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 $\{a(0),a(1),\dots ,a(n-1)\}$ denote the $\in$-increasing enumeration of the elements of $a$.

Lemma. In the generic extension by forcing with $\mathbb{C}$, there exists a coloring $f:[{\omega }_1{]}^2\to \omega$ such that for every finite function $g:n\times n\to \omega$, 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 $d_{\alpha }:\alpha \to \omega$ 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 $f:[{\omega }_1{]}^2\to \omega$ 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, $p:m\to \omega$, 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 $M\prec (H({\omega }_2),\in ,⊲)$ 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 $$\gamma :=max\{max(d_{b(i)}^{-1}[k+m])\mid i<n\}+1.$$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 $$f(a(i),b(j))=g(i,j)\text{ for all }i,j<n.\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$$

Proof of main theorem. Fix a coloring $f:[{\omega }_1{]}^2\to \omega$ 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 $I\cap J=\mathrm{\varnothing }$. 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 $X:=⟨{\omega }_1,\tau ⟩$ 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 $\bigcap _{\gamma \in \mathrm{\varnothing }}{u}_{\gamma }={\omega }_1$), 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 $\psi :{\omega }_1\to Y$ so that $\psi (\beta )\notin \overline{\psi [\beta ]}$ 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 $$\alpha \notin v_{\beta }\text{ for all }\alpha \in Z\cap \beta .$$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 $Z^{\prime }\subseteq Z$, we may assume the existence of  $A\subseteq \omega$, and $ϵ<{\omega }_1$ such that:

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

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

Put $\mathcal{A}:=\{a_{\beta }\mid \beta \in Z^{\prime }\}$, and define $g:(n+m)\times (n+m)\to 2$ by stipulating that $g(i,j)=0$ iff $j\in A$. By the choice of $f$, we may now pick $\alpha <\beta$ in $Z^{\prime }$ 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 $\gamma \in J_{\beta }\setminus ϵ$.
Recalling that $\alpha \notin v_{\beta }\supseteq \bigcap \{u_{\gamma }\mid \gamma \in I_{\beta }\}\setminus \bigcup \{u_{\gamma }\mid \gamma \in J_{\beta }\}$, 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 $\gamma \in J_{\beta }\setminus ϵ$, we get that $\gamma \in J_{\beta }\cap ϵ$. So $\alpha \in u_{\gamma }\subseteq \gamma +1\subseteq ϵ$, contradicting the fact that $\alpha \in Z^{\prime }$ and item (1). $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$