Syndetic colorings with applications to S and L

Notation. Write $\mathcal{Q}(A):=\{a\subseteq A\mid a\text{ is finite},a\ne \mathrm{\varnothing }\}$.

Definition. An L-space is a regular hereditarily Lindelöf topological space which is not hereditarily separable.
Definition. We say that a coloring $c:[{\omega }_1{]}^2\to \omega$ is L-syndetic if the following holds.
For every uncountable family $\mathcal{A}\subseteq \mathcal{Q}({\omega }_1)$ of mutually disjoint sets, every uncountable $B\subseteq {\omega }_1$, and every $n<\omega$, there exist $a\in \mathcal{A}$, and $\{{\beta }_i\mid i<n\}\subseteq B$ such that $max(a)<\underset{i<n}{min}{\beta }_i$ and $$c(\alpha ,{\beta }_i)=c(\alpha ,{\beta }_0)+i\text{ for all }\alpha \in a.$$

Proposition (folklore). The existence of an L-syndetic coloring entails an L-space.
Proof.  We build the L-space $X$ as a subspace of the product space $2^{{\omega }_1}$.
Recall that a basic open set in $2^{{\omega }_1}$ has the form $[\sigma ]:=\{x\in 2^{{\omega }_1}\mid \sigma \subseteq x\}$ for some $\sigma :a\to 2$ with $a\in \mathcal{Q}({\omega }_1)$.

Let $c:[{\omega }_1{]}^2\to \omega$ be an L-syndetic coloring. Let $⟨f_{\alpha }\mid \alpha <{\omega }_1⟩$ be a b-universal binary sequence. For every $\beta <{\omega }_1$, define $x_{\beta }:{\omega }_1\to 2$ by letting for all $\alpha <{\omega }_1$:
$$x_{\beta }(\alpha ):=\{\begin{array}{ll}0,& \alpha >\beta \\ 1,& \alpha =\beta \\ f_{\alpha }(c(\alpha ,\beta )),& \alpha <\beta \end{array}.$$
Clearly, we have $\overline{\{x_{\beta }\mid \beta \le \gamma \}}\cap \{x_{\beta }\mid \gamma <\beta <{\omega }_1\}=\mathrm{\varnothing }$ for all $\gamma <{\omega }_1$. In particular, $X:=\{x_{\beta }\mid \beta <{\omega }_1\}$ is not separable.

Subclaim. If $X$ is not hereditarily Lindelöf, then $X$ contains an uncountable discrete subspace.
Proof. Suppose that $Y$ is a subspace of $X$ which is not Lindelöf. Then, we may find an open cover $⟨U_i\mid i<{\omega }_1⟩$ of $Y$ that does not admit a countable subscover. In particular, we may define strictly increasing functions $\psi :{\omega }_1\to {\omega }_1$ and $\phi :{\omega }_1\to {\omega }_1$ such that for all $\alpha <{\omega }_1$:
$$x_{\phi (\alpha )}\in (U_{\psi (\alpha )}\cap Y)\setminus (\bigcup \{U_{\psi (i)}\mid i<\alpha \}\cup \{x_{\beta }\mid \beta <sup(\phi [\alpha ])\}).$$ For all $\alpha <{\omega }_1$, write $$V_{\alpha }:=U_{\psi (\alpha )}\setminus \overline{\{x_{\beta }\mid \beta <sup(\phi [\alpha ])\}}.$$ Then $V_{\alpha }$ is open. Write $Z:=\{x_{\beta }\mid \beta \in \phi [{\omega }_1]\}$. Then $V_{\alpha }\cap Z$ is a singleton for all $\alpha <{\omega }_1$, and hence $Z$ is a discrete subspace of $X$. End of proof of subclaim.

