Let $\mathcal I$ be a $\sigma$-ideal with Borel base on an uncountable Polish space $X=\bigcup\mathcal I$.

Let $\mathrm{cov}(\mathcal I)$ (resp. $\mathrm{cov}_\sqcup(\mathcal I)$) be the smallest cardinality of a cover of $X$ by (parvise disjoint) Borel sets that belong to the ideal $\mathcal I$.

The cardinal $\mathrm{cov}(\mathcal I)$ is one of four classical cardinal characteristics of an ideal. What about its disjoint modification? Was it considered in the literature?

Observe the following trivial relations between the cardinals $\mathrm{cov}(\mathcal I)$ and $\mathrm{cov}_\sqcup(\mathcal I)$:

1) $\mathrm{cov}(\mathcal I)\le\mathrm{cov}_\sqcup(\mathcal I)$;

2) If $\mathrm{cov}(\mathcal I)\le\omega_1$, then $\mathrm{cov}(\mathcal I)=\mathrm{cov}_\sqcup(\mathcal I)$.

Problem 1. Is $\mathrm{cov}(\mathcal I)=\mathrm{cov}_\sqcup(\mathcal I)$?

This problem is especially interesting for the ideal $\mathcal M$ of meager subsets of the real line and for the ideal $\mathcal N$ of Lebesgue null sets in $\mathbb R$.

Problem 2. Are the strict inequalities $\mathrm{cov}(\mathcal M)<\mathrm{cov}_\sqcup(\mathcal M)$ and $\mathrm{cov}(\mathcal N)<\mathrm{cov}_\sqcup(\mathcal N)$ consistent?

The negative answer to the second part of problem 2 would give an affirmative answer to Problem 8 of this MO post.

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    $\begingroup$ If $\mathcal I$ is closed under taking arbitrary subsets, then any family $\mathcal A$ of sets in $\mathcal I$ can be made into a disjoint family $\mathcal A'$ of set in $\mathcal I$ with $\bigcup \mathcal A = \bigcup \mathcal A'$. (Just enumerate $\mathcal A = \{A_\alpha : \alpha < \kappa\}$ and put $A'_\alpha = A_\alpha \setminus (\bigcup_{\beta < \alpha}A_\beta)$.) $\endgroup$ – Will Brian Nov 16 '17 at 19:24
  • $\begingroup$ In the linked post, what makes things so hard (and interesting!) is that you're only looking at the closed sets in $\mathcal I$, so this trick doesn't work. $\endgroup$ – Will Brian Nov 16 '17 at 19:26
  • $\begingroup$ @WillBrian Please note that I am requiring the pieces to be Borel! But your trick does not produce Borel sets (especially for $\kappa>\omega_1$). $\endgroup$ – Taras Banakh Nov 16 '17 at 19:28
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    $\begingroup$ @WillBrian What I realized is that $\acute{\mathfrak n}$ is equal to the smallest cadinality of a disjoint compact cover of any non $\sigma$-compact absolute Borel space (just use the fact that each absolute Borel space is a continuous injective image of a Polish space). So, in a sense $\acute{\mathfrak n}$ is a universal "constant". It would be interesting to know what happens with $\acute{\mathfrak n}$ for analytic non-Borel spaces. $\endgroup$ – Taras Banakh Nov 16 '17 at 19:37
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    $\begingroup$ @WillBrian On the other hand, the cardinal $\acute{\mathfrak m}$ is equal to the product $\acute{\mathfrak n}\cdot\mathrm{cov}_\sqcup(\mathcal N)$, so is reducible in a sense. This was a motivation for my question on $cov_\sqcup(\mathcal I)$. $\endgroup$ – Taras Banakh Nov 16 '17 at 19:39

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