Proof fact 2|M|≤|V(G)| in a graph G Definition of Independent set I of G is a subset of V(G) such that it induces no edges  in G. A matching M of G is a subset of E(G) such that no two distinct edges in M have the same endpoints.
The fact is "Any matching M in graph G satisfies 2|M|≤|V(G)|"
How to prove the fact ?
 A: An edge in a matching saturates its two endpoints.  A vertex is saturated by a matched edge if it is one of that edge's endpoints.
Each vertex can be saturated by at most one edge in a matching and each edge in a matching saturates exactly two vertices (directly from the definition of a matching and the definition of an edge).
Trivially, the set of vertices that are saturated by the matching is a subset of the whole vertex set.  Furthermore, since each edge in the matching contributes at most (in fact exactly) two vertices to the set of saturated vertices, and those vertices by definition of a matching cannot be saturated by any other matched edge, we see that:
$2|M| \leq |\{v~:~v~\text{is saturated}\}|\leq |V(G)|$
Here I used the fact that if $A\subseteq B$ then $|A|\leq |B|$.

Seen from a different perspective, if $|M|>\frac{|V|}{2}$, you would notice that by the pigeonhole principle there must be at least one vertex that is matched by more than one edge, contradicting the definition of a matching.  (no two matched edges should share a vertex in common)