Thus, to complete the proof that $X$ is an $L$-space, let us suppose towards a contradiction that $\{x_{\beta }\mid \beta \in B\}$ is discrete, for a fixed uncountable $B\subseteq {\omega }_1$.
Then, there exists a collection of finite functions $\{{\sigma }_{\beta }:a_{\beta }\to 2\mid \beta \in B\}$ such that $\{\alpha \in B\mid x_{\alpha }\in [{\sigma }_{\beta }]\}=\{\beta \}$ for all  $\beta \in B$.
By passing to an uncountable subset of $B$ (via pigeonholes and $\mathrm{\Delta }$-system lemma), we may already assume the existence of $r\in \mathcal{Q}(\omega )$  such that:

•  $a_{\alpha }\cap a_{\beta }=r$ for all $\alpha <\beta$ in $B$;
• $\{n(a_{\beta }\setminus r)\mid \beta \in B\}$ contains a single element, say, $n$;
• $\{{\sigma }_{\beta }\upharpoonright r\mid \beta \in B\}$ contains a single element, say, $\sigma$;
• $\alpha <min(a_{\delta }\setminus r)\le max(a_{\delta }\setminus r)<\beta$ for all $\alpha <\delta <\beta$ in $B$.

Put $\mathcal{A}:=\{a_{\beta }\setminus r\mid \beta \in B\}$. Then, we may find $a\in \mathcal{A}$, and $\{{\beta }_i\mid i<n\}\subseteq B$ such that $max(a)<\underset{i<n}{min}{\beta }_i$ and $$c(\alpha ,{\beta }_i)=c(\alpha ,{\beta }_0)+i\text{ for all }\alpha \in a.$$

Define $h:a\to \omega$ by letting $h(\alpha ):=c(\alpha ,{\beta }_0)$ for all $\alpha \in a$. Since $n(a)=n$, we have $\{h^i\mid i<n\}={}^{a}2$. Let $\delta \in B$ be such that $a_{\delta }\setminus r=a$, and pick $i<n$ such that $h^i={\sigma }_{\delta }\upharpoonright a$. Then for all $\alpha \in a$:
$${\sigma }_{\delta }(\alpha )=h^i(\alpha )=f_{\alpha }(h(\alpha )+i)=f_{\alpha }(c(\alpha ,{\beta }_0)+i)=f_{\alpha }(c(\alpha ,{\beta }_i)).$$
So $x_{{\beta }_i}\upharpoonright a={\sigma }_{\delta }\upharpoonright a$. Recalling that $x_{{\beta }_i}\in [{\sigma }_{{\beta }_i}]$, we get in particular, that $x_{{\beta }_i}\upharpoonright r={\sigma }_{{\beta }_i}\upharpoonright r=\sigma ={\sigma }_{\delta }\upharpoonright r$.
Altogether, ${\sigma }_{\delta }\subseteq x_{{\beta }_i}$, that is, $x_{{\beta }_i}\in [{\sigma }_{\delta }]$, contradicting the fact that ${\beta }_i\ne \delta$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.1.0/svg/25fb.svg">}$

Definition. An S-space is a regular hereditarily separable topological space which is not hereditarily Lindelöf.
Definition. We say that a coloring $c:[{\omega }_1{]}^2\to \omega$ is S-syndetic if the following holds.
For every uncountable $A\subseteq {\omega }_1$, every uncountable family $\mathcal{B}\subseteq \mathcal{Q}({\omega }_1)$ of mutually disjoint sets, and every $n<\omega$, there exist $\{{\alpha }_i\mid i<n\}\subseteq A$ and $b\in \mathcal{B}$ such that $\underset{i<n}{max}{\alpha }_i<min(b)$ and $$c({\alpha }_i,\beta )=c({\alpha }_0,\beta )+i\text{ for all }\beta \in b.$$

Proposition (folklore). The existence of an S-syndetic coloring entails an S-space.
Proof.  We build the S-space $X$ as a subspace of the product space $2^{{\omega }_1}$.
Let $c:[{\omega }_1{]}^2\to \omega$ be an S-syndetic coloring. Let $⟨f_{\beta }\mid \beta <{\omega }_1⟩$ be a b-universal binary sequence. For every $\alpha <{\omega }_1$, define $x_{\alpha }:{\omega }_1\to 2$ by letting for all $\beta <{\omega }_1$:
$$x_{\alpha }(\beta ):=\{\begin{array}{ll}0,& \beta <\alpha \\ 1,& \beta =\alpha \\ f_{\beta }(c(\beta ,\alpha )),& \beta >\alpha \end{array}.$$
Clearly, we have $\{x_{\alpha }\mid \alpha \le \gamma \}\cap \overline{\{x_{\alpha }\mid \gamma <\alpha <{\omega }_1\}}=\mathrm{\varnothing }$ for all $\gamma <{\omega }_1$. In particular, $X:=\{x_{\alpha }\mid \alpha <{\omega }_1\}$ is not Lindelöf.

