An $S$-space from a Cohen real

Recall that an $S$-space is a regular hereditarily separable topological space which is not hereditarily Lindelöf. In this post, we shall establish the consistency of the existence of such a space.

Theorem (Roitman, 1979). Let $\mathbb C=({}^{<\omega}\omega,\subseteq)$ be the notion of forcing for adding a Cohen real. Then, in the generic extension by forcing with $\mathbb C$, there exists an $S$-space.

The proof we give here is not Roitman’s original argument; I am not sure where did I occur into the version that will be given here.

We commence with a lemma. Its statement requires a piece of notation, as follows. For a set of ordinals $a$ of finite size $n$, we let $\{ a(0),a(1),\ldots,a(n-1)\}$ denote the $\in$-increasing enumeration of the elements of $a$.

Lemma. In the generic extension by forcing with $\mathbb C$, there exists a coloring $f:[\omega_1]^2\rightarrow\omega$ such that for every finite function $g:n\times n\rightarrow\omega$, and an uncountable $\mathcal A\subseteq[\omega_1]^{n}$ whose elements are pairwise disjoint, there exists some $a,b\in\mathcal A$ such that $\max(a)<\min(b)$ and $$f(a(i),b(j))=g(i,j)\text{ for all }i,j< n.$$

Proof. Work in $V$. Let $\{ A_\alpha\mid \alpha<\omega_1\}$ be the injective enumeration of some almost-disjoint family in $[\omega]^\omega$. For all $\alpha<\omega$, let $d_\alpha:\alpha\rightarrow\omega$ be arbitrary. For all infinite $\alpha<\omega_1$, let $d_\alpha:\alpha\leftrightarrow A_\alpha$ be some bijection. Next, let $G$ be $\mathbb C$-generic over $V$, and work in $V[G]$. Put $c:=\bigcup G$. Then, define $f:[\omega_1]^2\rightarrow\omega$ by letting for all $\alpha<\beta<\omega_1$:$$f(\alpha,\beta):=c(d_\beta(\alpha)).$$

Next, suppose that $n,g$ and $\mathcal A$ are as in the statement of the lemma. As $g$ is finite, $g$ belongs to $V$. As $\mathbb C$ is countable, $\mathcal A$ contains an uncountable subcollection that lies in $V$. Note also that $f$ has a simple (“canonical”) $\mathbb C$-name, hence, we shall prove via a density argument.
Fix an aribtrary $\mathbb C$-condition, say, $p:m\rightarrow\omega$, together with ground model $n, g$ and $\mathcal A$. Fix an infinite ordinal $\delta<\omega_1$ such that $\sup\{ \min(a)\mid a\subseteq\delta\}=\delta$ (e.g., if $\delta=M\cap\omega_1$ for a countable elementary submodel $M\prec (H(\omega_2),\in,\lhd)$ with $\mathcal A\in M$). Pick $b\in\mathcal A$ such that $\min(b)>\delta$. Pick a large enough $k<\omega$ such that $A_{b(i)}\cap A_{b(j)}\subseteq k$ for all $i<j<n$. Put $$\gamma:=\max\{\max(d^{-1}_{b(i)}[k+m])\mid i<n\}+1.$$So, in simple words, $\gamma$ is a large enough ordinal below $\delta$ that satisfies:

• the elements of $\{ d_{b(i)}[\delta\setminus\gamma]\mid i<n\}$ are pairwise disjoint;
• $f_{b(i)}(\alpha)>k$ whenever $i<n$ and $\gamma\le\alpha<\delta$.

By the choice of $\delta$, we can now pick $a\in\mathcal A$ such that $\gamma<\min(a)\le\max(a)<\delta$. Then, the map $(\alpha,\beta)\mapsto d_\beta(\alpha)$ is injective over $a\times b$, and its image is disjoint from $\text{dom}(p)$. Consequently, it is possible to extend $p$ to a condition $q$ so that $$q(d_{b(j)}(a(i))=g(i,j)\text{ for all }i,j<n.$$Evidently, any such $q$ forces that $$f(a(i),b(j))=g(i,j)\text{ for all }i,j<n. \square$$

Proof of main theorem. Fix a coloring $f:[\omega_1]^2\rightarrow\omega$ as in the lemma. For every $\beta<\omega_1$, put $$u_\beta:=\{\alpha<\beta\mid c(\alpha,\beta)=0\}\cup\{\beta\}.$$ We say that $(I,J)$ is good pair if $I,J\in[\omega_1]^{<\omega}$ and $I\cap J=\emptyset$. Denote$$U(I,J):=\bigcap\{ u_\gamma\mid \gamma\in I\}\setminus \bigcup\{ u_\gamma\mid \gamma\in J\}.$$

