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Difference between maximals and minimum
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Given a partially ordered set $(U,leq )$, its maximum is always a maximal, but none of its maximals may necessarily be a maximum, right?
In which cases is none of the maximals a maximum?
I understand the concept and I know that they are not the same, what I don't see is in which cases the maximals are not maximum. It's very clear with the order relationship of $|$ (divide a), but I don't understand it very well with the canonical relationship of the naturals ($leq $).
order-theory
asked Jul 27 at 15:40
a4956279
1
 |Â
show 4 more comments
up vote
0
down vote
favorite
Given a partially ordered set $(U,leq )$, its maximum is always a maximal, but none of its maximals may necessarily be a maximum, right?
In which cases is none of the maximals a maximum?
I understand the concept and I know that they are not the same, what I don't see is in which cases the maximals are not maximum. It's very clear with the order relationship of $|$ (divide a), but I don't understand it very well with the canonical relationship of the naturals ($leq $).
order-theory
asked Jul 27 at 15:40
a4956279
1
Note that for the naturals with $leq$ you actually define a total order
– ippiki-ookami
Jul 27 at 15:44
You can talk about maximum for chain only. For example, if the order relation is divisibility in $mathbb N$, what would be the maximum of $1,2,3,4,5,6$ ?
– Surb
Jul 27 at 15:46
@Surb , There is no maximum, but yes several maximals.
– a4956279
Jul 27 at 15:53
You should check the definition of "maximum". I agree that there is a supremum, and several upper bound, but there is no maximum.
– Surb
Jul 27 at 15:55
@Surb Because given a partially ordered set $(U, <)$ (where $<$ is an arbitrary order relationship), and a subset $A$ of $U$, the maximum is a value $M$ such that for all elements $x$ of $A$, $x < M$.
– a4956279
Jul 27 at 16:01
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a partially ordered set $(U,leq )$, its maximum is always a maximal, but none of its maximals may necessarily be a maximum, right?
In which cases is none of the maximals a maximum?
I understand the concept and I know that they are not the same, what I don't see is in which cases the maximals are not maximum. It's very clear with the order relationship of $|$ (divide a), but I don't understand it very well with the canonical relationship of the naturals ($leq $).
order-theory
asked Jul 27 at 15:40
a4956279
1
Given a partially ordered set $(U,leq )$, its maximum is always a maximal, but none of its maximals may necessarily be a maximum, right?
In which cases is none of the maximals a maximum?
I understand the concept and I know that they are not the same, what I don't see is in which cases the maximals are not maximum. It's very clear with the order relationship of $|$ (divide a), but I don't understand it very well with the canonical relationship of the naturals ($leq $).
order-theory
asked Jul 27 at 15:40
a4956279
1
asked Jul 27 at 15:40
a4956279
1
asked Jul 27 at 15:40
a4956279
1
asked Jul 27 at 15:40
a4956279
1
1
Note that for the naturals with $leq$ you actually define a total order
– ippiki-ookami
Jul 27 at 15:44
You can talk about maximum for chain only. For example, if the order relation is divisibility in $mathbb N$, what would be the maximum of $1,2,3,4,5,6$ ?
– Surb
Jul 27 at 15:46
@Surb , There is no maximum, but yes several maximals.
– a4956279
Jul 27 at 15:53
You should check the definition of "maximum". I agree that there is a supremum, and several upper bound, but there is no maximum.
– Surb
Jul 27 at 15:55
@Surb Because given a partially ordered set $(U, <)$ (where $<$ is an arbitrary order relationship), and a subset $A$ of $U$, the maximum is a value $M$ such that for all elements $x$ of $A$, $x < M$.
– a4956279
Jul 27 at 16:01
 |Â
show 4 more comments
Note that for the naturals with $leq$ you actually define a total order
– ippiki-ookami
Jul 27 at 15:44
You can talk about maximum for chain only. For example, if the order relation is divisibility in $mathbb N$, what would be the maximum of $1,2,3,4,5,6$ ?
– Surb
Jul 27 at 15:46
@Surb , There is no maximum, but yes several maximals.
– a4956279
Jul 27 at 15:53
You should check the definition of "maximum". I agree that there is a supremum, and several upper bound, but there is no maximum.
– Surb
Jul 27 at 15:55
@Surb Because given a partially ordered set $(U, <)$ (where $<$ is an arbitrary order relationship), and a subset $A$ of $U$, the maximum is a value $M$ such that for all elements $x$ of $A$, $x < M$.
– a4956279
Jul 27 at 16:01
Note that for the naturals with $leq$ you actually define a total order
– ippiki-ookami
Jul 27 at 15:44
Note that for the naturals with $leq$ you actually define a total order
– ippiki-ookami
Jul 27 at 15:44
You can talk about maximum for chain only. For example, if the order relation is divisibility in $mathbb N$, what would be the maximum of $1,2,3,4,5,6$ ?
– Surb
Jul 27 at 15:46
You can talk about maximum for chain only. For example, if the order relation is divisibility in $mathbb N$, what would be the maximum of $1,2,3,4,5,6$ ?
– Surb
Jul 27 at 15:46
@Surb , There is no maximum, but yes several maximals.
– a4956279
Jul 27 at 15:53
@Surb , There is no maximum, but yes several maximals.
– a4956279
Jul 27 at 15:53
You should check the definition of "maximum". I agree that there is a supremum, and several upper bound, but there is no maximum.
– Surb
Jul 27 at 15:55
You should check the definition of "maximum". I agree that there is a supremum, and several upper bound, but there is no maximum.
– Surb
Jul 27 at 15:55
@Surb Because given a partially ordered set $(U, <)$ (where $<$ is an arbitrary order relationship), and a subset $A$ of $U$, the maximum is a value $M$ such that for all elements $x$ of $A$, $x < M$.
– a4956279
Jul 27 at 16:01
@Surb Because given a partially ordered set $(U, <)$ (where $<$ is an arbitrary order relationship), and a subset $A$ of $U$, the maximum is a value $M$ such that for all elements $x$ of $A$, $x < M$.
– a4956279
Jul 27 at 16:01
 |Â
show 4 more comments
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Note that for the naturals with $leq$ you actually define a total order
– ippiki-ookami
Jul 27 at 15:44
You can talk about maximum for chain only. For example, if the order relation is divisibility in $mathbb N$, what would be the maximum of $1,2,3,4,5,6$ ?
– Surb
Jul 27 at 15:46
@Surb , There is no maximum, but yes several maximals.
– a4956279
Jul 27 at 15:53
You should check the definition of "maximum". I agree that there is a supremum, and several upper bound, but there is no maximum.
– Surb
Jul 27 at 15:55
@Surb Because given a partially ordered set $(U, <)$ (where $<$ is an arbitrary order relationship), and a subset $A$ of $U$, the maximum is a value $M$ such that for all elements $x$ of $A$, $x < M$.
– a4956279
Jul 27 at 16:01