# Continuous Lattice

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.

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 $$\operatorname{im} p \cong L$$.
• Thank you for your attention.Could i ask an other question?On the page 120 of the same book,'By the characterization of the way-below relation in c(L) contained in I-2.2,there is a u$\in$K(L) with u$\leq$a such that a $\leq$c(u)$\ll$a.'why u is in K(L) not in L? Commented Sep 23, 2021 at 6:45