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Let $\kappa\geq \aleph_0$ be a cardinal. We say a collection ${\cal E} \subseteq {\cal P}(\kappa)$ is diverse if $|(A \setminus B) \cup (B \setminus A)| = \kappa$ whenever $A\neq B\in {\cal E}$. A standard application of Zorn's Lemma shows that every diverse family ${\cal E}\subseteq {\cal P}(\kappa)$ is contained in a maximal diverse family ${\cal E'}\subseteq {\cal P}(\kappa)$ with respect to set inclusion.

Question. If $\kappa$ is an infinite cardinal and $\cal E_1,\cal E_2 \subseteq {\cal P}(\kappa)$ are maximal diverse, do $\cal E_1,\cal E_2$ necessarily have the same cardinality?

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    $\begingroup$ I'm assuming with "$\mathcal E_{1,2}$" you mean two different collections $\mathcal E_1,\mathcal E_2$? $\endgroup$
    – Wojowu
    Commented Apr 8 at 9:53
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    $\begingroup$ The relation $|X\operatorname{\triangle}Y|\lt\kappa$ is an equivalence relation. A "maximal diverse family" $\mathcal E$ is just a set of representatives for that equivalence relation, so the cardinality of $\mathcal E$ is just the number of equivalence classes, which is $2^\kappa$. $\endgroup$
    – bof
    Commented Apr 8 at 10:52
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    $\begingroup$ @bof: That comment seems like a complete solution; why not put it as an actual answer? Giving answers in comments is awkward — it discourages others from posting the same or similar solutions as an actual answer, so the question gets stuck unanswered. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Apr 8 at 15:04
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    $\begingroup$ @JoelDavidHamkins As you can see, I posted an answer before I realized how utterly simple the question is when looked at the right way. Then I deleted my answer because answering off-topic questions is discouraged. I will be glad to post an answer if you really think it's a "research level question". After all you're an expert on set theory and I'm just a dilettante. Almost disjoint families are different because the relation $|X\cap Y|=\kappa$ is not an equivalence relation. $\endgroup$
    – bof
    Commented Apr 8 at 21:00
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    $\begingroup$ @bof Thanks for posting. I get the hesitancy, since I've often felt the same way. Nevertheless, I also think that if a question remains open, then everything works best when one just posts the answer, even if the answer makes things seem easy. In this case, the question clearly arises from motivating analogy with the mad family case, but the point is that they are not actually analogous, and that could be confusing at first, but is completely clarified when you point out that the symmetric difference case corresponds to an equivalence relation, but almost disjointness does not. $\endgroup$
    – Joel David Hamkins
    Commented Apr 8 at 23:19

1 Answer 1

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If $\kappa$ is an infinite cardinal, then the relation $|X\bigtriangleup Y|\lt\kappa$ (where $X\bigtriangleup Y=(X\setminus Y)\cup(Y\setminus X)$) is an equivalence relation on $\mathcal P(\kappa)$. A collection $\mathcal E\subseteq\mathcal P(\kappa)$ is "diverse" if it contains at most one element of each equivalence class. $\mathcal E$ is a "maximal diverse family" if it contains exactly one element of each equivalence class, in which case $|\mathcal E|$ is equal to the number of equivalence classes, which is $2^\kappa$.

To see that the number of equivalence classes is $2^\kappa$, observe that $\{X\times\kappa:X\subseteq\kappa\}$ is a "diverse" collection of subsets of $\kappa\times\kappa$.

This argument does not apply to maximal almost disjoint families, because the relation $|X\cap Y|=\kappa$ is not an equivalence relation.

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