Set sizes in linear set systems on $\mathbb{N}$ containing some disjoint sets

Question

Is there a set $E\subseteq {\cal P}(\mathbb{N})$ of subsets of $\mathbb{N}$ with the following properties?

1. $|e| > 2$ for all $e\in E$,
2. $e_1\neq e_2 \in E$ implies $|e_1 \cap e_2| \leq 1$,
3. for all $m,n\in\mathbb{N}$ there is $e\in E$ with $\{m,n\}\subseteq e$,
4. there are $e_1, e_2 \in E$ with $e_1\cap e_2 = \emptyset$, and
5. there are $e_1, e_2 \in E$ with $|e_1| \neq |e_2|$.

Answer 1

Yes. Let $[\mathbb N]^2=\{p_n:n\in\mathbb N\}$ with $p_1=\{1,2\}$. Define $e_n$ recursively as follows. $e_1=\{1,2,3,4\}$. For $n\gt1$, if $p_n\subseteq e_k$ for some $k\lt n$, let $e_n=e_k$; otherwise let $e_n=p_n\cup\{x\}$ where $x$ is the least element of $\mathbb N\setminus(e_1\cup\cdots\cup e_{n-1}\cup p_n)$. The set $E=\{e_n:n\in\mathbb N\}\subseteq[\mathbb N]^3\cup[\mathbb N]^4$ satisfies your requirements.

Answer 2

Yes. Fix a partition $\left<A_n\right>_{n\leq\omega}$ of $\mathbb{N}$ such that $A_n$ is infinite for each $n<\omega$, and $A_\omega$ has cardinality 3. For $n<\omega$ let $B_\ell=\bigcup_{n\leq\ell\leq\omega}B_\ell$. So $B_\omega=\bigcap_{n<\omega}B_n$, for example. Given $L\subseteq\mathbb{N}\times\mathbb{N}$, say $L$ is a \emph{full line} if it is just a line in the usual geometric sense; that is, there are points $(a,b),(c,d)\in\mathbb{N}\times\mathbb{N}$ with $(a,b)\neq(c,d)$ and for $(x,y)\in\mathbb{N}$, we have $(x,y)\in L$ iff either

• $(x,y)\in\{(a,b),(c,d)\}$, or
• $b=d=y$, or
• $y\neq b\neq d$ and $\frac{x-a}{y-b}=\frac{c-a}{d-b}$.

For each $n<\omega$, let $\pi_n:B_n\to\mathbb{N}\times\mathbb{N}$ be a bijection such that $\pi_n``B_{n+1}$ is a full line.

Now define our family $E$ as follows: Given $e\subseteq\mathbb{N}$, put $e\in E$ iff either:

(a) there is $n<\omega$ such that $e\subseteq B_n$ and $\pi_n``e$ is a full line which is distinct from $\pi_n``B_{n+1}$, or

(b) $e=B_\omega$.

If clause (a) above holds, say that $e$ is a stage $n$ line.

This family has the desired properties:

Each $e\in E$ is either infinite or has cardinality $3$, so in particular, $\mathrm{card}(e)>2$.

If $e_1,e_2\in E$ with $e_1\neq e_2$, then $\mathrm{card}(e_1\cap e_2)\leq 1$, since in fact either:

• there is $n<\omega$ such that $e_1,e_2$ are both stage $n$ lines, in which case $\mathrm{card}(e_1\cap e_2)\leq 1$; or

• there is $n<\omega$ such that $e_1$ is a stage $n$ line and $e_2\subseteq B_{n+1}$, in which case $\pi_n``e_2\subseteq\pi_n``B_{n+1}$, and $\pi_n``B_{n+1}$ is a line distinct from $\pi_n``e_1$, so $\mathrm{card}(e_1\cap e_2)\leq 1$; or

• vice versa (of last point), so $\mathrm{card}(e_1\cap e_2)\leq 1$.

For all $a,b\in\mathbb{N}$ with $a\neq b$, there is $e\in E$ with $a,b\in e$: If $a,b\in B_\omega$ then $e=B_\omega$ witnesses this; otherwise let $m$ be largest such that $a\in B_m$, and $n$ largest such that $b\in B_n$; we may assume $m\leq n$. Now just consider the full line through $\pi_m(a)$ and $\pi_m(b)$.

There are $e_1,e_2\in E$ with $e_1\cap e_2=\emptyset$; just consider some parallel lines.

And there $e_1,e_2\in E$ with $\mathrm{card}(e_1)\neq\mathrm{card}(e_2)$, since in fact we get infinite sets in $E$ and also $B_\omega$.