Here is the question I am trying to solve (Matroid Theory, Second edition, Chapter 1, section 4 ):
Let $M$ be a matroid, and $r$ and $cl$ be its rank function and its closure operator. Prove the following:
The intersection of all of the flats containing $X$ equals $cl(X).$
I found this question here Intersection of flats containing X but I am not very comfortable with this way of solving it.
Here are thoughts about a way of solving it:
Let $G(X)$ be the intersection of all the flats containing $X.$ We need to show that $G(X) = cl(X).$
$First.$
Let $x \in X.$ Since we know that $X \subseteq cl(X)$ and since, by the given definition of $G(X),$ it is the intersection of all of the flats containing $X,$ then for any $x \in X,$ if $x \in cl(X),$ then $x \in G(X).$ So we proved that $ cl(X) \subseteq G(X).$
$Second.$
Let $x \in G(X),$ we want show that $x \in cl(X).$ We know that by definition of a flat of a matroid $M,$ we know that any flat of a matroid $M$ is a subset $X$ of $E(M)$ for which $cl(X) = X$ and we also know that $G(X)$ is the intersection of all the flats containing $X.$
But I do not know how can I conclude the required. I was also thinking about taking $x \in G(X)$ but it is not in $cl(X)$ and arriving to a contradiction, but I failed also to reach a solution.
Could anyone clarify the correct solution to me please?
In first you show that $X\subseteq G(X)$, which is not enough as $cl(X)$ contains more elements unless $X$ is a flat itself (so $G(X)=X$). To finish this implication, just notice that if $X\subseteq F$, then $cl(X)\subseteq cl(F)=F$ if $F$ is a flat.Hence $cl(X)$ is inside every flat that contains $X$ and so its in the intersection.
In second just, as you kind of write, $cl(X)$ is a flat (because $cl(cl(X))=cl(X)$) containing $X$ and so $G(X)\subseteq cl(X)$ because $G(X)$ is the intersection of all such flats.