Given a set $ X$ and a closure operator $ \text{cl}:2^X\to 2^X$ on $ X$ , if we define $ \psi:2^X\to 2^X$ s.t. $ \psi(Q)=\{q\in Q:q\in\text{cl}(Q\setminus\{q\})\}$ then is it true that $ \forall S\subseteq X(\psi(\psi(S))=\psi(S))$ (i.e. that $ \psi$ is idempotent) implies $ \text{cl}$ is the closure operator of some matroid? I know if $Read more