alpha }(i)\le f_{\beta }(i)$. End of subclaim
  

End of Claim $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

In a previous post, we showed that PID + $\mathfrak{p}>{\omega }_1$ implies that there are no $S$-spaces. The above shows that we should indeed at least assume $\mathfrak{b}>{\omega }_1$. A minor modification of the original argument, yields a relative.

Theorem (Todorcevic, 2012). PID + $\mathfrak{b}>{\omega }_1$ implies that there are no Fréchet–Urysohn comapct $S$-spaces.
Proof. We go along the lines of the former proof. Suppose that $\mathbb{X}=⟨X,\tau ⟩$ is a regular, not hereditarily Lindelöf space. We may assume that ${\omega }_1$ — the set of all countable ordinals — is a subset of the underlying set $X$, and that there exists a sequence of open sets $⟨U_{\beta }\mid \beta <{\omega }_1⟩$ such that for all $\beta <{\omega }_1$:$$\beta \in U_{\beta }\cap {\omega }_1\subseteq \beta +1.$$

Then, by regularity of $\mathbb{X}$, for all $\beta <{\omega }_1$, let us pick an open set $V_{\beta }$ such that $$\beta \in V_{\beta }\subseteq \overline{V_{\beta }}\subseteq U_{\beta }.$$ We claim that $\mathbb{X}$ is not hereditarily separable. Write $I_{\beta }:=\overline{V_{\beta }}\setminus \{\beta \}$, and put $$\mathcal{I}:=\{Y\in [{\omega }_1{]}^{\le {\mathrm{\aleph }}_0}\mid |\{\beta <{\omega }_1\mid Y\cap I_{\beta }\text{ is finite}\}|\le {\mathrm{\aleph }}_0\}.$$By $\mathfrak{b}>{\omega }_1$, we see that $\mathcal{I}$ is a P-ideal. Thus, 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 there exists an uncountable $B\subseteq A$ such that $\alpha \notin V_{\beta }$ for all $\alpha <\beta$ in $B$. So $B$ is an uncountable discrete subspace of $\mathbb{X}$, and hence the latter is not hereditarily separable.

$▸$ There exists an uncountable $Z\subseteq {\omega }_1$ such that $[Z{]}^{{\mathrm{\aleph }}_0}\cap \mathcal{I}=\mathrm{\varnothing }$.
Since $\mathbb{X}$ is compact, our set $Z$ admits a complete accumulation point, say, $z$. Since $\mathbb{X}$ is Fréchet–Urysohn, we may pick a countable $a\subseteq Z$ that converges to $z$. Since $a\in [Z{]}^{{\mathrm{\aleph }}_0}$ and the latter is disjoint from $\mathcal{I}$, there must exist some $\beta <{\omega }_1$ such that $a\cap I_{\beta }$ is infinite. In particular, $\overline{V_{\beta }}$ contains a subsequence of $a$, and hence $z\in \overline{V_{\beta }}\subseteq U_{\beta }$. So, $U_{\beta }\cap Z$ must be uncountable, contradicting the fact that $U_{\beta }\cap Z\subseteq \beta +1$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$