Suppose there is a Matroids \ (M = (S, the I) \) , its maximum independent set of rights \ (the U-\) . Now to \ (S \) was added an element \ (X \) , and \ (the I \) was added Involving the \ (X \) set to give Matroid \ (M '= (S' , I ' ) \) . Then \ (M '\) a maximum independent set \ (U' \) is either \ (the U-\ Cup X \) , either added \ (X \) removing a minimum weight element after satisfying remove this element after What remains is an independent set.
Proof: If \ (U \ cup x \) is an independent set of clearly established.
Otherwise, join \ (x \) after deletes one \ (U \) elements \ (the y-\) . Suppose because deleting \ (Y \) Effect of an otherwise not \ (the U-\) of \ (Z \) joined \ (the U-'\) , provided the original \ (the U-\) in weight \ (\ ge val_z \) set of elements constituting removed \ (x, y, z \ ) outside is \ (Q \) . Then \ (Q \ cup x \ cup z \) is an independent set, \ (Q \ cup Y \) is independent sets, according to the switching nature of the \ (x, z \) there is at least one element may be added \ (Q \ cup the y-\) . If this element is \ (the X-\) , so \ (Q \ cup x \ cup y \) is an independent set, obviously not delete \ (the y-\) , a contradiction. Otherwise the element is \ (Z \) , described \ (Q \ cup y \ cup z \) is an independent set, the process according to the greedy algorithm, which is\ (z \) is not \ (U \) in contradictions.
With this property may prove something, such as joint bipartite graph can prove Matroid yesterday T3 conclusions.