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 $\overrightarrow f=\langle f_\alpha\mid\alpha<\omega_1\rangle$ witness that $\mathfrak b=\omega_1$, that is:
- $f_\alpha\in{}^\omega\omega$ is an increasing function for all $\alpha<\omega_1$;
- $\{ n<\omega\mid f_\alpha(n)\ge f_\beta(n)\}$ is finite, whenever $\alpha<\beta<\omega_1$;
- 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 $$U(f,n):=\left\{ g\in X\mid \bigwedge_{m<n}g(m)=f(m)\wedge\bigwedge_{m\ge n}g(m)\le f(m)\right\}.$$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 $\mathbb X:=\langle X,\tau\rangle$ is Hausdorff, simply notice that if $f,g$ are distinct elements of $X$ and $n<\omega$ is such that $f(n)\neq g(n)$, then $f\in U(f,n+1), g\in U(g,n+1)$, while $U(f,n+1)\cap U(g,n+1)=\emptyset$.
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 $n'<\omega$ such that $g\restriction n’\not\subseteq h$ for all $h\in U(f,n)$. Then $U(g,n’)\cap U(f,n)=\emptyset$.
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 $\psi:\omega_1\rightarrow Y$ such that $Z:=\psi[Y]$ forms an uncountable discrete subspace of $\mathbb X$. Fix $\varphi:Z\rightarrow\omega$ such that $\{ U(f_\alpha,\varphi(\alpha))\mid \alpha\in Z\}$ witnesses that $Z$ is discrete. Find $n^*<\omega$ such that $\varphi^{-1}\{n^*\}$ is uncountable, and then pick an uncountable $B\subseteq \varphi^{-1}\{n^*\}$ for which $\{ f_\beta\restriction n^*\mid \beta\in B\}$ is a singleton.
It follows that for all $\alpha<\beta$ in $B$, there exists some $n\ge n^*$ such that $f_\alpha(n)>f_\beta(n)$. In particular, for all $\alpha<\beta$ in $B$, we have $\Delta(\alpha,\beta)>0$, where $$\Delta(\alpha,\beta):=\min\{ m<\omega\mid m\le n<\omega\rightarrow(f_\alpha(n)\le f_\beta(n))\}.$$
Thus, to meet the desired contradiction, it suffices to establish the following.
Subclaim. There exists $\alpha<\beta$ in $B$ such that $\Delta(\alpha,\beta)=0$.
Proof. Fix a countable elementary submodel $\mathcal N\preceq(H(\omega_2),\in)$ with $B,\overrightarrow f\in\mathcal N$. Let $\delta:=\min(B\setminus\mathcal N)$. Write $B_n:=\{\beta\in B\setminus(\delta+1)\mid \Delta(\delta,\beta)=n\}$.
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 $B^t_n:=\{ \beta\in B_n\mid t\subseteq f_\beta\}$. By minimality of $m$, the set $\{ t\in{}^m\omega\mid B^t_n\neq\emptyset\}$ is finite, so we can easily find some $t\in{}^m\omega$ such that $$\sup\{ f_\beta(m)\mid \beta\in B^t_n\}=\omega.$$ 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 $k:=\Delta(\alpha,\delta)$, and then pick $\beta\in B^t_n$ such that $f_\beta(m)>f_\alpha(k+n)$. Of course, $\alpha<\delta<\beta$. We claim that $\Delta(\alpha,\beta)=0$.
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 $\Delta(\alpha,\delta)=k<i$ and $f_\alpha(i)\le f_\delta(i)$, as well as $\Delta(\delta,\beta)=n<i$ and $f_\delta(i)\le f_\beta(i)$. Altogether, $f_\alpha(i)\le f_\beta(i)$. End of subclaim
End of Claim $\square$
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 $\mathbb X=\langle X,\tau\rangle$ 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 $\langle U_\beta\mid \beta<\omega_1\rangle$ 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 $$\mathcal I:=\{ Y\in[\omega_1]^{\le\aleph_0}\mid |\{\beta<\omega_1\mid Y\cap I_\beta\text{ is finite}\}|\le\aleph_0\}.$$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:
$\blacktriangleright$ There exists an uncountable $A\subseteq\omega_1$ such that $[A]^{\aleph_0}\subseteq\mathcal I$.
Then there exists an uncountable $B\subseteq A$ such that $\alpha\not\in V_\beta$ for all $\alpha<\beta$ in $B$. So $B$ is an uncountable discrete subspace of $\mathbb X$, and hence the latter is not hereditarily separable.
$\blacktriangleright$ There exists an uncountable $Z\subseteq\omega_1$ such that $[Z]^{\aleph_0}\cap\mathcal I=\emptyset$.
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 $a\in[Z]^{\aleph_0}$ 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 $U_\beta\cap Z\subseteq\beta+1$. $\square$
