The supremum of the sum is at least as big as the sum of the suprema

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Consider a (additively written) magma $mathbfM = (M,+)$ with a compatible partial order $leq$, i.e. for every $m, n, p in M$ such that $m leq n$, we have
$$
m + p leq n + p,quadquad p + m leq p + n.
$$
Suppose that $mathbfM$ is complete w.r.t. arbitrary joins, i.e. for every subset $S subseteq M$ there is a least upper bound $sup S in M$. For every $A, B subseteq M$ define $A + B := a+b : ain A, bin B$.

It is always true that
$$
forall A, B subseteq M. sup(A+B) leq (sup A) + (sup B).
$$

Moreover, when $M = [0,infty]$, we also have
$$
forall A,B subseteq M. sup(A+B) geq (sup A) + (sup B).tag*
$$

However, I see no reason why the latter inequality should hold in general.

Has property $(*)$ been given a name in the literature? Have algebraic structures that manifest this property been given a name (maybe not p.o. magmas, but p.o. semi-modules or p.o. vector spaces)?

abstract-algebra terminology order-theory

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edited Aug 1 at 16:51

asked Aug 1 at 15:45

Evan Aad

5,36011748

add a comment | 

up vote
2
down vote

favorite

Consider a (additively written) magma $mathbfM = (M,+)$ with a compatible partial order $leq$, i.e. for every $m, n, p in M$ such that $m leq n$, we have
$$
m + p leq n + p,quadquad p + m leq p + n.
$$
Suppose that $mathbfM$ is complete w.r.t. arbitrary joins, i.e. for every subset $S subseteq M$ there is a least upper bound $sup S in M$. For every $A, B subseteq M$ define $A + B := a+b : ain A, bin B$.

It is always true that
$$
forall A, B subseteq M. sup(A+B) leq (sup A) + (sup B).
$$

Moreover, when $M = [0,infty]$, we also have
$$
forall A,B subseteq M. sup(A+B) geq (sup A) + (sup B).tag*
$$

However, I see no reason why the latter inequality should hold in general.

Has property $(*)$ been given a name in the literature? Have algebraic structures that manifest this property been given a name (maybe not p.o. magmas, but p.o. semi-modules or p.o. vector spaces)?

abstract-algebra terminology order-theory

share|cite|improve this question

edited Aug 1 at 16:51

asked Aug 1 at 15:45

Evan Aad

5,36011748

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Consider a (additively written) magma $mathbfM = (M,+)$ with a compatible partial order $leq$, i.e. for every $m, n, p in M$ such that $m leq n$, we have
$$
m + p leq n + p,quadquad p + m leq p + n.
$$
Suppose that $mathbfM$ is complete w.r.t. arbitrary joins, i.e. for every subset $S subseteq M$ there is a least upper bound $sup S in M$. For every $A, B subseteq M$ define $A + B := a+b : ain A, bin B$.

It is always true that
$$
forall A, B subseteq M. sup(A+B) leq (sup A) + (sup B).
$$

Moreover, when $M = [0,infty]$, we also have
$$
forall A,B subseteq M. sup(A+B) geq (sup A) + (sup B).tag*
$$

However, I see no reason why the latter inequality should hold in general.

Has property $(*)$ been given a name in the literature? Have algebraic structures that manifest this property been given a name (maybe not p.o. magmas, but p.o. semi-modules or p.o. vector spaces)?

abstract-algebra terminology order-theory

share|cite|improve this question

edited Aug 1 at 16:51

asked Aug 1 at 15:45

Evan Aad

5,36011748

Consider a (additively written) magma $mathbfM = (M,+)$ with a compatible partial order $leq$, i.e. for every $m, n, p in M$ such that $m leq n$, we have
$$
m + p leq n + p,quadquad p + m leq p + n.
$$
Suppose that $mathbfM$ is complete w.r.t. arbitrary joins, i.e. for every subset $S subseteq M$ there is a least upper bound $sup S in M$. For every $A, B subseteq M$ define $A + B := a+b : ain A, bin B$.

