Q:

For any finite simple graph $G$, can we always find an intersection graph of some set such that it is isomorphic to $G$?

Let $G=(V,E)$ be a finite simple graph. Does there exist a finite set
  $S$ and $S_G\subseteq P(S)$ such that the intersection graph $H$ of
  $S_G$ is isomorphic to $G$, i.e. $H\cong G$.

Here's my idea:
If I can find such an intersection graph $H$ that $H\cong G$, then the proposition is proved.
Let $V=\{v_1,\dots,v_n\}$ and $E=\{e_1,\dots,e_m\}$, where $n,m$ are both positive integers since $G$ is finite. Now, let $S=V$, and $S_G=E\cup \{\{v_i\}\}$, where $v_i$ satisfies the condition that $v_i\neq e_j\cap e_k$ for all $j,k\le m$. That is, $v_i$ is a vertex with degree 1. We define the set of edges $F$ by
$$f=\begin{cases}
\{v_i,e_i\}&\text{ if }v_i\in S_G\ \text{and}\ v_i\text{ is incident with }e_i\\
\{e_j,e_k\}&\text{ for all }e_j,e_k\in E\text{ with }j\neq k
\end{cases}$$
Then $H=(S_G,F)\cong G$.
I find my idea works well for closed graph, but not so well for others. There must be some problems in my idea. Anyone has any idea or can modify my solution?

A:

I don't understand your solution. Here's how I'd do it. Let $G=(V,E)$ be a simple graph, finite or infinite. Let $S=V\cup E.$ (Of course, if $G$ is finite, then $S$ is finite.) For each vertex $v\in V$ let $f(v)=\{v\}\cup\{e\in E:e\text{ is incident with }v\}\in P(S)$ and let $S_G=\{f(v):v\in V\}\subseteq P(S).$ Then $f:V\to S_G$ is bijective and, for distinct vertices $v_1,v_2\in V,$ we have $v_1$ adjacent to $v_2$ if and only $f(v_1)\cap f(v_2)\ne\emptyset.$ That is, $f$ is an isomorphism from the graph $G$ to the intersection graph of $S_G.$

