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If $\kappa$ is an infinite cardinal and $\mathcal E$ is a maximal diverse subset of $\mathcal P(\kappa)$, then $|\mathcal E|=2^\kappa$.

Proof. Suppose $\mathcal E\subseteq\mathcal P(\kappa)$ is diverse and $|\mathcal E|\lt2^\kappa$; I will show that $\mathcal E$ is not maximal.

Let $\kappa=\bigcup_{\alpha\in\kappa}S_\alpha$ where the $S_\alpha$ are pairwise disjoint sets of cardinality $\kappa$. Define $f:\mathcal E\to\mathcal P(\kappa)$ by setting $f(X)=\{\alpha\in\kappa:|X\cap S_\alpha|=\kappa\}$.

Since $|\mathcal E|\lt2^\kappa$, $f$ is not surjective. Choose a set $A\subseteq\kappa$ which is not in the range of $f$, and let $Y=\bigcup_{\alpha\in A}S_\alpha$, so that $f(Y)=A$ and $Y\notin\mathcal E$.

If $X\in\mathcal E$ then $f(X)\ne f(Y)$, so there is some $\alpha\in\kappa$ such that either $|X\cap S_\alpha|\lt\kappa=|Y\cap S_\alpha|$ or else $|Y\cap S_\alpha|\lt\kappa=|X\cap S_\alpha|$, so $|(X\setminus Y)\cup(Y\setminus X)|=\kappa$, so $\mathcal E\cup\{Y\}$ is diverse, showing that $\mathcal E$ is not maximal.

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.

If $\kappa$ is an infinite cardinal, every maximal diverse family of subsets of $\kappa$ has cardinality $2^\kappa$.

Proof. Let me use the usual notation $X\operatorname{\triangle}Y=(X\setminus Y)\cup(Y\setminus X)$. 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 $2^\kappa$.

If $\kappa$ is an infinite cardinal and $\mathcal E$ is a maximal diverse subset of $\mathcal P(\kappa)$, then $|\mathcal E|=2^\kappa$.

Proof. Suppose $\mathcal E\subseteq\mathcal P(\kappa)$ is diverse and $|\mathcal E|\lt2^\kappa$; I will show that $\mathcal E$ is not maximal.

Let $\kappa=\bigcup_{\alpha\in\kappa}S_\alpha$ where the $S_\alpha$ are pairwise disjoint sets of cardinality $\kappa$. Define $f:\mathcal E\to\mathcal P(\kappa)$ by setting $f(X)=\{\alpha\in\kappa:|X\cap S_\alpha|=\kappa\}$.

Since $|\mathcal E|\lt2^\kappa$, $f$ is not surjective. Choose a set $A\subseteq\kappa$ which is not in the range of $f$, and let $Y=\bigcup_{\alpha\in A}S_\alpha$, so that $f(Y)=A$ and $Y\notin\mathcal E$.

If $X\in\mathcal E$ then $f(X)\ne f(Y)$, so there is some $\alpha\in\kappa$ such that either $|X\cap S_\alpha|\lt\kappa=|Y\cap S_\alpha|$ or else $|Y\cap S_\alpha|\lt\kappa=|X\cap S_\alpha|$, so $|(X\setminus Y)\cup(Y\setminus X)|=\kappa$, so $\mathcal E\cup\{Y\}$ is diverse, showing that $\mathcal E$ is not maximal.

If $\kappa$ is an infinite cardinal and $\mathcal E$ is a maximal diverse subset of $\mathcal P(\kappa)$, then $|\mathcal E|=2^\kappa$.

Proof. Suppose $\mathcal E\subseteq\mathcal P(\kappa)$ is diverse and $|\mathcal E|\lt2^\kappa$; I will show that $\mathcal E$ is not maximal.

Let $\kappa=\bigcup_{\alpha\in\kappa}S_\alpha$ where the $S_\alpha$ are pairwise disjoint sets of cardinality $\kappa$. Define $f:\mathcal E\to\mathcal P(\kappa)$ by setting $f(X)=\{\alpha\in\kappa:|X\cap S_\alpha|=\kappa\}$.

Since $|\mathcal E|\lt2^\kappa$, $f$ is not surjective. Choose a set $A\subseteq\kappa$ which is not in the range of $f$, and let $Y=\bigcup_{\alpha\in A}S_\alpha$, so that $f(Y)=A$ and $Y\notin\mathcal E$.

If $X\in\mathcal E$ then $f(X)\ne f(Y)$, so there is some $\alpha\in\kappa$ such that either $|X\cap S_\alpha|\lt\kappa=|Y\cap S_\alpha|$ or else $|Y\cap S_\alpha|\lt\kappa=|X\cap S_\alpha|$, so $|(X\setminus Y)\cup(Y\setminus X)|=\kappa$, so $\mathcal E\cup\{Y\}$ is diverse, showing that $\mathcal E$ is not maximal.

If $\kappa$ is an infinite cardinal, every maximal diverse family of subsets of $\kappa$ has cardinality $2^\kappa$.

Proof. Let me use the usual notation $X\operatorname{\triangle}Y=(X\setminus Y)\cup(Y\setminus X)$. 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 $2^\kappa$.