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Why a least element is a distinct property?
Clash Royale CLAN TAG#URR8PPP
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A Well-Ordered Set always have a least element iff it's non-empty.
Then why is this a distinct property that has a lot of attention while having a greatest element property (like $mathbbZ^-$) has no attention.
set-theory
asked Jul 14 at 14:55
Zyad Yasser
19910
add a comment |Â
up vote
2
down vote
favorite
A Well-Ordered Set always have a least element iff it's non-empty.
Then why is this a distinct property that has a lot of attention while having a greatest element property (like $mathbbZ^-$) has no attention.
set-theory
asked Jul 14 at 14:55
Zyad Yasser
19910
3
Having a least element is not a particularly interesting property. The fact that any non-empty subset has a least element is noteworthy.
– Saucy O'Path
Jul 14 at 14:57
1
There's nothing preventing you from flipping everything upside down and defining a reverse notion of well ordering in terms of all nonempty subsets having a greatest element, other than common sense and convention: the set of natural numbers is, almost by definition, a more natural exemplar of well ordering than the set of negative integers. (If you do want to define a reverse notion, perhaps it should be call "ill" ordering.)
– Barry Cipra
Jul 14 at 15:17
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A Well-Ordered Set always have a least element iff it's non-empty.
Then why is this a distinct property that has a lot of attention while having a greatest element property (like $mathbbZ^-$) has no attention.
set-theory
asked Jul 14 at 14:55
Zyad Yasser
19910
A Well-Ordered Set always have a least element iff it's non-empty.
Then why is this a distinct property that has a lot of attention while having a greatest element property (like $mathbbZ^-$) has no attention.
set-theory
asked Jul 14 at 14:55
Zyad Yasser
19910
asked Jul 14 at 14:55
Zyad Yasser
19910
asked Jul 14 at 14:55
Zyad Yasser
19910
asked Jul 14 at 14:55
Zyad Yasser
19910
19910
3
Having a least element is not a particularly interesting property. The fact that any non-empty subset has a least element is noteworthy.
– Saucy O'Path
Jul 14 at 14:57
1
There's nothing preventing you from flipping everything upside down and defining a reverse notion of well ordering in terms of all nonempty subsets having a greatest element, other than common sense and convention: the set of natural numbers is, almost by definition, a more natural exemplar of well ordering than the set of negative integers. (If you do want to define a reverse notion, perhaps it should be call "ill" ordering.)
– Barry Cipra
Jul 14 at 15:17
add a comment |Â
3
Having a least element is not a particularly interesting property. The fact that any non-empty subset has a least element is noteworthy.
– Saucy O'Path
Jul 14 at 14:57
1
There's nothing preventing you from flipping everything upside down and defining a reverse notion of well ordering in terms of all nonempty subsets having a greatest element, other than common sense and convention: the set of natural numbers is, almost by definition, a more natural exemplar of well ordering than the set of negative integers. (If you do want to define a reverse notion, perhaps it should be call "ill" ordering.)
– Barry Cipra
Jul 14 at 15:17
3
3
Having a least element is not a particularly interesting property. The fact that any non-empty subset has a least element is noteworthy.
– Saucy O'Path
Jul 14 at 14:57
Having a least element is not a particularly interesting property. The fact that any non-empty subset has a least element is noteworthy.
– Saucy O'Path
Jul 14 at 14:57
1
1
There's nothing preventing you from flipping everything upside down and defining a reverse notion of well ordering in terms of all nonempty subsets having a greatest element, other than common sense and convention: the set of natural numbers is, almost by definition, a more natural exemplar of well ordering than the set of negative integers. (If you do want to define a reverse notion, perhaps it should be call "ill" ordering.)
– Barry Cipra
Jul 14 at 15:17
There's nothing preventing you from flipping everything upside down and defining a reverse notion of well ordering in terms of all nonempty subsets having a greatest element, other than common sense and convention: the set of natural numbers is, almost by definition, a more natural exemplar of well ordering than the set of negative integers. (If you do want to define a reverse notion, perhaps it should be call "ill" ordering.)
– Barry Cipra
Jul 14 at 15:17
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
In principle we could repeat the entire theory of well-orders with the words and relation signs flipped such that we speak about "greatest" elements of arbitrary nonempty subsets instead of "least".
Since formal logic doesn't care about the words we use, we would have exactly the same theorems in the new formulations, just in the different direction. But there are some pragmatic arguments for spending more attention on the usual concept of well-orders:
Well-orders have an intuitively appealing main example in the form of the natural numbers. There's nothing really comparable for co-well-orders.
It would turn the intuition about induction awry. One main use for well-orders is to construct things by long induction over the well order. Here you imagine constructing your thing starting with the first elements of the order and then your induction hypothesis tells you that there's no way you won't get a unique result for everything. Whereas the underlying formalism doesn't care about our ideas of the construction being "spread out in time" (what the induction theorem says is that the whole thing we're defining just is there, period), it would certainly make it harder to think about if we had to imagine starting with the last elements for an inductive construction.
Note, by the way that if you combine the concepts and want to speak about ordered subsets where every nonempty subset has both a least and a greatest element, then you get a rather uninteresting theory out of it: Those orders are exactly the finite total orders. So it is essential with some asymmetry in the assumptions in order to get the rich theory of ordinals, even though it is formally irrelevant whether the asymmetry is to one side or the other.
answered Jul 14 at 16:06
Henning Makholm
226k16291520
3
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
1
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
In principle we could repeat the entire theory of well-orders with the words and relation signs flipped such that we speak about "greatest" elements of arbitrary nonempty subsets instead of "least".
