Set from powerset which gives nonempty intersection Have such question: Let us have $S=\\{1,2,\dots,n\\}$ and let there be powerset $P(S)$. Prove that there exists such subset $F\subseteq P(S)$ that $A\cap B\neq\emptyset$ for any $A\in F$ and $B\in F$ and $|F|=2^{n-1}$ I might be missing point here, but could one make argument that $S\in P(S)$, so if $S\subseteq F$ and $\emptyset \not\subseteq F$, we get ourself proper $F$, in which any two elements have nonempty intersection.

You're looking for a large family of subsets of $S$ whose members overlap. Notice that $2^{n-1}$ is a very nice number: it is the number of subsets of any set with $n-1$ elements. So, fix some $s$ in $S$, have a look at the collection $F_0 = \mathcal{P}(S\setminus\\{s\\})$, and throw $s$ back in to each of those sets: $F = \\{A \cup \\{s\\} : A \in F_0\\}$. That's the $F$ you want! Equivalently, let $F = \\{A \subseteq S : s \in S\\}$. Anyway, it is clear that $A \cap B \
e \varnothing$ for any two sets $A$, $B$ in $F$, because $A \cap B \supseteq \\{s\\}$.