If $X \subseteq Y$ and $r(X) = r(Y),$ then $cl(X) = cl(Y).$

Question

Here is the question I am trying to understand its solution:

Let $M$ be a matroid, and $r$ and $cl$ be its rank function and its closure operator. Prove the following:

$(e)$ If $X \subseteq Y$ and $r(X) = r(Y),$ then $cl(X) = cl(Y)$.

Here is my attempt:

If $X \subseteq Y,$ then $cl(X) \subseteq cl(Y).$ Now, it remains to show that $cl(Y) \subseteq cl(X).$ Since $r(X) = r(Y).$ Then $r(cl(X)) = r(cl(Y)).$ And then $r(cl(Y)) \leq r(cl(X))$ but does this implies that $cl(Y) \subseteq cl(X)$? could anyone clarify this to me in this please?

Answer 1

It doesn't from there, but from what you have done before you have that if $y\in cl(Y)$, then $r(Y\cup \{y\})=r(Y)=r(X)$ and so $r(X\cup \{y\})\leq r(Y\cup \{y\})=r(X)$ so $r(X\cup \{y\})=r(X)$ which implies that $y\in cl(X).$