Here is the question I am trying to prove:
Let $M$ be a matroid, and $r$ and $cl$ be its rank function and its closure operator. Prove the following:
$(d)$ $r(X \cup Y) = r(X \cup cl(Y)) = r(cl(X) \cup cl(Y)) = r(cl(X \cup Y))$
Here is my attempt:
Third, proving that $r(cl(X) \cup cl(Y)) = r(cl(X \cup Y)).$
I know that $cl(X) \subseteq cl(X \cup Y)$ and $cl(Y) \subseteq cl(X \cup Y),$ then $cl(X) \cup cl(Y) \subseteq cl( X \cup Y) $ and hence by the properties of the rank function we have $r(cl(X) \cup cl(Y)) \leq r(cl(X \cup Y)) $. Now, for proving the other direction I know that $cl(X \cup Y) \subseteq cl(cl(X) \cup cl(_Y)).$ But then I know that $r(X) = r(cl(X))$ but I do not know how to complete. Any help will be appreciated.