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 $k = 2$?



Family of sets with a covering property
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.


