Are the real numbers a lattice? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)understanding lattice in detailedDrawing a lattice for a set partially ordered by divisibilityLattice DefinitionHow is a complete lattice defined solely by a least-upper bound?Lattice of a POSET RelationLattice orders and number of elements in a setA Lattice that is not a Complete LatticeComplete LatticeHow to prove that Dn, a set of all positive divisors of n under the relation of divisibility, is a lattice?Formal Definition of a Lattice
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Are the real numbers a lattice?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)understanding lattice in detailedDrawing a lattice for a set partially ordered by divisibilityLattice DefinitionHow is a complete lattice defined solely by a least-upper bound?Lattice of a POSET RelationLattice orders and number of elements in a setA Lattice that is not a Complete LatticeComplete LatticeHow to prove that Dn, a set of all positive divisors of n under the relation of divisibility, is a lattice?Formal Definition of a Lattice
$begingroup$
A lattice is a set with a partial order, where every pair has a unique upper and lower bound.
As far as I can tell, there is nothing in the definition that forces the set to be discrete. In particular, the real numbers with their usual partial order and upper/lower bounds seems to fit the definition. But all the examples I've come across are discrete, which makes me wonder if I've missed something. (e.g. Wikipedia)
lattice-orders
asked Apr 8 at 20:20
prdnrprdnr
956
$endgroup$
add a comment |
$begingroup$
A lattice is a set with a partial order, where every pair has a unique upper and lower bound.
As far as I can tell, there is nothing in the definition that forces the set to be discrete. In particular, the real numbers with their usual partial order and upper/lower bounds seems to fit the definition. But all the examples I've come across are discrete, which makes me wonder if I've missed something. (e.g. Wikipedia)
lattice-orders
asked Apr 8 at 20:20
prdnrprdnr
956
$endgroup$
4
$begingroup$
You're not missing anything. Every totally ordered set is a lattice.
$endgroup$
– Wojowu
Apr 8 at 20:23
1
$begingroup$
You didn't miss anything, the real numbers form a lattice.
$endgroup$
– Javi
Apr 8 at 20:24
add a comment |
$begingroup$
A lattice is a set with a partial order, where every pair has a unique upper and lower bound.
As far as I can tell, there is nothing in the definition that forces the set to be discrete. In particular, the real numbers with their usual partial order and upper/lower bounds seems to fit the definition. But all the examples I've come across are discrete, which makes me wonder if I've missed something. (e.g. Wikipedia)
lattice-orders
asked Apr 8 at 20:20
prdnrprdnr
956
$endgroup$
A lattice is a set with a partial order, where every pair has a unique upper and lower bound.
As far as I can tell, there is nothing in the definition that forces the set to be discrete. In particular, the real numbers with their usual partial order and upper/lower bounds seems to fit the definition. But all the examples I've come across are discrete, which makes me wonder if I've missed something. (e.g. Wikipedia)
lattice-orders
lattice-orders
asked Apr 8 at 20:20
prdnrprdnr
956
asked Apr 8 at 20:20
prdnrprdnr
956
asked Apr 8 at 20:20
prdnrprdnr
956
asked Apr 8 at 20:20
prdnrprdnr
956
asked Apr 8 at 20:20
prdnrprdnr
956
956
4
$begingroup$
You're not missing anything. Every totally ordered set is a lattice.
$endgroup$
– Wojowu
Apr 8 at 20:23
1
$begingroup$
You didn't miss anything, the real numbers form a lattice.
$endgroup$
– Javi
Apr 8 at 20:24
add a comment |
4
$begingroup$
You're not missing anything. Every totally ordered set is a lattice.
$endgroup$
– Wojowu
Apr 8 at 20:23
1
$begingroup$
You didn't miss anything, the real numbers form a lattice.
$endgroup$
– Javi
Apr 8 at 20:24
4
4
$begingroup$
You're not missing anything. Every totally ordered set is a lattice.
$endgroup$
– Wojowu
Apr 8 at 20:23
$begingroup$
You're not missing anything. Every totally ordered set is a lattice.
