Solution
= Solution
List $\sigma(Y)\setminus Y=\{x_1,\ldots,x_m\}$ and put $Y_j=Y\cup\{x_1,\ldots,x_j\}$. Each $x_j\in\sigma(Y)$ also belongs to $\sigma(Y_{j-1})$ by (iii). Thus the definition of the <closure of a matroid> gives $r(Y_j)=r(Y_{j-1})$ at every step. Iteration yields
$$
\boxed{r(\sigma(Y))=r(Y_m)=r(Y).}
$$
In particular, collecting all elements that individually fail to increase the <rank of a matroid> does not create a collective increase.