A base of a matroid contains its independent subset and is disjoint of an independent subset of its base.

Question

Let $A$ and $B$ be two disjoint subsets of the matroid $M$. Let $A$ be independent in $M$, and $B$ be independent in $M$'s dual.

I would like to ask for help in proving that $M$ has a base which contains $A$ and is disjoint from $B$.

Thank you!

Answer 1

By definition, for $B$ to be independant in $M^*$, the set $M-B$ must contain a basis of $M$. Select such a basis $S$. Assuming that $A$ is not already a basis of $M$, we have that $r(S)>r(A)$, so there is some element of $s\in S$ such that $A\cup \{s\}$ is independant. Repeating this procedure, one may choose $S_0\subseteq S$, such that $A\cup S_0$ is a base. As both $A$ and $S_0$ are disjoint from $B$, $A\cup S_0$ is as well.