Proving that the intersection of two closed sets is closed in a matroid

Question

I am stuck on a little homework problem I have. Here, $M$ is a matroid with rank function $R$. I am given this definition: In a matroid $M$, a set $A$ is closed if $R(A \cup e) > R(A)$ for all $e \in E \setminus A$. I want to show that the intersection of two closed sets is again closed. My approach thus far is to take $((A \cap B ) \cup e))$ and stick it into the semimodular law for the rank function, aiming to show that $R((A \cap B ) \cup e)) > R(A \cap B)$ for $e \notin A \cap B$, but nothing of value seems to come out. What am I missing? I know I will need to use the fact that $A$ and $B$ are closed, but I am not sure how to work that in.

Answer 1

Notation: $A+f$ means $A\cup \{ f\}$ for a set $A$ and element $f$.

Proof with submodularity

Recall, that submodularity for rank function $R$ means that for any $X, Y \subset E$ $$R(X)+R(Y) \geq R(X\cup Y)+R(X\cap Y)$$ Let $f\in E\setminus A$ be arbitrary, $X:=A$, $Y:=A\cap B+f$. $$R(A)+R(A\cap B+f) \geq R(A+f) + R(A\cap B)>R(A)+R(A\cap B)$$ $$R(A\cap B+f)>R(A\cap B)$$ The same is true for every $f\in E\setminus B$, except with $X:=B$. Thus the equation holds for every $f\in E\setminus (A\cap B)$, which was the statement.

$$\tag*{$\blacksquare$}$$

Proof without submodularity

Define $AB_{max} \subset A_{max} \subset A$ such, that $AB_{max}$ and $A_{max}$ are maximal independent subsets in $A \cap B$, and $A$ respectively. (recall, that every subset of an "$A_{max}$" is an "$AB_{max}$", and every "$AB_{max}$" can be expanded to become an "$A_{max}$")

If $A \cap B$ is NOT closed, there exists an $f$ element outside of $A \cap B$ with which $AB_{max} + f$ is dependent, for an arbitrary choice of $AB_{max}$.
(this is equivalent to $R(A\cap B) = R(A\cap B+f)$, the proposition to be contradicted)

Where could such an $f$ be?

If it's NOT in $A$, then for arbitrary $A_{max}$, $A_{max}+f$ would be dependent as well, since it contains an "$(AB_{max}+f)$", which is dependent by the definition of $f$. This contradicts the closeness of $A$. (it's the equivalent of $R(A)=R(A+f)$)

Similarly if it's NOT in $B$, then it contradicts the closeness of $B$.

Such $f$ therefore cannot exist and thus $A\cap B$ is a closed set.

$$\tag*{$\blacksquare$}$$