Subclaim. If $X$ is not hereditarily separable, then $X$ contains an uncountable discrete subspace.
Proof. Suppose that $Y$ is a subspace of $X$ which is not separable. Then, we may define a strictly increasing function $\phi :{\omega }_1\to {\omega }_1$ such that for all $\alpha <{\omega }_1$:
$$x_{\phi (\alpha )}\in Y\setminus \overline{\{x_{\beta }\mid \beta <sup(\phi [\alpha ])\}}.$$ Write $Z:=\{x_{\phi (\alpha )}\mid \alpha <{\omega }_1\}$. Then $Z$ is a discrete subspace of $X$. End of proof of subclaim.

Thus, to complete the proof that $X$ is an $S$-space, let us suppose towards a contradiction that $\{x_{\alpha }\mid \alpha \in A\}$ is discrete, for a fixed uncountable $A\subseteq {\omega }_1$.
Then, there exists a collection of finite functions $\{{\sigma }_{\alpha }:b_{\alpha }\to 2\mid \alpha \in A\}$ such that $\{\alpha \in A\mid x_{\alpha }\in [{\sigma }_{\beta }]\}=\{\alpha \}$ for all  $\alpha \in A$. By thinning out $A$, we may assume the existence of $r\in \mathcal{Q}(\omega )$  such that:

•  $b_{\alpha }\cap b_{\beta }=r$ for all $\alpha <\beta$ in $A$;
• $\{n(b_{\alpha }\setminus r)\mid \alpha \in A\}$ contains a single element, say, $n$;
• $\{{\sigma }_{\alpha }\upharpoonright r\mid \alpha \in A\}$ contains a single element, say, $\sigma$;
• $\alpha <min(b_{\delta }\setminus r)\le max(b_{\delta }\setminus r)<\beta$ for all $\alpha <\delta <\beta$ in $A$.

Put $\mathcal{B}:=\{b_{\alpha }\setminus r\mid \alpha \in A\}$. Then, we may find $b\in \mathcal{B}$, and $\{{\alpha }_i\mid i<n\}\subseteq A$ such that $\underset{i<n}{max}{\alpha }_i<min(b)$ and $$c({\alpha }_i,\beta )=c({\alpha }_0,\beta )+i\text{ for all }\beta \in b.$$

Define $h:b\to \omega$ by letting $h(\beta ):=c({\alpha }_0,\beta )$ for all $\beta \in b$. Since $n(b)=n$, we have $\{h^i\mid i<n\}={}^{b}2$. Let $\delta \in A$ be such that $b_{\delta }\setminus r=b$, and pick $i<n$ such that $h^i={\sigma }_{\delta }\upharpoonright b$. Then for all $\beta \in b$:
$${\sigma }_{\delta }(\beta )=h^i(\beta )=f_{\beta }(h(\beta )+i)=f_{\beta }(c({\alpha }_0,\beta )+i)=f_{\beta }(c({\alpha }_i,\beta )).$$
So $x_{{\alpha }_i}\upharpoonright b={\sigma }_{\delta }\upharpoonright b$. Recalling that $x_{{\alpha }_i}\in [{\sigma }_{{\alpha }_i}]$, we get in particular, that $x_{{\alpha }_i}\upharpoonright r={\sigma }_{{\alpha }_i}\upharpoonright r=\sigma ={\sigma }_{\delta }\upharpoonright r$.
Altogether, ${\sigma }_{\delta }\subseteq x_{{\alpha }_i}$, that is, $x_{{\alpha }_i}\in [{\sigma }_{\delta }]$, contradicting the fact that ${\alpha }_i\ne \delta$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.1.0/svg/25fb.svg">}$

Todorcevic proved the consistency of the nonexistence of an S-space, consequently, S-syndetic coloring may not exist. On the other hand, it is easy to see that an S-syndetic coloring is available in the forcing extension by adding a single Cohen real (see here).

Moore proved that an L-syndetic coloring exists (that is, in ZFC). However, the following remains open:

Question (Moore). Does there exist an $L$-space $X$ such that $X^2$ is still an L-space?