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In a 3-page note on Bernstein set,

Suppose, $\operatorname{cf}(\mathfrak{c})= \mathfrak{c}$, $\operatorname{non}({\bf{L}})= \min\{|X|: X \subset \Bbb{R}, X \text{ is not a Lebesgue measure zero subset of } \Bbb{R}\} = \mathfrak{c}$ Let $\operatorname{PD}(B)$ denote the family of all pairwise disjoint subsets of $B$, $\bf L$ denote the $\sigma$-ideal of Lebesque measure zero subsets of the real line.Then:$$\operatorname{sat}({\cal{P}}(\Bbb{R}) / {\bf{L}})= \min\{ \kappa: \forall {X} \in \operatorname{PD}({\cal{P}}(\Bbb{R}) / {\bf{L}})(|X|< \kappa)\}$$

I can't understand Theorem $2$ states that, given above conditions, $\operatorname{sat}({\cal{P}}(\Bbb{R}) / {\bf{L}}) > \mathfrak{c}^{+}$.

By lemma $1$ in the note, we could find a continuum of sets of cardinality continuum, $\mathcal A$ which consists of a partition of $\Bbb R$,such that for any perfect set $B$, there exists $A \in \cal A$ such that $A \subseteq B$, then we could have "$\cal S$ be a maximal with respect to inclusion a family of selectors of $\cal A$ such that $$X, Y \in {\cal{ S}} \land X \neq Y \to |X \cap Y | < \mathfrak{c}"$$

In particular, I don't understand why "it's easy to check that $|\mathcal{S}|= \mathfrak{c}^{+} $". Moreover, I don't know what $[S]_{\bf L}$ means.

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@user14111:Thanks. Fixed it. –  Metta World Peace Jun 14 '13 at 6:32

1 Answer 1

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Maybe part of the confusion is that you used $\mathcal{P}(\mathbb{R}) \setminus \bf{L}$ instead of $\mathcal{P}(\mathbb{R}) / \bf{L}$, which is the set of equivalence classes of subsets of $\mathbb{R}$ where two sets are equivalent if their symmetric difference is in $\bf{L}$; the class of a set $S$ is then denoted by $[S]_{\bf{L}}$.

This is a special case of a quotient of a Boolean algebra modulo an ideal; the result is again a Boolean algebra (where the class of $\emptyset$, i.e. all sets of $\bf{L}$, which are all equivalent, is the $0$-element), so $\operatorname{sat}$ is defined for it.

Corollary 2 in the note follows because the equivalence classes of the $\mathfrak{c}$ many pairwise disjoint Bernstein sets that Corollary 1 gives us are all non-equivalent and non-zero classes in $\mathcal{P}(\mathbb{R})/\bf{L}$ that are still disjoint, and so $\operatorname{sat}( \mathcal{P}(\mathbb{R})/\bf{L} )$ cannot be $\mathfrak{c}$ (or less) any more. Hence is it at least $\mathfrak{c}^{+}$.

As to the proof of Theorem 2: Note that $\operatorname{non}(\bf{L}) = \mathfrak{c}$ just is a reformulation of "every subset $A$ of $\mathbb{R}$ such that $|A| < \mathfrak{c}$ has Lebesgue measure $0$ (i.e. is in $\bf{L}$)". So if we have family $\mathcal{S}$ of selectors for $\mathcal{A}$ with the property that any 2 different selectors $X$ and $Y$ have $| X \cap Y | < \mathfrak{c}$, then if we take the classes of these sets modulo $\bf{L}$ then their intersection is in $\bf{L}$ (all sets smaller than continuum have measure $0$), so the classes indeed form a disjoint family in the quotient Boolean algebra. If we also know that their are at least $\mathfrak{c}^{+}$ many, then the $\operatorname{sat}$ of this algebra is strictly larger than $\mathfrak{c}^{+}$, as claimed (see the definition of $\operatorname{sat}$).

It is clear by a Zorn lemma argument that there indeed is a maximal (by inclusion) family of selectors for $\mathcal{A}$ with the intersection property.

I think we only need that $\mathcal{A}$ is a partition of $\mathbb{R}$ of size $\mathfrak{c}$, all of whose members have size $\mathfrak{c}$. In that case $\operatorname{cf}(\mathfrak{c}) = \mathfrak{c}$ should imply that a family of "almost disjoint" (in the sense that their intersection has size strictly less than $\mathfrak{c}$) that has size $\le \mathfrak{c}$ can be extended by a new set, and so cannot be maximal. Hence the existing maximal set must have larger size than $\mathfrak{c}$, as claimed.

To see this: let $S_{\xi}, \xi < \mathfrak{c}$ be a family of selectors with smaller than continuum pairwise intersection. Define $T_\xi = S_\xi \setminus \cup_{\eta < \xi} S_\eta$, Then by regularity of $\mathfrak{c}$, the $T_\xi$ are pairwise disjoint, and all of size $\mathfrak{c}$. Picking one element from each of them gives us a new selector that has intersection size $< \mathfrak{c}$ with all the $S_\xi$.

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Thank you very much for your well-writen and helpful answer and also for kindly pointing out my confusion over \ and /. –  Metta World Peace Jun 13 '13 at 19:36
    
Glad to be able to help! –  Henno Brandsma Jun 14 '13 at 3:36

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