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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?

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1 Answer 1

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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.

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  • $\begingroup$ Thanks, this is a nice simple argument. $\endgroup$
    – Jakim
    Apr 19 at 19:06

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