The union of two cuts is a cut?

Question

Every poset $\langle P, \leq \rangle$ has a Dedekind-MacNeille completion, a complete lattice that embeds $\langle P, \leq \rangle$.

For $A \subseteq P$, the upset $U(A) = \{p \in P\ |\ \forall a \in A:\ a \leq p\}$. Likewise, the downset $D(A)= \{p \in P\ |\ \forall a \in A:\ p \leq a\}$. Then a cut is any set $A$ such that $A=D(U(A))$.

Now we take the set of cuts in $P$:

$$DM(P) = \{A\subseteq P\ |\ A=D(U(A))\}$$

The poset $ \langle DM(P), \subseteq\rangle$ consisting of the set of cuts in $P$, $ DM(P) $, ordered by set inclusion, $\subseteq$, forms a complete lattice.

I'm trying to prove this, and but can't seem to show even that for every two elements $A, B \in DM(P)$, their union is in the lattice, i.e. $A\cup B = D(U(A\cup B))$.

It's easy to show that $A\cup B \subseteq D(U(A\cup B))$: Suppose $x \in A\cup B$. Then $x \leq y$ for every $y \in U(A\cup B)$. Then $x \in D(U(A\cup B))$.

But I can't manage to prove the other direction, i.e. that $D(U(A\cup B)) \subseteq A\cup B$. The dozen or so textbooks on lattice and order theory I've looked at (and MacNeille's original paper) do not go through the proof.

Answer 1

When people say that cuts form a complete lattice under $\subseteq$, they don't necessarily mean that the lattice operations are the ordinary union and intersection. So verifying this claim does not require checking that $D(U(A\cup B)) = A\cup B$ when $A$ and $B$ are cuts (which is a good thing: this isn't true, if our lattice is the four element lattice $\{0,1\}^2$). Instead we have to figure out what the lattice operations on the cuts actually are, and check that they form a legitimate lattice - how do we do this?

In this case, something nice happens: the operation $C: A \mapsto D(U(A))$ is a closure operation, i.e.

• $A \subseteq C(A)$,
• $C(C(A)) = C(A)$, and
• if $A \subseteq B$, then $C(A) \subseteq C(B)$.

In general, for any closure operation $C$, the set of "closed" sets $A$ such that $C(A) = A$ always forms a complete lattice, where the lattice meet is the usual intersection, while the lattice join is the closure of the union.

Answer 2

So the point is that what is denoted as a union $A \cup B$ is not to be interpreted as a set-theoretic union, but as a join within the poset of cuts.

The proof you're after is just a special case of a much more general proposition, that if you have a closure operator on a complete lattice (which in this case is the power set of the poset $P$), then the poset of fixed points of the closure operator is again a complete lattice. So in the present case, what one needs is that $D \circ U$ is a closure operator. This is not hard because there is a Galois connection between $D$ and $U$; I can explain if need be. Let me take it for granted for now.

Now suppose given a complete lattice $L$ and a closure operator $\phi: L \to L$. If $c_i$ is a collection of fixed points, then their meet in $L$ is a fixed point. We just need that $\phi(\bigwedge_i c_i) \leq \bigwedge_i c_i$, but this follows immediately by observing

$$\phi(\bigwedge_i c_i) \leq \phi(c_i) \leq c_i$$

for all $i$. This alone is enough to guarantee existence of arbitrary joins (any poset with arbitrary meets has arbitrary joins), but to be more specific about your problem, the join of two closed elements $c, d$ in the poset of closed elements is $c \vee d = \phi(c \cup d)$ where $\cup$ denotes the join in $L$. It's clear that $c, d \leq \phi(c \cup d)$, so for $e$ closed, $\phi(c \cup d) \leq e$ implies $c \leq e$ and $d \leq e$. In the converse direction, given $c \leq e$ and $d \leq e$, we have $c \cup d \leq e$ and therefore $\phi(c \cup d) \leq \phi(e) = e$.