Prove that a collection of pairwise non-disjoint subsets of a set $M$ with $|M| = n$ can be extended to $2^{n-1}$ pairwise non-disjoint subsets 
Let $M$ be a set of size $n$. Prove that any collection of pairwise non-disjoint subsets of $M$ can be extended to a collection of $2^{n-1}$ pairwise non-disjoint subsets of $M$.

I've tried using Ray-Chaudhuri-Wilson as well as induction and I was able to solve it for the case the union of all subsets is not equal to $M$ and the case there are elements $a,b \in M$ with $a \neq b$ and $a \in C \Leftrightarrow b \in C$ for any subset $C$, but I wasn't able to prove the general case. How does one go on about that?
 A: Suppose we are given such a collection $C$, maximal for the inclusion relation. One can show that for all $a \subset M$, either $a$ or $\overline a \in C$ (and thus $|C| \geq 2^{n-1}$).
Indeed, suppose $a \notin C$ and $\overline a \notin C$. By maximality there exist $b, c \in C$ with $a \cap b = \varnothing$ and $\overline a \cap c = \varnothing$. That is, $b \subset \overline a$ and $c \subset a$. But then $b \cap c = \varnothing$, a contradiction.
A: Let $\{A_1,\cdots,A_p\}$ a collection of (distinct and) pairwise non-disjoint subsets of a set $X$ with $\vert X\vert=n$.
Let's denote by $\overline Y$ the complement of any subset $Y$ of $X$.
It is clear that $\{A_1,\cdots,A_p\}\cap\{\overline{A_1},\cdots,\overline{A_p}\}=\emptyset$, hence :
$$2p=\vert\{A_1,\cdots,A_p\}\cup\{\overline{A_1},\cdots,\overline{A_p}\}\vert\leqslant2^n$$and so we conclude that $p\leqslant2^{n-1}$.
It remains to prove that $p=2^{n-1}$ can be achieved.
Pick any element $x\in X$ et consider the subsets $\{x\}\cup Y$ where $Y$ is any subset of $X-\{x\}$ : there are $2^{n-1}$ such subsets and they are pairwise non-disjoint (the intersection of any two of them must contain $\{x\}$).
EDIT : I realize that I didn't answer the question... Sorry. Anyway, this could be helpful ... so I prefer not to remove that "answer".
