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The closure of a matroid adds all elements that fail to increase its rank of a matroid. It abstracts the linear span in a linear matroid. It is extensive, preserves inclusion, preserves the rank of a matroid, and is an idempotent function:Proof. Extensivity follows from $A+x=A$ when $x\in A$. If $A\subseteq B$ and $x\notin B$, submodularity of matroid rank gives $r(B+x)-r(B)\le r(A+x)-r(A)$. Thus an element in ${\mathrm{cl}}_M(A)$ also lies in ${\mathrm{cl}}_M(B)$; the case $x\in B$ is automatic. Add the finitely many elements of ${\mathrm{cl}}_M(A)\setminus A$ one at a time. Inclusion preservation makes each addition have zero marginal increment of the rank of a matroid, proving rank preservation.

For completeness, nested subsets $A\subseteq B$ of equal rank of a matroid have the same closure of a matroid. If $x\notin B$, apply submodularity of matroid rank to $B$ and $A+x$ to obtain $r(B+x)\le r(A+x)$; monotonicity gives the reverse bound. The defining tests for membership in their closures are therefore equal. If $x\in B$, $r(A)\le r(A+x)\le r(B)=r(A)$, so both closures contain $x$. Apply this equal-rank result to $A\subseteq {\mathrm{cl}}_M(A)$ to prove idempotence. In particular, every maximal matroid independent set of $B$ has the same closure of a matroid as $B$.

We first justify the submodularity of matroid rank, which will also support later arguments. Choose a maximal matroid independent set $I_0$ in $A\cap B$, extend it to a maximal matroid independent set $I_A$ in $A$, and then to one $I_U$ in $A\cup B$. These extensions exist by successive additions, and their sizes are the respective values of the rank of a matroid. No element of $I_U\setminus I_A$ lies in $A$: such an element could have been added to $I_A$, contradicting maximality in $A$. Therefore $I_0\cup (I_U\setminus I_A)$ is an independent subset of $B$, givingNow let $Y\subseteq Z$ and $x\in \sigma (Y)$. If $x\in Z$, then $x\in \sigma (Z)$ by (ii). Otherwise apply submodularity of matroid rank to $Y+x$ and $Z$, whose intersection is $Y$:The monotonicity of the rank of a matroid supplies the opposite bound $r(Z+x)\ge r(Z)$, so equality holds. Thus $x\in \sigma (Z)$, proving $\sigma (Y)\subseteq \sigma (Z)$.