Family of sets with a covering property Suppose $X$ is an infinite set and $\mathcal{A}$ is a family of subsets of $X$.
(1) We say that $\mathcal{A}$ is $k$-cover-free if for every distinct $a_0, a_1, \cdots, a_k \in \mathcal{A}$, $a_0 \nsubseteq \bigcup_{1 \leq i \leq k} a_i$.
(2) We say that $\mathcal{A}$ is $k$-good if for every distinct $a_1, \cdots, a_k \in \mathcal{A}$, there exists $a_0 \in \cal{A} \setminus \{a_1, \cdots, a_k\}$ such that $a_0 \subseteq \bigcup_{1 \leq i \leq k} a_i$.
Note that the family $\mathcal{A}$ of all subsets of $X$ of size $2$ is both $1$-cover-free and $2$-good. So there are $1$-cover-free and $2$-good families of all possible infinite cardinalities. I would like to know if there are similar examples for $k \geq 2$.
Question: Let $k \geq 2$. Are there families of arbitrarily large cardinalities which are both $k$-cover-free and $(k+1)$-good? What if $k = 2$?
 A: For each natural number $n\in\omega$, let $U_n\subseteq\mathbb{R}^k$ be a randomly selected (according to some reasonable distribution) affine hyperplane. Suppose now that $a_0,\dots,a_k$ are distinct natural numbers. Then with probability $1$, we know that

*

*$|U_{a_1}\cap\dots\cap U_{a_k}|=1$,


*$U_{a_1}\cap\dots\cap U_{a_k}\not\subseteq U_{a_0}$, and


*$U_{a_0}\cap\dots\cap U_{a_k}=\emptyset$.
Therefore, if we set $V_{n}=U_n^c$, then whenever $a_0,\dots,a_k$ are distinct natural numbers, we have $V_{a_0}\cup\dots\cup V_{a_k}=\mathbb{R}^k$ and
$V_{a_0}\subseteq V_{a_1}\cup\dots\cup V_{a_k}$ with probability $1$. Therefore, the set $(V_n)_{n\in\omega}$ is $k+1$-good but $k$-cover free with probability $1$.
Controlling the cardinality
One can control the cardinality of the cardinality of $k$-cover free, $k+1$-good families using the Lowenhein-Skolem theorem and by rephrasing the intersection version of this problem in terms of lattices.
Suppose that $X$ is a meet-semilattice and $A\subseteq X$ is a subset that generates $X$ where if $a_0,a_1,\dots,a_k$ are distinct elements of $A$, then $a_0\wedge\dots\wedge a_k=0$ and $a_1\wedge\dots\wedge a_k\not\leq a_0$. For $a\in A$, let $L_a=\{x\in X\setminus\{0\}\mid x\leq a\}$ for $a\in A$. Then observe that $L_{a_1}\cap\dots\cap L_{a_k}\not\subseteq L_{a_0}$, but $L_{a_0}\cap\dots\cap L_{a_k}=0.$
Now, one can control the size of the set $X$ simply by using the Lowenhein-Skolem theorem. Since $|A|=|X|$, we have control over the size of the set $A$ as well.
A: Yes. Let $S$ be a set of cardinality $|S|\ge k+2$. Let $X=\binom Sk$, the set of all $k$-element subsets of $S$. For each $s\in S$ let $a_s=\{x\in X:s\notin x\}$, and let $A=\{a_s:s\in S\}$. It is easy to verify that $A$ is $k$-cover-free and $(k+1)$-good.
Given distinct elements $s_0,s_1,\dots,s_k\in S$, we have $a_{s_0}\not\subseteq a_{s_1}\cup\cdots\cup a_{s_k}$ because $\{s_1,\dots,s_k\}\in a_{s_0}\setminus(a_{s_1}\cup\cdots\cup a_{s_k})$.
Given distinct elements $s_1,\dots,s_{k+1}\in S$, take any $s_0\in S\setminus\{s_1,\dots,s_{k+1}\}$; we have $a_{s_0}\subseteq a_{s_1}\cup\cdots\cup a_{s_{k+1}}$ because $a_{s_1}\cup\cdots\cup a_{s_{k+1}}=X$.
P.S. In a comment the OP asked:

Is possible to get such a family whose members are all finite?

The answer is negative; for $k\ge2$ a family which is $k$-cover-free and $(k+1)$-good can't contain uncountably many finite sets. In fact:
Theorem. For $k\ge2$ a family which is $k$-cover-free and $(k+1)$-good can't contain infinitely many finite sets of bounded size.
Proof. Assume for a contradiction that $k\ge2$ and that $A$ is a $k$-cover-free $(k+1)$-good family containing infinitely many finite sets of bounded size. Then $A$ contains an infinite quasidisjoint family of finite sets, i.e., there exist an infinite sequence of distinct finite sets $a_i\in A$ and a set $d$ such that $a_i\cap a_j=d$ whenever $i\ne j$. Let $a'_i=a_i\setminus d$ and let $s=a_1\cup\cdots\cup a_k$.
Since $A$ is $(k+1)$-good, for each $i\gt k$ we can choose a set $b_i\in A$ so that $b_i\subseteq s\cup a_i$ and $b_i\notin\{a_1,\dots,a_k\}\cup\{a_i\}$. Since $A$ is $k$-cover-free we have $b_i\not\subseteq s$, whence $b_i\cap a'_i\ne\varnothing$, so the sets $b_i$ ($i\gt k$) are distinct.
Since $s$ is finite, we can choose $i,j$ with $k\lt i\lt j$ and $b_i\cap s=b_j\cap s$.Now $b_i,b_j,a_j$ are distinct elements of $A$, and $b_j\subseteq b_i\cup a_j$, showing that $a$ is not $2$-cover-free. Since $k\ge2$, this contradicts our assumption that $A$ is $k$-cover-free.
