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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.

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.

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).

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.

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 is 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).

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).