I am reading $\textit{On the Foundations of Combinatorial Theory II. Combinatorial Geometries}$. They give a definition.
A closure relation on a set $S$ is a function $A\mapsto \bar{A}$ defined for all subsets > $A\subseteq S$, satisfying $$ \overline{A}\subseteq S,\quad A\subseteq \overline{A},\quad A\subseteq \overline{B}\text{ implies }\overline{A}\subseteq \overline{B} $$
They then write that we can prove that $\overline{A\cap B}=\overline{A}\cap\overline{B}$. I can show that $\overline{A\cap B}\subseteq \overline{A}\cap\overline{B}$, but I cannot figure out the reverse containment. Is this even true in general?
Unless there are some background assumptions that you’ve not stated, the assertion is false. Let $S=\Bbb R$, and for $A\subseteq\Bbb R$ let $\overline{A}$ be the usual topological closure of $A$ in $\Bbb R$. Then $A\subseteq\overline{A}\subseteq\Bbb R$ for all $A\subseteq\Bbb R$, and $A\subseteq\overline{B}$ implies that $\overline{A}\subseteq\overline{B}$, but if we set $A=(0,1)$ and $B=(1,2)$, then
$$\overline{A\cap B}=\overline{(0,1)\cap(1,2)}=\overline{\varnothing}=\varnothing\subsetneqq\{1\}=[0,1]\cap[1,2]=\overline{A}\cap\overline{B}\;.$$