Continuous Lattice

Question

Let $L$ be a lattice.Then the following statements are equivalent.

1. $L$ is contionuous.
2. There are an arithmetic lattice $A$ and a surjective map $r:A\rightarrow L$ preserving arbitrary infs and directed sups.
3. There are algebraic lattice $A$ and a surjective map $r:A\rightarrow L$ preserving infs and directed sups.
4. There are a set $X$ and a projection map $p:2^{X}\rightarrow 2^{X}$ preserving directed sups such that $L\cong \operatorname{im} p$.

This is the corollary on the page 123 of the Continuous Lattices and Domains author by G.GIERZ at all. I don't understand the sentence

By I-4.13,there is a closure opertor $c$ preserving directed sups on some $2^{X}$ such that the algebraic lattice $A$ is isomorphic to the image of $c$.

What exactly is map $c$? Why define $p:2^{X}\rightarrow 2^{X}$ by $p=c_{\circ}k c^{\circ}$? I would appreciate your suggestions and comments.

Answer 1

I think it should refer to Theorem I-4.16, instead of I-4.13. That theorem tells you that any algebraic lattice is isomorphic to the image of a closure operator $c:2^X \to 2^X$ which preserves directed unions (for some set $X$). Applying this for the algebraic lattice $A$ gives you the map $c$ of the proof.

Now, the map $c$ has image isomorphic to $A$, while you want a projection $p$ with image isomorphic to $L$. This is accomplished by defining $p=c_\circ k c^\circ$, where $k=dr$ and $d:L \to A$ is the lower adjoint of $r$. Indeed, since $r$ is surjective, $d$ is injective, so $\operatorname{im} k \cong \operatorname{im} r \cong L$. Moreover, $k$ is idempotent. Using this together with the fact that $c$ is a closure operator (hence also idempotent), it follows that $p:2^X \to 2^X$ is indeed the desired projection map with $\operatorname{im} p \cong L$.