$ω(G − S) = ω(G)$ Proposition: Let $S$ be a stable set in a minimally imperfect graph $G$. Then $ω(G − S) = ω(G)$.
Proof: 
We have the following string of inequalities.
$ω(G − S) ≤ ω(G) ≤ χ(G) − 1 ≤ χ(G − S) = ω(G − S)$
Can someone help and say, why this inequalities in the proof hold?
 A: Recall that for any graph $H$, we have $\omega(H) \le \chi(H)$, since coloring a clique of size $\omega(H)$ already requires $\omega(H)$ different colors. The graph $H$ is perfect if:


*

*equality holds: $\omega(H)= \chi(H)$, and

*moreover, equality holds for every induced subgraph of $H$.


If the graph $G$ is minimally imperfect, then all proper induced subgraphs $H$ of $G$ are perfect and satisfy $\omega(H) = \chi(H)$. If we had $\omega(G) = \chi(G)$, then $G$ would be perfect, so we must have $\omega(G) < \chi(G)$. Therefore $\omega(G) \le \chi(G)-1$, and we have proven one of the inequalities (the hardest).
The other inequalities are relatively easy to show.


*

*$\omega(G-S) \le \omega(G)$ because removing vertices can't help there be a larger clique.

*$\omega(G) \le \chi(G)-1$ as shown above.

*$\chi(G)-1 \le \chi(G-S)$ because if we color $G-S$ with $k$ colors, we can color $G$ with $k+1$ colors by giving all of $S$ a color not used for $G-S$.

*$\chi(G-S) = \omega(G-S)$ because $G-S$ is a proper induced subgraph of $G$, and therefore perfect.

