equal unions and intersections Let $N$ be a $n$-element set and $k\ge n+2$.
The sets $P_1,\dots,P_k$ are nonempty and their union is $N$.
Then there exists disjoint nonempty sets $I,J\subset\{1,\dots,k\}$ such that $\bigcup_{i\in I}P_i=\bigcup_{j\in J}P_j$ and $\bigcap_{i\in I}P_i=\bigcap_{j\in J}P_j$.
How to prove it? Or maybe this is a special case of some theorem?
Here I found a similar, but weaker problem:
Proving the existence of disjoint subsets
 A: I will build upon the answer given to the weaker problem and use the same notations for clarity - except for the subsets which are $P_i$ here instead of $A_i$.
Here the number of subsets is $n+2$, which means the incidence matrix $M$ has a kernel of dimension at least $2$ when regarded as a linear map $\mathbb{R}^{n+2}\rightarrow \mathbb{R}^n$.
The subspace $V=\left\lbrace \vec c \in \mathbb{R}^{n+2} \ |\ c_1+\ldots + c_{n+2}=0\right\rbrace$ has codimension $1$ and hence must intersect $\operatorname{Ker}M$ along a subspace of dimension at least $1$. 
Take a nonzero vector $\vec c \in V\cap \operatorname{Ker}M$.
Define again $I=\left\lbrace i \ | \ c_i>0\right\rbrace$ and $J=\left\lbrace j \ | \ c_j<0\right\rbrace$. The conditions $I\cap J=\emptyset$ and $\cup_{i\in I}P_i = \cup_{j\in J}P_j$ are for free, according to Tad's answer. 
Now let $M_i$ denote the $i$-th column of $M$.
$\cap_{i\in I}P_i$ is the set of all indices $k$ such that the $k$-th coordinate of $\sum_{i\in I}c_i M_i$ is $\sum_{i\in I}c_i.$ Similarly $\cap_{j\in J}P_j$ is the set of all indices $k$ such that the $k$-th coordinate of $\sum_{j\in J}c_j M_j$ is $\sum_{j\in J}c_j.$
Since $\vec c $ was chosen in $V$, the intersections must agree as required.
