Let $\Omega$ be a set, $\mathcal{c}: \mathcal{P}(\Omega) \rightarrow \mathcal{P}(\Omega)$ be a closure operator (i.e., $\mathcal{c}$ satisfies $X \subseteq \mathcal{c}(X)$ and $\mathcal{c}(\mathcal{c}(X)) = \mathcal{c}(X)$ for all $X$, as well as $X \subseteq Y \Rightarrow \mathcal{c}(X) \subseteq \mathcal{c}(Y)$ for all $X, Y$). To avoid confusion, I’ll call sets $X$ with $X = \mathcal{c}(X)$ $\mathcal{c}$-closed. Now, let $\tau$ be the weakest topology on $\Omega$ s.t. all $\mathcal{c}$-closed sets are closed, i.e., $\tau$ is the topology generated by declaring $\mathcal{c}$-closed sets a closed subbasis. Let $S$ be a collection of $\mathcal{c}$-closed sets s.t. $S$ is closed under taking finite intersections and taking finite joins (where joins are understood in the lattice of $\mathcal{c}$-closed sets, i.e., $A \vee B = \mathcal{c}(A \cup B)$). If $S$ is a closed subbasis for $\tau$, does it follow that $S$ is a basis for $\mathcal{c}$, in the sense that all $\mathcal{c}$-closed sets are intersections of elements of $S$? One possible issue might be infinite intersections can be empty, so if it helps assume $(\Omega, \tau)$ is compact.
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