# The size of independent set in Claw-free graph

A claw-free graph is a graph that does not have a claw as an induced subgraph or contains no induced subgraph isomorphic to $$K_{1,3}$$. Let $$\alpha(G)$$ denote the maximum size of an independent set in G. Let $$i(G)$$ denote the minimum size of an independent dominating set. I wonder if $$\alpha(G) \leq 2i(G)$$?

Intuitively, each vertex of $$G$$ can be dominated by up to two adjacent vertices. I'm not sure that's a rigorous analysis.

Let $$I^*$$ be a maximum independent set and $$I$$ be a maximal independent set (so also dominating). You just need to realize that any vertex of $$I$$ is adjacent (including the vertex itself in the neighborhood) to at most two vertices in $$I^*$$, and that each vertex of $$I^*$$ is adjacent to at least one vertex of $$I$$. So by a double counting argument, when counting once the vertices of $$I^*$$, we can choose one adjacent vertex in $$I$$, and we will have counted at most twice the vertices of $$I$$