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.


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.

| cite | improve this answer | |
  • $\begingroup$ 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 Dec 15 '15 at 7:00
  • $\begingroup$ I used to think that the symbol for reaping number was $\tau$! $\endgroup$ – Ashutosh Dec 15 '15 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 Dec 21 '15 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.