Closure function of a matroid

Question

I need some help to understand:

If $M$ is matroid and e is an element in that matroid, what is the closure function of $M\setminus e$?

And what is the closure function of $ M/e$?

Any help would be much appreciated!

Answer 1

$\newcommand{\cl}{\operatorname{cl}}$Let $r$ be the rank function on $M$, and let $E$ be the underlying set of $M$. The rank function $r_{M\setminus e}$ on $M\setminus e$ is just the restriction of $r$ to $\wp(E\setminus\{e\})$, so for every $S\subseteq E\setminus\{e\}$, $r_{M\setminus e}(S)=r(S)$. The closure function on $M\setminus e$ is given by

$$\cl_{M\setminus e}S=\{x\in E\setminus\{e\}:r_{M\setminus e}(S)=r_{M\setminus e}(S\cup\{x\})\}$$

for $S\subseteq E\setminus\{e\}$. But if $S\subseteq E\setminus\{e\}$ and $x\in E\setminus\{e\}$, then $r_{M\setminus e}(S\cup\{x\})=r(S\cup\{x\})$ and $r_{M\setminus e}(S)=r(S)$, so

$$\cl_{M\setminus e}S=\{x\in E\setminus\{e\}:r(S)=r(S\cup\{x\})\}=\cl S\;,$$

where $\cl$ is the closure function in $M$.

The rank function $r_{M/e}$ on $M/e$ is defined by

$$r_{M/e}(S)=r(S\cup\{e\})-r(\{e\})$$

for $S\subseteq E\setminus\{e\}$, and of course

$$\cl_{M/e}S=\{x\in E\setminus\{e\}:r_{M/e}(S)=r_{M/e}(S\cup\{x\})\}\;.$$

Thus, if $S\subseteq E\setminus\{e\}$ and $x\in E\setminus\{e\}$, then $x\in\cl_{M/e}S$ if and only if

$$R(S\cup\{e\})-r(\{e\})=r(S\cup\{x,e\})-r(\{e\})$$

and hence if and only if

$$R(S\cup\{e\})=r(S\cup\{x,e\})\;.\tag{1}$$

But $(1)$ is exactly the condition that puts $x\in\cl(S\cup\{e\})$, so

$$\cl_{M/e}S=\big(\cl(S\cup\{e\})\big)\setminus\{e\}\;.$$