# Reconstructing a closure operator from a set of fixed points

Let $$L$$ be a lattice, not necessarily complete. We define a closure operator as a function $$f\colon L\to L$$ which is:

1. idempotent, $$f(f(x)) = f(x)$$,
2. isotone, $$x\leq y \Rightarrow f(x) \leq f(y)$$,
3. extensive, $$x \leq f(x)$$

for each $$x,y \in L$$. Then the set of fixed points of $$f$$ is the set of closed elements.

If we have an arbitrary set $$C\subseteq L$$, we can construct a closure operator $$f_C$$ such the set of fixed points is exactly $$C$$ by setting:

$$f(x) = \bigwedge (x{\uparrow}\cap C) = \bigwedge \{{c\in C: x\leq c\}}$$

provided that either:

1. $$C$$ is closed under arbitrary meets, or
2. for each $$x\in L$$, the $$x{\uparrow} \cap C$$ has a meet which belongs to $$C$$.

The second condition seems to be weaker than the first one. Are these equivalent for complete lattices?

Yes, just apply condition 2 to the meet of the given subset of $$C$$.
In more detail: Suppose $$L$$ is a complete lattice and $$C\subseteq L$$. Assume $$C$$ satisfies condition $$2$$. We would like to show that $$C$$ is closed under arbitrary meets. So let $$X\subseteq C$$. Since $$L$$ is complete, $$X$$ has a meet $$x$$, and we want to show $$x\in C$$. Let $$Y = x{\uparrow} \cap C$$. By condition 2, $$Y$$ has a meet $$y\in C$$. It remains to show that $$x = y$$.
For all $$z\in Y$$, $$z\in x{\uparrow}$$, so $$x\leq z$$. Thus $$x\leq y$$. Conversely, for all $$z\in X$$, $$x\leq z$$, so $$z\in x{\uparrow}$$, and since $$X\subseteq C$$, $$z\in C$$, so $$z\in Y$$, and $$y\leq z$$. Thus $$y\leq x$$, so $$x = y\in C$$ as desired.