# graph theory /combinatorics committee existence

I'm having trouble figuring out the problem below. I've laid out my approach and it seems combinatorics formulas might help solve this. If anyone can point to me to the right direction i would greatly appreciate it. Not looking for a direct answer, just a path I can follow.

Thanks.

PROVE that In any group of $n \ge 1$ people, there exists a committee with the following two properties:

(a) No $2$ members of the committee are friends and

(b) Every person not included in the committee is a friend of at least $1$ member of the committee.

$n = 10$

$c$ – Committee size

$n-c$ -> every person not included in the committee

each committee member has $c-1$ enemies

$n-c$ people have at least $1$ friend who is a member of $c$

Form a graph $G$ with vertex set $V$ by letting the people be the vertices of the graph and connecting two people by an edge iff they are friends. Let $C$ be a maximal independent set of vertices: that is, no two vertices of $C$ are joined by an edge, and if $C\subsetneqq C'\subseteq V$, then $C'$ is not independent. Now use the maximality of $C$ to show that it satisfies condition (b) as well as condition (a).
• @JavaBeans: Apart from the very small missing last step, what I gave qualifies as a formal proof at the usual level of formality. Are you worried about the existence of a maximal independent set? If $v\in V$, $\{v\}$ is trivially independent, so there is at least one independent set of vertices. $V$ is finite, so among all independent sets of vertices there is at least one with the largest possible cardinality; pick one such and call it $C$. If $v\in V\setminus C$, $C\cup\{v\}$ is bigger than $C$ and therefore is not independent, so $v$ must be joined by an edge to a vertex in $C$. – Brian M. Scott Dec 8 '12 at 22:53
• @JavaBeans: There is only one vertex in the set $\{v\}$, and the graph has no loops $-$ we don’t consider anyone a friend of himself $-$ so the set is necessarily independent. $A\subseteq V$ is independent if there are no edges between members of $A$; this says nothing about edges between vertices in $A$ and vertices outside of $A$. – Brian M. Scott Dec 9 '12 at 1:12