Let $[n]=\{1,\dots,n\}$, and consider sets $A_1,\dots,A_n$. For each non-empty $S\subseteq[n]$ let
$U_S=\bigcap_{k\in S}A_k\setminus\bigcup_{k\in[n]\setminus S}A_k\;;$
$U_S$ is the set of $x$ that belong to $A_k$ if and only if $k\in S$.
To get a clearer picture of this, consider first the case $n=2$. Then $U_{\{1\}}=A_1\setminus A_2$, $U_{\{2\}}=A_2\setminus A_1$, and $A_{\{1,2\}}=A_1\cap A_2$. These are the three regions of a Venn diagram for two sets that are inside at least one of the two sets. A Venn diagram for three sets has eight regions, and you should check that the seven that are in at least one of the three sets are precisely the seven sets $U_S$ such that $\varnothing\ne S\subseteq[3]$.
The relationships amongst the sets $A_1,\dots,A_n$ are completely determined by which of the sets $U_S$ are empty and which are not. Some of these are pretty uninteresting: for instance, if $U_S=\varnothing$ for every non-empty $S\subseteq[n]$, then $A_1=\ldots=A_n=\varnothing$. However, if you’re willing to accept that the emptiness or not of each of the sets $A_k$ is part of their relationship, then we can count the possibilities quite easily: $[n]$ has $2^n-1$ non-empty subsets $S$, so $\{U_S:\varnothing\ne S\subseteq[n]\}$ is a family of $2^n-1$ sets. This family has $2^{2^n-1}$ subfamilies, so there are $2^{2^n-1}$ different ways to choose which of the sets $U_S$ are to be empty. In this sense, then, there are $2^{2^n-1}$ possible relationships amongst $n$ sets.