# Do closure operators on arbitrary posets give rise to complete lattices?

The notion of a closure operator is defined here for an arbitrary partially ordered set.

Now consider an arbitrary set $x$, and call its powerset $P.$ Furthermore, let us order $P$ by inclusion, thereby obtaining a complete lattice. Under these circumstances, it well known that given a closure operator $\mathrm{cl} : P \rightarrow P$, we can obtain a complete lattice $Q \subseteq P$ as follows.

1. The elements of $Q$ are precisely the $a \in Q$ such that $\mathrm{cl}(a)=a.$
2. The meet of $A \subseteq Q$ is simply $\bigcap A.$ (By intersection, I simply mean the meet operation of $P$. Thus if $A$ is empty, its meet is simply $x$).
3. The join of $A \subseteq Q$ is $\mathrm{cl}\left(\bigcup A\right).$

So I'm wondering. What happens in more general situations? e.g. What if $P$ is an arbitrary poset, or an arbitrary lattice, or even an arbitrary complete lattice?

Edit: For example, if $P$ is a lattice, do we obtain a new lattice? (Clearly it needn't be complete - as Carl Mummert explains in the comments, we can always take $\mathrm{cl}$ to be the identity operator). And what if $P$ is a join-semilattice, or a meet-semilattice?

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If $P$ is an arbitrary poset and $C$ is the identity operator, then the fixed point lattice for $C$ on $P$ is just $P$, and so will be complete if and only if $P$ is. – Carl Mummert Jun 7 '13 at 16:11