When can any graph $G$ be expressed as a union of $\alpha(G)$ complete graphs? If for any graph $H$ we define $\alpha(H)$ to be the cardinality of any maximum size indepedent set in $H$. Then under what conditions can any graph $G$ be expressed as a union of $\alpha(G)$ complete graphs?
Or equivalently if $[S]_2$ is the set of all two element subsets of $S$, then for which class of graphs $G$ will there exist $\alpha(G)$ sets $V_{1},V_2,\ldots V_{\alpha(G)}$ that satisfy $V(G)=\bigcup_{n=1}^{\alpha(G)}V_n$ and $E(G)=\bigcup_{n=1}^{\alpha(G)}[V_n]_2$?
Are there any reasonably simple necessary and sufficient conditions for categorizing these graphs?


Lastly also note if there are $n$ complete graphs $C_1,C_2,\ldots C_n$ with $G=\bigcup_{k=1}^n C_k$ and $n<\alpha(G)$ there must exist an independent set $I$ of $G$ such that $|I|=n+1$ however by assumption we have that $I\subseteq V(G)=\bigcup_{k=1}^nV(C_k)$ therefore by the pigeonhole principal there must exists two distinct vertices $u,v\in V(C_j)$ for some integer $1\leq j\leq n$ yet since $V(C_j)$ is a clique of $G$ this means we have $\{u,v\}\in E(G)$ contradicting the fact $I$ is an independent set of $G$, which proves $n\geq \alpha(G)$. Thus a a corollary if we let $\theta(G)$ be the intersection number of $G$ then by definition we have that:
$$G\text{ is a union of }\alpha(G)\text{ complete graphs}\iff \alpha(G)=\theta(G)$$
However I'm unable to simplify these equivalent conditions into anything nice either.
 A: I don't think there is a pretty answer to this question.  A graph is in this class iff it is the union of a set of cliques such that each of the cliques has a vertex not in any of the other cliques.
Given a maximum independent set $S$, you can identify said cliques as the closed neighbourhoods of the vertices in $S$. Conversely given such a set of cliques, the special vertices are a maximum independent set.
A: I know I already accepted an answer nine months ago, though these came up again in a problem I was working on involving digraph Kernels and upon further investigation these are actually called "bound graphs" as a matter of fact we have the following chain of equivalences:
$$G\text{ is a bound graph}\iff\alpha(G)=\theta(G)\iff \vartheta(G)=\theta(G)\\\iff\text{For all spanning subgraphs }H\text{ of }G\text{ we have }\theta(H)\geq\theta(G)\\\iff \text{For all subgraphs }H\text{ of }G\text{ we have }\theta(H)\geq \theta(G[V(H)])$$
Thus the property of being a "bound graph" is hereditary which means for any bound graph $G$ we see every induced subset $H$ of $G$ is a bound graph and therefore note for every induced subset $H$ of $G$ we have the identity $\alpha(H)=\theta(H)=\vartheta(H)$ though this means every induced subgraph $H$ of $G^{\complement}$ satisfies $\chi(H)=\omega(H)$ thus $G^{\complement}$ must be a perfect graph which by the Perfect graph theorem this means $G$ is a perfect graph. Thus this proves bound graphs are a particular class of perfect graphs.