We then define $\mathcal B:=\{ U(I,J)\mid (I,J)\text{ good pair}\}$ as a basis to a topology $\tau$ on $\omega_1$. Note that $X:=\langle \omega_1,\tau\rangle$ is Hausdorff, since $\alpha<\beta<\omega_1$ implies that $\beta\in u_\beta$ and $\alpha\in\omega_1\setminus u_\beta$. Also $\mathbb X$ is zero-dimensional (given the convention that $\bigcap_{\gamma\in\emptyset}u_\gamma=\omega_1$), and hence regular.
Note that since $u_\beta\subseteq\beta+1$ for all $\beta<\omega_1$, the cover $\{ u_\beta\mid \beta<\omega_1\}$ witnesses that $\mathbb X$ is not Lindelöf.
Thus, towards a contradiction, suppose that $\mathbb X$ contains a non-separable subspace $Y$. Then, one can recursively construct an increasing function $\psi:\omega_1\rightarrow Y$ so that $\psi(\beta)\not\in\overline{\psi[\beta]}$ for all $\beta<\omega_1$. Put $Z:=\psi[\omega_1]$, and write $v_\beta:=u_\beta\setminus(\overline{Z\cap\beta})$ for all $\beta\in Z$. Then $v_\beta$ is an open neighbourhood around $\beta$, and we have $$\alpha\not\in v_\beta\text{ for all }\alpha\in Z\cap\beta.$$Fix $\beta\in Z$. As $v_\beta$ is an open neighbourhood around $\beta$, and by definition of $\tau$, we may fix a good pair $(I_\beta,J_\beta)$ such that $$\beta\in \bigcap\{u_\gamma\mid \gamma\in I_\beta\}\setminus \bigcup\{u_\gamma\mid \gamma\in J_\beta\}\subseteq v_\beta.$$As $v_\beta\subseteq u_\beta$, we may also assume that $\min(I_\beta)=\beta$. Next, by passing to an uncountable subcollection $Z’\subseteq Z$, we may assume the existence of  $A\subseteq\omega$, and $\epsilon<\omega_1$ such that:

1. $Z’\cap \epsilon=\emptyset$;
2. $\{ I_\beta\mid \beta\in Z’\}$ consists of mutually disjoint sets;
3. $\{J_\beta\setminus \epsilon\mid \beta\in Z’\}$ consists of mutually disjoint sets;
4. $\{J_\beta\cap\epsilon\mid \beta\in Z’\}$ consists of a single element;
5. $\{\langle |I_\beta|,|J_\beta\setminus\epsilon|\rangle\mid \beta\in Z’\}$ consists of a single element, say $\langle n,m\rangle$;
6. Denote $a_\beta:=I_\beta\cup(J_\beta\setminus\epsilon)$. Then $\alpha<\min(a_\beta)$ for all $\alpha<\beta$ in $Z’$;
7. For all $\beta\in Z’$ and $j<n+m$: $$a_\beta(j)\in I_\beta\text{ iff }j\in A.$$

Note that items (3),(4) simply say that $\{ J_\beta\mid \beta\in Z’\}$ is a $\Delta$-system, and item (7) asserts that $I_\beta$ embeds in $a_\beta$ in a certain fixed pattern for all $\beta\in Z’$.

Put $\mathcal A:=\{ a_\beta\mid \beta\in Z’\}$, and define $g:(n+m)\times(n+m)\rightarrow 2$ by stipulating that $g(i,j)=0$ iff $j\in A$. By the choice of $f$, we may now pick $\alpha<\beta$ in $Z’$ such that  $$f(a_\alpha(i),a_\beta(j))=g(i,j)\text{ for all }i,j< n+m.$$In particular, for all $j<n+m$: $$f(\alpha,a_\beta(j))=0\text{ iff }j\in A.$$So, $f(\alpha,\gamma)=0$ for all $\gamma\in I_\beta$, and $f(\alpha,\gamma)=1$ for all $\gamma\in J_\beta\setminus\epsilon$.
Recalling that $\alpha\not\in v_\beta\supseteq \bigcap\{u_\gamma\mid \gamma\in I_\beta\}\setminus \bigcup\{u_\gamma\mid \gamma\in J_\beta\}$, and the definition of $u_\gamma$, we get that $\alpha\in u_\gamma$ for some $\gamma\in J_\beta$. As $f(\alpha,\gamma)=1$ for all $\gamma\in J_\beta\setminus\epsilon$, we get that $\gamma\in J_\beta\cap\epsilon$. So $\alpha\in u_\gamma\subseteq\gamma+1\subseteq\epsilon$, contradicting the fact that $\alpha\in Z’$ and item (1). $\square$