Let be a finite matroid with rank function . For all ,
Submodularity
Fix . We construct one independent set in whose intersection with already has the largest possible size.
Fill the intersection, then extend to the union
Choose an independent set with . Extend by adding elements of while preserving independence, until no further element can be added. Call the resulting set .
By exchangeability, a maximal independent subset of has maximum size: a larger independent subset would supply another element to add. Hence
Why the intersection stays exactly
Suppose extending to added some . Since , we would have
By heredity, deleting elements from the independent set leaves an independent set. In particular, delete everything except and :
Schematic: and are outside . The orange region contains the retained elements , not necessarily all of . Deleting exposes an independent subset that is too large; the same argument works for any size of .
The remaining set would therefore satisfy all three conditions:
But an independent subset of has size at most . This would force
a contradiction. Adding elements outside cannot repair the dependent subset . Thus extending to adds no further element of :
Bound the two portions
Both and are independent, since they are subsets of . By the definition of rank,
Count the overlap twice
The two portions and cover and overlap exactly in . Adding their sizes counts every element of once, and every element of once more:
Shown for and : each column is one element, and each dot is one occurrence in the sum. The same count gives, for arbitrary and ,
This is the required inequality.
Function
Definition
Rank Function (Matroid)
The notion of ranks extends to arbitrary subsets. The rank function is