Bound the size of the union of intersecting family Let $F \subset 2^{[n]}$, where $[n] = \{ 1, 2, \dots , n \}$. $F$ is an intersecting family if $\forall A,B \in F, A \cap B \ne \emptyset$. If $F_{i} \subset 2^{[n]} : i \in [t]$ are intersecting families, show that $\left| \cup_{i \in [t]} F_{i} \right| \leq 2^{n} - 2^{n-t}$. I was thinking   that this bound implies that if the union were greater, then a set and its complement must be contained in the union but I dont know how to proceed. 
 A: This is the main result in Daniel J. Kleitman, Families of Non-disjoint subsets, Journal of Combinatorial Theory $\mathbf{1}$ $(1966)$, $153$-$155$. The proof uses the following lemma.

Lemma. Let $\mathscr{U}\subseteq\wp([n])$ be closed under supersets and $\mathscr{L}\subseteq\wp([n])$ be closed under subsets; then $2^n|\mathscr{U}\cap\mathscr{L}|\le|\mathscr{U}||\mathscr{L}|$.

The answer to this question gives Kleitman’s proof of the lemma.
We also need the following observation.

Observation. Let $\mathscr{F}\subseteq\wp([n])$ be an intersecting family of cardinality less than $2^{n-1}$; then $\mathscr{F}$ is not maximal among intersecting families in $\wp([n])$.
Proof. If $|\mathscr{F}|<2^{n-1}$, there is an $A\subseteq[n]$ such that neither $A$ nor $[n]\setminus A$ belongs to $\mathscr{F}$. If $A$ meets every member of $\mathscr{F}$, let $\mathscr{F}'=\mathscr{F}\cup\{A\}$; clearly $\mathscr{F}'$ is an intersecting family strictly containing $\mathscr{F}$. Otherwise let $\mathscr{F}'=\mathscr{F}\cup\{[n]\setminus A\}$; there is some $F\in\mathscr{F}$ disjoint from $A$, so $F\subseteq[n]\setminus A$, and again $\mathscr{F}'$ is an intersecting family strictly containing $\mathscr{F}$. $\dashv$
An immediate corollary is that if $\mathscr{F}\subseteq\wp([n])$ is an intersecting family, then there is a maximal intersecting family $\mathscr{F}'\supseteq\mathscr{F}$ in $\wp([n])$, which is necessarily of cardinality $2^{n-1}$.

The actual theorem is then proved by induction on $t$. We know that it holds for $t=1$, since any family of subsets of $[n]$ with more than $2^n-2^{n-1}=2^{n-1}$ members must contain a pair of complementary sets and therefore cannot be an intersecting family. Now suppose that $t>1$, and the result holds for $t-1$. Let $\mathscr{F}_1,\ldots,\mathscr{F}_t\subseteq\wp([n])$ be intersecting families. For each $k\in[t]$ there is a maximal intersecting family $\mathscr{F}_k'\subseteq\wp([n])$ such that $\mathscr{F}_k\subseteq\mathscr{F}_k'$.
Let $\mathscr{U}=\bigcup_{k=1}^{t-1}\mathscr{F}_k'$; by the induction hypothesis $|\mathscr{U}|\le 2^n-2^{n-(t-1)}$. Each $\mathscr{F}_k'$, being maximal, must be closed under supersets, so $\mathscr{U}$ is closed under supersets. Let $\mathscr{L}=\wp([n])\setminus\mathscr{F}_t'$; clearly $|\mathscr{L}|=2^n-|\mathscr{F}_t'|=2^n-2^{n-1}=2^{n-1}$. Moreover, $\mathscr{F}_t'$ is closed under supersets, so $\mathscr{L}$ is closed under subsets. The lemma therefore implies that
$$\begin{align*}
\left|\bigcup_{k=1}^t\mathscr{F}_k\right|&\le|\mathscr{U}\cup\mathscr{F}_t'|\\
&=|\mathscr{U}\cap\mathscr{L}|+|\mathscr{F}_t'|\\
&\le 2^{-n}|\mathscr{U}||\mathscr{L}|+|\mathscr{F}_t'|\\
&\le 2^{-n}\left(2^n-2^{n-(t-1)}\right)\cdot 2^{n-1}+2^{n-1}\\
&=\left(2^{n-1}-2^{n-t}\right)+2^{n-1}\\
&=2^n-2^{n-t}\,,
\end{align*}$$
and the induction step is complete.
