Suppose that $L\subseteq {\cal P}(\omega)$ has the following properties:

1. for $e\in L$ we have $|e|\geq 2$;
2. if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$;
3. if $m,n\in \omega$ there is $e\in L$ such that $m,n\in e$.

It is not hard to see that $L$ is countable. Is there a bijection $p:\omega\to L$ such that for all $n\in\omega$ we have $p(n)\cap p(n+1) \neq \emptyset$?

Suppose that $L\subseteq {\cal P}(\omega)$ has the following properties:

1. $\omega \notin L$, and for $e\in L$ we have $|e|\geq 2$;
2. if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$;
3. if $m,n\in \omega$ there is $e\in L$ such that $\{m,n\}\subseteq e$.

It is not hard to see that $L$ is countable. Is there a bijection $p:\omega\to L$ such that for all $n\in\omega$ we have $p(n)\cap p(n+1) \neq \emptyset$?

Suppose that $L\subseteq {\cal P}(\omega)$ has the following properties:

1. for $e\in L$ we have $|e|\geq 2$;
2. if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$;
3. if $m,n\in \omega$ there is $e\in L$ such that $x,y\in L$.

It is not hard to see that $L$ is countable. Is there a bijection $p:\omega\to L$ such that for all $n\in\omega$ we have $p(n)\cap p(n+1) \neq \emptyset$?

Suppose that $L\subseteq {\cal P}(\omega)$ has the following properties:

1. for $e\in L$ we have $|e|\geq 2$;
2. if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$;
3. if $m,n\in \omega$ there is $e\in L$ such that $m,n\in e$.

It is not hard to see that $L$ is countable. Is there a bijection $p:\omega\to L$ such that for all $n\in\omega$ we have $p(n)\cap p(n+1) \neq \emptyset$?