Question about Intersection Graph in Graph Theory I am reading "Intersection Graphs" from chapter 2 in Frank Harary Graph Theory Book

Let $S$ be a set and $F=\{S_1,\ldots,S_p\}$ a nonempty family of distinct nonempty subsets of $S$ whose union is $S$. The intersection graph of $F$ is denoted $\Omega(F)$ and defined by $V(\Omega(F))=F$, with $S_i$ and $S_j$ adjacent whenever $i\neq j$ and $S_i\cap S_j\neq\emptyset$. Then a graph $G$ is an intersection graph on $S$ if there exists a family $F$ of subsets of $S$ for which $G$ and $\Omega(F)$ are isomorphic graphs.

Now, coming to my questions:

*

*What is the set S in this context? Since we are talking about graphs is it "$S = ( V, E )$", where $V$ and $E$ themselves are sets?


*Or is $S$ a set of all vertices and edges in a graph.
In either of the case, it does not make sense, cause the number of subsets, which I believe here mean a vertex in the resultant graph, would be more than the total vertex in the original graph and then the original and the resultant graph won't be isomorphic.
 A: In general, the set $S$ can be anything you like, and the set $F$ can be any family of subsets of $S$ you like. There's no question of the "original graph" and "resultant graph" being isomorphic, because you're not necessarily starting with any "original graph": you are constructing a graph $\Omega(F)$ out of the information in $F$.
For example, suppose I am a big fan of the set family $\{\{1,2,3\}, \{2,3,4\}, \{3,4,5\}, \{4,5,6\}\}$. Then I could create the intersection graph $\Omega(\{\{1,2,3\}, \{2,3,4\}, \{3,4,5\}, \{4,5,6\}\})$, which would look like this:

Further, we say that a graph $G$ is an intersection graph on $S$ if $G$ is isomorphic to $\Omega(F)$ for some $F$ that's a family of subsets of $S$. The diagram above shows that if you delete an edge from $K_4$ and call that $G$, then $G$ is an intersection graph on $\{1,2,3,4,5,6\}$.
When you're figuring out if we can represent a graph as an intersection graph, then we have a question of an "original graph" $G$ which we're comparing to $\Omega(F)$. But we're still not told what $S$ must be. We must pick $S$ and $F$ ourselves in some clever way that makes $\Omega(F)$ isomorphic to $G$.
For example, we can prove (and I think Harary does this) that every graph $G$ is an intersection graph on $E(G)$: the set of all edges in $G$. Specifically:

*

*Let $S = E(G)$;

*For every vertex $v \in V(G)$, let $S_v$ be the subset of all edges incident to $v$;

*Let $F = \{S_v : v \in V(G)\}$.

Then $\Omega(F)$ is isomorphic to $G$.
Some special families of graphs also have other representations as intersection graphs, with a different $S$ and $F$.
