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Tag fix (order lattices, not subgroups of space) & MacNeille spelling & link
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Jukka Kohonen
Jukka Kohonen
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Every poset $\langle P, \leq \rangle$ has a Dedekind-Macneille CompletionDedekind-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'sMacNeille's original paper) do not go through the proof.

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.

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.

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gnarizard
gnarizard
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The union of two cuts is a cut?

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.