It is always true that
$$
forall A, B subseteq M. sup(A+B) leq (sup A) + (sup B).
$$

Moreover, when $M = [0,infty]$, we also have
$$
forall A,B subseteq M. sup(A+B) geq (sup A) + (sup B).tag*
$$

However, I see no reason why the latter inequality should hold in general.

Has property $(*)$ been given a name in the literature? Have algebraic structures that manifest this property been given a name (maybe not p.o. magmas, but p.o. semi-modules or p.o. vector spaces)?

abstract-algebra terminology order-theory

share|cite|improve this question

edited Aug 1 at 16:51

asked Aug 1 at 15:45

Evan Aad

5,36011748

share|cite|improve this question

share|cite|improve this question

edited Aug 1 at 16:51

edited Aug 1 at 16:51

edited Aug 1 at 16:51

asked Aug 1 at 15:45

Evan Aad

5,36011748

asked Aug 1 at 15:45

Evan Aad

5,36011748

asked Aug 1 at 15:45

Evan Aad

5,36011748

5,36011748

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

According to Theorem 3.10 and Corollary 3.11 in
Residuated Lattices: An Algebraic Glimpse at Substructural Logics,
pages 148-150, the property you call (*) holds iff the magma is residuated.

In they're multiplicative notation this means
$$x cdot y leq z Leftrightarrow y leq xbackslash z Leftrightarrow x leq z/y,$$
for some operations $/$ and $backslash$ that can be defined.

Also from the same theorem, these are
$$x backslash z = max y : xy leq z $$
and
$$z/y = max x : xy leq z .$$
So these are just divisions on each side.

With your additive notation, the correspondent are subtractions on each side.

If these can be defined, then the join is compatible with the operation;
otherwise, it isn't.

Edit

After a comment from the OP, I realised it isn't exactly as stated.

The correct form of the result above is, from Theorem 3.10, if the magma is residuated, then property (*) holds, while the converse is true, by Corollary 3.11 if additionally the magma is complete.

share|cite|improve this answer

edited Aug 1 at 20:22

answered Aug 1 at 19:08

amrsa

3,2432518

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
  – Evan Aad
  Aug 1 at 19:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
  – amrsa
  Aug 1 at 20:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Good enough for me.
  – Evan Aad
  Aug 1 at 20:17
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

According to Theorem 3.10 and Corollary 3.11 in
Residuated Lattices: An Algebraic Glimpse at Substructural Logics,
pages 148-150, the property you call (*) holds iff the magma is residuated.

In they're multiplicative notation this means
$$x cdot y leq z Leftrightarrow y leq xbackslash z Leftrightarrow x leq z/y,$$
for some operations $/$ and $backslash$ that can be defined.

Also from the same theorem, these are
$$x backslash z = max y : xy leq z $$
and
$$z/y = max x : xy leq z .$$
So these are just divisions on each side.

With your additive notation, the correspondent are subtractions on each side.

If these can be defined, then the join is compatible with the operation;
otherwise, it isn't.

Edit

After a comment from the OP, I realised it isn't exactly as stated.

The correct form of the result above is, from Theorem 3.10, if the magma is residuated, then property (*) holds, while the converse is true, by Corollary 3.11 if additionally the magma is complete.

share|cite|improve this answer

edited Aug 1 at 20:22

answered Aug 1 at 19:08

amrsa

3,2432518

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
  – Evan Aad
  Aug 1 at 19:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
  – amrsa
  Aug 1 at 20:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Good enough for me.
  – Evan Aad
  Aug 1 at 20:17
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

According to Theorem 3.10 and Corollary 3.11 in
Residuated Lattices: An Algebraic Glimpse at Substructural Logics,
pages 148-150, the property you call (*) holds iff the magma is residuated.

In they're multiplicative notation this means
$$x cdot y leq z Leftrightarrow y leq xbackslash z Leftrightarrow x leq z/y,$$
for some operations $/$ and $backslash$ that can be defined.

Also from the same theorem, these are
$$x backslash z = max y : xy leq z $$
and
$$z/y = max x : xy leq z .$$
So these are just divisions on each side.

With your additive notation, the correspondent are subtractions on each side.

If these can be defined, then the join is compatible with the operation;
otherwise, it isn't.

