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This question assumes familiarity with combinatorial cardinal characteristics of the continuum.

Let $\mathcal{E}$ be the $\sigma$-ideal generated by closed measure zero subsets of the real line. It is known that $$\operatorname{cov}(\mathcal{N})\cdot\operatorname{cov}(\mathcal{M})\le \operatorname{cov}(\mathcal{E})\le \operatorname{cov}(\mathcal{N})\cdot \mathfrak{d},$$ and that the first inequality is consistently strict (Bartoszynski-Shelah).

Is the second inequality consistently strict?

I conjecture that the answer is positive, and known, but did not find it in the cited paper (or elsewhere, but I didn't search thoroughly).

Update: This problem is solved below by Ashutosh, and the solution suggests a follow-up question.

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Since we can cover the reals by at most $\mathfrak{r}$ (reaping number) closed null sets, we can do a countable support iteration of rational perfect set forcing over a model of CH to get a model of $\omega_1 = \mathfrak{r} < \mathfrak{d} = \omega_2$.

To see that $\operatorname{cov}(\mathcal{E}) \leq \mathfrak{r}$, just note that for every infinite $A \subseteq \omega$, $i \in \{0, 1\}$, the set $N_{A, i} = \{x \in 2^{\omega} : (\forall n \in A) (x(n) = i)\}$ is closed null.

$\mathfrak{d} = \omega_2$ in this model because Miller real is not dominated by any ground model real. Also the ground model p-points are preserved so $\mathfrak{r} = \mathfrak{u} = \omega_1$. The proof of this fact can be found in chapter 23 of Halbeisen's Combinatorial set theory.

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  • $\begingroup$ I think I can see what you mean, that is, why $\operatorname{cov}(\mathcal{E})\le\mathfrak{r}$. Then the forcing part can be replaced by the consistency of $\mathfrak{r}<\mathfrak{d}$. But to be on the safe side, and for other readers, could you provide more details for the former assertion, and if exists, also a reference? (Also, any reason why the reaping number is denoted here $\tau$ and not $\mathfrak{r}$ as is more customary?) $\endgroup$
    – Boaz Tsaban
    Commented Dec 15, 2015 at 7:00
  • $\begingroup$ I used to think that the symbol for reaping number was $\tau$! $\endgroup$
    – Ashutosh
    Commented Dec 15, 2015 at 7:18
  • $\begingroup$ Thanks for the answer! It immediately raises another question, posed seperately so as not to deprive you of answering the first question. $\endgroup$
    – Boaz Tsaban
    Commented Dec 21, 2015 at 12:24

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