# Syndetic colorings with applications to S and L

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

Definition. An L-space is a regular hereditarily Lindelöf topological space which is not hereditarily separable.
Definition.
We say that a coloring $c:[\omega_1]^2\rightarrow\omega$ 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 $\max(a)<\min_{i<n}\beta_i$ 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 $[\sigma]:=\{ x\in2^{\omega_1}\mid \sigma\subseteq x\}$ for some $\sigma:a\rightarrow2$ with $a\in\mathcal Q(\omega_1)$.

Let $c:[\omega_1]^2\rightarrow\omega$ be an L-syndetic coloring. Let $\langle f_\alpha\mid\alpha<\omega_1\rangle$ be a b-universal binary sequence. For every $\beta<\omega_1$, define $x_\beta:\omega_1\rightarrow 2$ by letting for all $\alpha<\omega_1$:
$$x_\beta(\alpha):=\begin{cases}0,&\alpha>\beta\\1,&\alpha=\beta\\f_\alpha(c(\alpha,\beta)),&\alpha<\beta\end{cases}.$$
Clearly, we have $\overline{\{ x_\beta\mid \beta\le \gamma\}}\cap\{x_\beta\mid \gamma< \beta<\omega_1\}=\emptyset$ 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 $\langle U_i\mid i<\omega_1\rangle$ of $Y$ that does not admit a countable subscover. In particular, we may define strictly increasing functions $\psi:\omega_1\rightarrow\omega_1$ and $\varphi:\omega_1\rightarrow\omega_1$ such that for all $\alpha<\omega_1$:
$$x_{\varphi(\alpha)}\in (U_{\psi(\alpha)}\cap Y)\setminus\left(\bigcup\{ U_{\psi(i)}\mid i<\alpha\}\cup\{ x_\beta\mid \beta<\sup(\varphi[\alpha])\}\right).$$ For all $\alpha<\omega_1$, write $$V_\alpha:=U_{\psi(\alpha)}\setminus\overline{\{ x_\beta\mid \beta<\sup(\varphi[\alpha])\}}.$$ Then $V_\alpha$ is open. Write $Z:=\{ x_\beta\mid \beta\in\varphi[\omega_1]\}$. 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 $\{ \sigma_\beta:a_\beta\rightarrow 2\mid \beta\in B\}$ 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 $\Delta$-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, $n$;
• $\{ \sigma_\beta\restriction r\mid \beta\in B\}$ 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 $\max(a)<\min_{i<n}\beta_i$ and $$c(\alpha,\beta_i)=c(\alpha,\beta_0)+i\text{ for all }\alpha\in a.$$

Define $h:a\rightarrow\omega$ by letting $h(\alpha):=c(\alpha,\beta_0)$ for all $\alpha\in a$. Since $n(a)=n$, we have $\{ h^i\mid i<n\}={}^a2$. Let $\delta\in B$ be such that $a_\delta\setminus r=a$, and pick $i<n$ such that $h^i=\sigma_\delta\restriction a$. 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 $x_{\beta_i}\restriction a=\sigma_\delta\restriction a$. Recalling that $x_{\beta_i}\in [\sigma_{\beta_i}]$, we get in particular, that $x_{\beta_i}\restriction r=\sigma_{\beta_i}\restriction r=\sigma=\sigma_\delta\restriction r$.
Altogether, $\sigma_\delta\subseteq x_{\beta_i}$, that is, $x_{\beta_i}\in[\sigma_\delta]$, contradicting the fact that $\beta_i\neq\delta$. $\square$

Definition. An S-space is a regular hereditarily separable topological space which is not hereditarily Lindelöf.
Definition.
We say that a coloring $c:[\omega_1]^2\rightarrow\omega$ 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 $\max_{i<n}\alpha_i<\min(b)$ 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 $c:[\omega_1]^2\rightarrow\omega$ be an S-syndetic coloring. Let $\langle f_\beta\mid\beta<\omega_1\rangle$ be a b-universal binary sequence. For every $\alpha<\omega_1$, define $x_\alpha:\omega_1\rightarrow 2$ by letting for all $\beta<\omega_1$:
$$x_\alpha(\beta):=\begin{cases}0,&\beta<\alpha\\1,&\beta=\alpha\\f_\beta(c(\beta,\alpha)),&\beta>\alpha\end{cases}.$$
Clearly, we have $\{ x_\alpha\mid \alpha\le \gamma\}\cap\overline{\{x_\alpha\mid \gamma< \alpha<\omega_1\}}=\emptyset$ 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 $\varphi:\omega_1\rightarrow\omega_1$ such that for all $\alpha<\omega_1$:
$$x_{\varphi(\alpha)}\in Y\setminus\overline{\{ x_\beta\mid \beta<\sup(\varphi[\alpha])\}}.$$ Write $Z:=\{ x_{\varphi(\alpha)}\mid \alpha<\omega_1\}$. 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 $\{ \sigma_\alpha:b_\alpha\rightarrow 2\mid \alpha\in A\}$ 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, $n$;
• $\{ \sigma_\alpha\restriction r\mid \alpha\in A\}$ 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 $\max_{i<n}\alpha_i<\min(b)$ and $$c(\alpha_i,\beta)=c(\alpha_0,\beta)+i\text{ for all }\beta\in b.$$

Define $h:b\rightarrow\omega$ by letting $h(\beta):=c(\alpha_0,\beta)$ for all $\beta\in b$. Since $n(b)=n$, we have $\{ h^i\mid i<n\}={}^b2$. Let $\delta\in A$ be such that $b_\delta\setminus r=b$, and pick $i<n$ such that $h^i=\sigma_\delta\restriction b$. 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 $x_{\alpha_i}\restriction b=\sigma_\delta\restriction b$. Recalling that $x_{\alpha_i}\in [\sigma_{\alpha_i}]$, we get in particular, that $x_{\alpha_i}\restriction r=\sigma_{\alpha_i}\restriction r=\sigma=\sigma_\delta\restriction r$.
Altogether, $\sigma_\delta\subseteq x_{\alpha_i}$, that is, $x_{\alpha_i}\in[\sigma_\delta]$, contradicting the fact that $\alpha_i\neq\delta$. $\square$

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?

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### One Response to Syndetic colorings with applications to S and L

1. saf says:

Update: a few months ago, Yinhe Peng and Liuzhen Wu answered the above-mentioned question in the affirmative. Their proof will appear in a paper entitled “A Lindelof topological group with non-Lindelof square”.

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