When can one atom be below the join of two distinct atoms?
$begingroup$
Consider three distinct atoms of a lattice $a,b,c$. When can we rule out the possibility that $cle alor b$?
So, to be clear, the question is: What is the weakest natural property of a lattice that rules out the above possibility?
It seems like we have this configuration for the standard "diamond" lattice example of a non-distributive lattice, so is this something that can only happen in non-distributive lattices? (This certainly can't happen for lattices with a unique irredundant join of join-irreducible elements representation...)
order-theory lattice-orders universal-algebra
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
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add a comment |
$begingroup$
Consider three distinct atoms of a lattice $a,b,c$. When can we rule out the possibility that $cle alor b$?
So, to be clear, the question is: What is the weakest natural property of a lattice that rules out the above possibility?
It seems like we have this configuration for the standard "diamond" lattice example of a non-distributive lattice, so is this something that can only happen in non-distributive lattices? (This certainly can't happen for lattices with a unique irredundant join of join-irreducible elements representation...)
order-theory lattice-orders universal-algebra
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
$endgroup$
add a comment |
$begingroup$
Consider three distinct atoms of a lattice $a,b,c$. When can we rule out the possibility that $cle alor b$?
So, to be clear, the question is: What is the weakest natural property of a lattice that rules out the above possibility?
It seems like we have this configuration for the standard "diamond" lattice example of a non-distributive lattice, so is this something that can only happen in non-distributive lattices? (This certainly can't happen for lattices with a unique irredundant join of join-irreducible elements representation...)
order-theory lattice-orders universal-algebra
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
$endgroup$
Consider three distinct atoms of a lattice $a,b,c$. When can we rule out the possibility that $cle alor b$?
So, to be clear, the question is: What is the weakest natural property of a lattice that rules out the above possibility?
It seems like we have this configuration for the standard "diamond" lattice example of a non-distributive lattice, so is this something that can only happen in non-distributive lattices? (This certainly can't happen for lattices with a unique irredundant join of join-irreducible elements representation...)
order-theory lattice-orders universal-algebra
order-theory lattice-orders universal-algebra
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
edited Dec 17 '18 at 12:22
Seamus
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
asked Dec 14 '18 at 12:18
SeamusSeamus
1,92321934
1,92321934
add a comment |
add a comment |
1 Answer
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In a distributive lattice, $cwedge (avee b) = (cwedge a)vee (cwedge b) = 0 vee 0 = 0ne c$ if $a,b,c$ are distinct atoms. So $cnotleq avee b$.
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
$endgroup$
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In a distributive lattice, $cwedge (avee b) = (cwedge a)vee (cwedge b) = 0 vee 0 = 0ne c$ if $a,b,c$ are distinct atoms. So $cnotleq avee b$.
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
$endgroup$
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
add a comment |
$begingroup$
In a distributive lattice, $cwedge (avee b) = (cwedge a)vee (cwedge b) = 0 vee 0 = 0ne c$ if $a,b,c$ are distinct atoms. So $cnotleq avee b$.
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
$endgroup$
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
add a comment |
$begingroup$
In a distributive lattice, $cwedge (avee b) = (cwedge a)vee (cwedge b) = 0 vee 0 = 0ne c$ if $a,b,c$ are distinct atoms. So $cnotleq avee b$.
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
$endgroup$
In a distributive lattice, $cwedge (avee b) = (cwedge a)vee (cwedge b) = 0 vee 0 = 0ne c$ if $a,b,c$ are distinct atoms. So $cnotleq avee b$.
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
answered Dec 14 '18 at 12:26
WuestenfuxWuestenfux
4,4901413
4,4901413
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
add a comment |
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
$begingroup$
Thanks, so this is part of an answer. The other part is: are there (interesting) kinds of non-distributive lattice where we don't have this occurring?
$endgroup$
– Seamus
Dec 14 '18 at 12:29
add a comment |
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