Isomorphism of a closure system and a topped complete lattice

Question

It is in general true that if a closure system $\mathcal{F}$ on a given ground set $\Omega$ is order isomorphic to a complete lattice $\mathcal{G}$ on $P(\Omega)$ having $P(\Omega)$ as its top element, then $\mathcal{G}$ could fail to be a closure system? As an example, I took $\Omega=\{a,b,c,d\}$ and consider the set systems $\mathcal{F}=\{\Omega,\{a,b,c\},\{b,c,d\},\{b,c\}\}$ and $\mathcal{G}=\{\Omega,\{a,b,c\},\{b,c,d\},\emptyset\}$: they are order isomorphic as complete lattices, but not as closure system. Is my argument correct?

Answer 1

IMO, your example indeed shows that these systems (under inclusion as order) are both isomorphic to the simple lattice with top and bottom and two incomparable elements inbetween (it probably has a standard name; maybe "diamond"?) and one is a closure system and the other is not. A lattice is more general, one could say. Preserving order is more "lenient" than preserving exact intersections, which is the appropriate way to define isomorphism/structure-preservingness for closure systems, one could say.