Does the closure of the set of all irreducible elements always equal the whole set? Let $X$ denote a set and $\mathrm{cl}$ denote a finitary closure operator on its powerset. Call $x \in X$ irreducible iff for all $A \subseteq X$ we have that if $x \in \mathrm{cl}(A)$, then $x \in A$.
Question. Does the closure of the set of all irreducible elements necessarily equal $X$?
I think the answer should be "trivially, yes" but for some reason I cannot prove it.

There may not be enough irreducible elements, as you've defined them.
Take the example mentioned at the article you linked, of forming ${cl}(A)$ as the span of a subset $A$ of a vector space $X$.  Then no $x \in X$ is irreducible as we can find a set $A$ depending on $x$ such that $x$ is not in $A$ but is in the span of $A$.
In this case the set of irreducible elements is empty, and the closure of the empty set is {$0$}, which need not be all of $X$.
