# co.combinatorics – Does the non-extreme point operator of a closure space being idempotent, imply said space is a matroid (in flat lattice form)?

Given any set $$X$$ and some closure operator $$text{cl}:2^Xto 2^X$$ on $$X$$, we define $$psi:2^Xto 2^X$$ so that $$psi(Q)={qin Q:qintext{cl}(Qsetminus{q})}$$ for all $$Qsubseteq X$$, now if $$forall Ssubseteq X(psi(psi(S))=psi(S))$$ (i.e. if $$psi$$ is idempotent) must $$text{cl}$$ be the closure operator of some matroid?

Sorry if this is either trivially true or trivially false as I am still new to matroid theory.

I know that if $$text{cl}$$ is the closure operator of a matroid then $$psi$$ must be idempotent so the converse statement holds. Further I have been able to reduce the problem to proving $$mathscr{C}={Csubseteq X:psi(C)neq emptysetlandbigcup_{Ssubset S}psi(C)=emptyset}$$ is such that for all $$C_1,C_2in mathscr{C}$$ we have $$ein C_1cap C_2implies exists C_3in mathscr{C}:C_3subseteq (C_1cup C_2)setminus {e}$$ i.e. that the family of sets $$mathscr{C}$$ satisfies the matroid circuit axioms. This is because if $$text{cl}$$ is the closure operator of a matroid $$M$$ grounded on $$X$$, then $$psi$$ maps every $$Ssubseteq X$$ to the maximal cyclic set of $$M$$ in $$S$$ or in other words $$psi(S)$$ is the union of all circuits of $$M$$ contained in $$S$$, so in particular the closure operator $$text{cl}^*:2^Xto 2^X$$ of the dual matroid $$M^*$$ of $$M$$ satisfies $$text{cl}^*(Q)=Xsetminuspsi(Xsetminus Q)$$ for all $$Qsubseteq X$$.

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