Steinhaus number of a group

Question

$\newcommand\Sn{\mathit{Sn}}$A subset $A$ of a group $X$ is called algebraic if $A=\{x\in X: a_0xa_1x\dotsm xa_n=1\}$ for some elements $a_0,a_1,\dotsc,a_n\in X$.

Let $\mathcal A_X$ be the family of all algebraic sets in $X$.

Definition. The Steinhaus number $\Sn(X)$ of an infinite group $X$ is the largest cardinal $\kappa$ such that for any cover $\mathcal C\subset\mathcal A_X$ of $X$ with cardinality $\lvert\mathcal C\rvert<\kappa$ there exists a set $C\in \mathcal C$ such that $FCC^{-1}F=X$ for some finite set $F\subseteq X$.

By the famous Steinhaus–Weil Theorem, for any closed set $F$ of positive Haar measure in a compact topological group $X$ the set $FF^{-1}$ is a neighborhood of the identity in $X$.

This theorem implies that each compact topological group has $$\operatorname{cov}(\overline{\mathcal N}_X)\le \Sn(X)\le \lvert X\rvert$$ where $\operatorname{cov}(\overline{\mathcal N}_X)$ is the smallest cardinality of a cover of $X$ by closed sets of Haar measure zero.

Under Martin's Axiom, $\Sn(X)=\mathfrak c$ for each infinite compact Polish group.

Problem. Is $\Sn(X)=\mathfrak c$ for any infinite compact Polish group in ZFC?

Remark. The answer is affirmative for commutative groups. This follows from the observation that each algebraic set in a commutative group is a coset of a subgroup $\{x\in X: x^n=1\}$.

Answer 1

The answer to this problem is negative: For the compact Polish group $X=S_3^\omega$ we have $Sn(X)\le\mathfrak r$ where $$\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq [\omega]^\omega\;\wedge\;\forall f\in 2^\omega\;\;\exists R\in\mathcal R\;\;|f[R]|=1\}.$$ By induction it can be shown $\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq[\omega]^\omega\;\wedge\;\forall f\in n^\omega\;\exists R\in\mathcal R\;\;|f[R]|=1\}$ for every $n\ge 2$. So, we can find a family $\mathcal R\subseteq[\omega]^\omega$ of cardinality $|\mathcal R|=\mathfrak r$ such that for every function $f:\omega\to S_3$ there exists $R\in\mathcal R$ such that $|f[R]|=1$. Then $$S_3^\omega=\bigcup_{R\in\mathcal R}\bigcup_{g\in S_3}A_{R,g},\quad\mbox{where $A_{R,g}=\{f\in S_3^\omega:\{g\}=f[R]\}$}.$$ Observe that for every $R\in\mathcal R$ and $g\in S_3$ the set $A_{R,g}A_{R,g}^{-1}=A_{R,e}$ is nowhere dense in $S_3^\omega$ and hence $FA_{R,g}A_{R,g}^{-1}F\ne S_3^\omega$ for any finite set $F\subseteq S_3^\omega$.

It remains to show that for every $R\in\mathcal R$ and $g\in S_3$, the set $A_{R,g}$ is algebraic in $S_3^\omega$.

Let $d$ and $t$ be elements of order $2$ and $3$ in the symmetric group $S_3$. It is easy to check that $S_3=\{x\in S_3:dxxdxx=e\}$ where $e$ is the identity of the group $S_3$. On the other hand, $\{x\in S_3:xtxxtx=t^2\}=\{e\}$ and hence $\{x\in S_3:g^{-1}xtg^{-1}xg^{-1}xtg^{-1}x=t^2\}=\{g\}$.

Then $A_{R,g}=\{x\in S_3^\omega:a_0xa_1xa_2xa_3x=b\}$ where $$a_0(i)=a_2(i)=\begin{cases}g^{-1}&\mbox{if $i\in R$};\\ d&\mbox{if $i\in\omega\setminus R$}, \end{cases}\quad a_1(i)=a_3(i)=\begin{cases}tg^{-1}&\mbox{if $i\in R$};\\ e&\mbox{if $i\in\omega\setminus R$}, \end{cases} $$ and $$ b(i)=\begin{cases}t^2&\mbox{if $i\in R$};\\ e&\mbox{if $i\in\omega\setminus R$}, \end{cases} $$