Let $F$ be a finite family of non-empty sets such that any two of them intersect. Consider the set $F'$ consisting of all sets that are a subset of at least one element of $F$. Prove $|F'|\geq 2|F|$.


I conjectured this yesterday but have been completely unable to prove it. I tried doing things like inclusion-exclusion on the poset of the subsets but I was not able to solve it, I also tried things similar to Dilworth's theorem.


Since this problem is so natural I think that it should have been attempted before, does anyone have any references?



For an intersecting family of $m$ sets there are at least $2m$ sets that are contained in at least one of them
Assume that all the sets are contained in some master set of $n$ elements. Let $G$ be the set of super sets of at least one element of $F$. We apply the Harris-Kleitman inequality:


**Lemma** (Harris-Kleitman): *For $F'$ a downward-closed and $G$ an upward-closed collection of subsets on a set with $n$ elements,* $$|F'\cap G|\leq \frac{|F'| |G| }{2^n}.$$


*Proof*. Induction on $n$. Let $G\_0$ be the set of elements of $G$ that do not contain the first element, $G\_1$ the set of elements that contain the first element, and so on. Then $$
\begin{aligned}
|F'\cap G|&= |F'\_0 \cap G\_0|+|F'\_1 \cap G\_1| \\
&\leq \frac{ |F'\_0||G\_0|}{2^{n-1}}+ \frac {|F'\_1||G\_1|}{2^{n-1}}\\
&= \frac{ (|F'\_0|-|F'\_1|) (|G\_0|-|G\_1|)}{2^n}+ \frac{(|F'\_0|+|F'\_1|)(|G\_0|+|G\_1|)}{2^n} \\
&\leq 0+ \frac{|F'||G|}{2^n}
\end{aligned}
$$
QED


Now by the assumption on $F$, for any element of $G$, its complement is not in $G$, so $|G|\leq 2^{n-1}$, and thus


$$|F| \leq |F' \cap G| \leq \frac{|F'||G|}{2^n} \leq \frac{|F'|}{2}$$ as desired.


