Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgroups?

Question

Let $\mathrm{cov}_H(C_2^\omega)$ be the smallest cardinality of a cover of the Boolean group $C_2^\omega=(\mathbb Z/2\mathbb Z)^\omega$ by closed subgroups of infinite index. It can be shown that $$\max\{\mathrm{cov}(\mathcal M),\mathrm{cov}(\mathcal N)\}\le \mathrm{cov}(\mathcal E)\le\mathrm{cov}_H(C_2^\omega)\le \mathfrak r,$$ where $\mathcal E$ is the $\sigma$-ideal generated by closed subsets of Haar measure zero in $C_2^\omega$ (the upper bound $\mathrm{cov}_H(C_2^\omega)\le\mathfrak r$ is proved, for example, here). It is known that $\max\{\mathrm{cov}(\mathcal N),\mathrm{cov}(\mathcal M)\}$ can be strictly smaller than $\mathfrak b$ (this happens, for example, in the Laver and Mathias models).

Problem. Is $\mathrm{cov}_H(C_2^\omega)<\mathfrak b$ consistent? What is the value of $\mathrm{cov}_H(C_2^\omega)$ in the Laver (or Mathias) model?

Answer 1

Lyubomyr Zdomskyy proved that in the Laver model $\mathrm{cov}_H(2^\omega)=\omega_1<\mathfrak b=\mathfrak c$.

His argument used the following known Laver property of the Laver model $V'$: for every function $f:\omega\to\omega$ in $V'$ upper bounded by some function $h:\omega\to\omega$ in the ground model $V$, there exists $H:\omega\to [\omega]^{<\omega}$ in $V$ such that $|H(n)|\leq n+1$ and $f(n)\in H(n)$ for all $n\in\omega$.

Choose any increasing sequence $(k_n)_{n\in\omega}\in \omega^\omega\cap V$ such that $k_0=0$ and $k_{n+1}>k_n+n+1$ for every $n\in\omega$. Consider the family $\mathcal H$ of closed nowhere dense subgroups of $2^\omega$ of the form $ \prod_{n\in\omega}H_n, $ where $H_n$ is a proper subgroup of $2^{k_{n+1}\setminus k_n}$ and $( H_n)_{n\in\omega}\in V$. We claim that for every $x\in 2^\omega$ there exists $H\in\mathcal H$ with $x\in H$. Indeed, since $x{\restriction} (k_{n+1}\setminus k_n)\in 2^{k_{n+1}\setminus k_n}$, the latter set has size $2^{k_{n+1}-k_n}$, and the sequence $\langle 2^{k_{n+1}-k_n}\rangle_{n\in\omega}$ lies in $V$, there exists a sequence $\langle X_n\rangle_{n\in\omega}\in V$ such that $$x{\restriction}(k_{n+1}\setminus k_n)\in X_n\in \big[ 2^{k_{n+1}\setminus k_n}\big]^{n+1}$$ for all $n\in\omega$. It suffices to denote by $H_n$ the subgroup of $2^{k_{n+1}\setminus k_n}$ generated by $X_n$ and note that $H_n\neq 2^{k_{n+1}\setminus k_n}$ because $|X_n|=n+1$ and $k_{n+1}-k_n>n+1$.

Therefore in the model $V'$ we have $\mathrm{cov}_H(2^\omega)\le|\mathcal H|\le|V\cap 2^\omega|=\omega_1$.