Edit

After a comment from the OP, I realised it isn't exactly as stated.

The correct form of the result above is, from Theorem 3.10, if the magma is residuated, then property (*) holds, while the converse is true, by Corollary 3.11 if additionally the magma is complete.

share|cite|improve this answer

edited Aug 1 at 20:22

answered Aug 1 at 19:08

amrsa

3,2432518

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
  – Evan Aad
  Aug 1 at 19:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
  – amrsa
  Aug 1 at 20:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Good enough for me.
  – Evan Aad
  Aug 1 at 20:17
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

According to Theorem 3.10 and Corollary 3.11 in
Residuated Lattices: An Algebraic Glimpse at Substructural Logics,
pages 148-150, the property you call (*) holds iff the magma is residuated.

In they're multiplicative notation this means
$$x cdot y leq z Leftrightarrow y leq xbackslash z Leftrightarrow x leq z/y,$$
for some operations $/$ and $backslash$ that can be defined.

Also from the same theorem, these are
$$x backslash z = max y : xy leq z $$
and
$$z/y = max x : xy leq z .$$
So these are just divisions on each side.

With your additive notation, the correspondent are subtractions on each side.

If these can be defined, then the join is compatible with the operation;
otherwise, it isn't.

Edit

After a comment from the OP, I realised it isn't exactly as stated.

The correct form of the result above is, from Theorem 3.10, if the magma is residuated, then property (*) holds, while the converse is true, by Corollary 3.11 if additionally the magma is complete.

share|cite|improve this answer

edited Aug 1 at 20:22

answered Aug 1 at 19:08

amrsa

3,2432518

According to Theorem 3.10 and Corollary 3.11 in
Residuated Lattices: An Algebraic Glimpse at Substructural Logics,
pages 148-150, the property you call (*) holds iff the magma is residuated.

In they're multiplicative notation this means
$$x cdot y leq z Leftrightarrow y leq xbackslash z Leftrightarrow x leq z/y,$$
for some operations $/$ and $backslash$ that can be defined.

Also from the same theorem, these are
$$x backslash z = max y : xy leq z $$
and
$$z/y = max x : xy leq z .$$
So these are just divisions on each side.

With your additive notation, the correspondent are subtractions on each side.

If these can be defined, then the join is compatible with the operation;
otherwise, it isn't.

Edit

After a comment from the OP, I realised it isn't exactly as stated.

The correct form of the result above is, from Theorem 3.10, if the magma is residuated, then property (*) holds, while the converse is true, by Corollary 3.11 if additionally the magma is complete.

share|cite|improve this answer

edited Aug 1 at 20:22

answered Aug 1 at 19:08

amrsa

3,2432518

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 20:22

edited Aug 1 at 20:22

edited Aug 1 at 20:22

answered Aug 1 at 19:08

amrsa

3,2432518

answered Aug 1 at 19:08

amrsa

3,2432518

answered Aug 1 at 19:08

amrsa

3,2432518

3,2432518

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
  – Evan Aad
  Aug 1 at 19:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
  – amrsa
  Aug 1 at 20:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Good enough for me.
  – Evan Aad
  Aug 1 at 20:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
  – Evan Aad
  Aug 1 at 19:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
  – amrsa
  Aug 1 at 20:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. Good enough for me.
  – Evan Aad
  Aug 1 at 20:17
  

  
  
  
  

I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
– Evan Aad
Aug 1 at 19:37

I've browsed over the theorems you mentioned. Correct me if I'm wrong, but the 'iff' claim holds provided that the magma is order-complete, right?
– Evan Aad
Aug 1 at 19:37

1

1

@EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
– amrsa
Aug 1 at 20:11

@EvanAad Yes, you are right. If it is not complete we still have that multiplication preserves existing joins. But the converse holds only for complete ones. Summing up, if it is not complete, the operation might preserve existing joins, even if it isn't residuated (but I'm not sure whether this is true or unknown...). So this is only half answer...
– amrsa
Aug 1 at 20:11

Thanks. Good enough for me.
– Evan Aad
Aug 1 at 20:17

Thanks. Good enough for me.
– Evan Aad
Aug 1 at 20:17

add a comment | 

 

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