$endgroup$
– Wojowu
Apr 8 at 20:23
1
1
$begingroup$
You didn't miss anything, the real numbers form a lattice.
$endgroup$
– Javi
Apr 8 at 20:24
$begingroup$
You didn't miss anything, the real numbers form a lattice.
$endgroup$
– Javi
Apr 8 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In a lattice, every pair must have a least upper bound (called join) and greatest lower bound (called meet), that is different from it being unique. There can be many upper bounds or lower bounds. The join or meet of a pair however, is unique.
The reals are in fact a particularly nice kind of lattice:
- They are totally ordered, which means that any two elements $a, b in mathbbR$ can be compared and therefore the join and meet can be computed very easily by using $max(a, b)$ and $min(a, b)$ respectively.
- When we take a closed interval (e.g. $[0,1]$), then that interval is complete. This means that every subset $A subseteq [0,1]$ has a least upper bound, in this case given by its supremum $sup A$. Similarly for a greatest lower bound and the infimum.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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$begingroup$
In a lattice, every pair must have a least upper bound (called join) and greatest lower bound (called meet), that is different from it being unique. There can be many upper bounds or lower bounds. The join or meet of a pair however, is unique.
The reals are in fact a particularly nice kind of lattice:
- They are totally ordered, which means that any two elements $a, b in mathbbR$ can be compared and therefore the join and meet can be computed very easily by using $max(a, b)$ and $min(a, b)$ respectively.
- When we take a closed interval (e.g. $[0,1]$), then that interval is complete. This means that every subset $A subseteq [0,1]$ has a least upper bound, in this case given by its supremum $sup A$. Similarly for a greatest lower bound and the infimum.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In a lattice, every pair must have a least upper bound (called join) and greatest lower bound (called meet), that is different from it being unique. There can be many upper bounds or lower bounds. The join or meet of a pair however, is unique.
The reals are in fact a particularly nice kind of lattice:
- They are totally ordered, which means that any two elements $a, b in mathbbR$ can be compared and therefore the join and meet can be computed very easily by using $max(a, b)$ and $min(a, b)$ respectively.
- When we take a closed interval (e.g. $[0,1]$), then that interval is complete. This means that every subset $A subseteq [0,1]$ has a least upper bound, in this case given by its supremum $sup A$. Similarly for a greatest lower bound and the infimum.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In a lattice, every pair must have a least upper bound (called join) and greatest lower bound (called meet), that is different from it being unique. There can be many upper bounds or lower bounds. The join or meet of a pair however, is unique.
The reals are in fact a particularly nice kind of lattice:
- They are totally ordered, which means that any two elements $a, b in mathbbR$ can be compared and therefore the join and meet can be computed very easily by using $max(a, b)$ and $min(a, b)$ respectively.
- When we take a closed interval (e.g. $[0,1]$), then that interval is complete. This means that every subset $A subseteq [0,1]$ has a least upper bound, in this case given by its supremum $sup A$. Similarly for a greatest lower bound and the infimum.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In a lattice, every pair must have a least upper bound (called join) and greatest lower bound (called meet), that is different from it being unique. There can be many upper bounds or lower bounds. The join or meet of a pair however, is unique.
The reals are in fact a particularly nice kind of lattice:
- They are totally ordered, which means that any two elements $a, b in mathbbR$ can be compared and therefore the join and meet can be computed very easily by using $max(a, b)$ and $min(a, b)$ respectively.
- When we take a closed interval (e.g. $[0,1]$), then that interval is complete. This means that every subset $A subseteq [0,1]$ has a least upper bound, in this case given by its supremum $sup A$. Similarly for a greatest lower bound and the infimum.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
answered Apr 8 at 20:33
Mark KamsmaMark Kamsma
3617
3617
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mark Kamsma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
You're not missing anything. Every totally ordered set is a lattice.
$endgroup$
– Wojowu
Apr 8 at 20:23
1
$begingroup$
You didn't miss anything, the real numbers form a lattice.
$endgroup$
– Javi
Apr 8 at 20:24