Since formal logic doesn't care about the words we use, we would have exactly the same theorems in the new formulations, just in the different direction. But there are some pragmatic arguments for spending more attention on the usual concept of well-orders:
Well-orders have an intuitively appealing main example in the form of the natural numbers. There's nothing really comparable for co-well-orders.
It would turn the intuition about induction awry. One main use for well-orders is to construct things by long induction over the well order. Here you imagine constructing your thing starting with the first elements of the order and then your induction hypothesis tells you that there's no way you won't get a unique result for everything. Whereas the underlying formalism doesn't care about our ideas of the construction being "spread out in time" (what the induction theorem says is that the whole thing we're defining just is there, period), it would certainly make it harder to think about if we had to imagine starting with the last elements for an inductive construction.
Note, by the way that if you combine the concepts and want to speak about ordered subsets where every nonempty subset has both a least and a greatest element, then you get a rather uninteresting theory out of it: Those orders are exactly the finite total orders. So it is essential with some asymmetry in the assumptions in order to get the rich theory of ordinals, even though it is formally irrelevant whether the asymmetry is to one side or the other.
answered Jul 14 at 16:06
Henning Makholm
226k16291520
3
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
1
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
add a comment |Â
up vote
8
down vote
accepted
In principle we could repeat the entire theory of well-orders with the words and relation signs flipped such that we speak about "greatest" elements of arbitrary nonempty subsets instead of "least".
Since formal logic doesn't care about the words we use, we would have exactly the same theorems in the new formulations, just in the different direction. But there are some pragmatic arguments for spending more attention on the usual concept of well-orders:
Well-orders have an intuitively appealing main example in the form of the natural numbers. There's nothing really comparable for co-well-orders.
It would turn the intuition about induction awry. One main use for well-orders is to construct things by long induction over the well order. Here you imagine constructing your thing starting with the first elements of the order and then your induction hypothesis tells you that there's no way you won't get a unique result for everything. Whereas the underlying formalism doesn't care about our ideas of the construction being "spread out in time" (what the induction theorem says is that the whole thing we're defining just is there, period), it would certainly make it harder to think about if we had to imagine starting with the last elements for an inductive construction.
Note, by the way that if you combine the concepts and want to speak about ordered subsets where every nonempty subset has both a least and a greatest element, then you get a rather uninteresting theory out of it: Those orders are exactly the finite total orders. So it is essential with some asymmetry in the assumptions in order to get the rich theory of ordinals, even though it is formally irrelevant whether the asymmetry is to one side or the other.
answered Jul 14 at 16:06
Henning Makholm
226k16291520
3
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
1
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
In principle we could repeat the entire theory of well-orders with the words and relation signs flipped such that we speak about "greatest" elements of arbitrary nonempty subsets instead of "least".
Since formal logic doesn't care about the words we use, we would have exactly the same theorems in the new formulations, just in the different direction. But there are some pragmatic arguments for spending more attention on the usual concept of well-orders:
Well-orders have an intuitively appealing main example in the form of the natural numbers. There's nothing really comparable for co-well-orders.
It would turn the intuition about induction awry. One main use for well-orders is to construct things by long induction over the well order. Here you imagine constructing your thing starting with the first elements of the order and then your induction hypothesis tells you that there's no way you won't get a unique result for everything. Whereas the underlying formalism doesn't care about our ideas of the construction being "spread out in time" (what the induction theorem says is that the whole thing we're defining just is there, period), it would certainly make it harder to think about if we had to imagine starting with the last elements for an inductive construction.
Note, by the way that if you combine the concepts and want to speak about ordered subsets where every nonempty subset has both a least and a greatest element, then you get a rather uninteresting theory out of it: Those orders are exactly the finite total orders. So it is essential with some asymmetry in the assumptions in order to get the rich theory of ordinals, even though it is formally irrelevant whether the asymmetry is to one side or the other.
answered Jul 14 at 16:06
Henning Makholm
226k16291520
In principle we could repeat the entire theory of well-orders with the words and relation signs flipped such that we speak about "greatest" elements of arbitrary nonempty subsets instead of "least".
Since formal logic doesn't care about the words we use, we would have exactly the same theorems in the new formulations, just in the different direction. But there are some pragmatic arguments for spending more attention on the usual concept of well-orders:
Well-orders have an intuitively appealing main example in the form of the natural numbers. There's nothing really comparable for co-well-orders.
It would turn the intuition about induction awry. One main use for well-orders is to construct things by long induction over the well order. Here you imagine constructing your thing starting with the first elements of the order and then your induction hypothesis tells you that there's no way you won't get a unique result for everything. Whereas the underlying formalism doesn't care about our ideas of the construction being "spread out in time" (what the induction theorem says is that the whole thing we're defining just is there, period), it would certainly make it harder to think about if we had to imagine starting with the last elements for an inductive construction.
Note, by the way that if you combine the concepts and want to speak about ordered subsets where every nonempty subset has both a least and a greatest element, then you get a rather uninteresting theory out of it: Those orders are exactly the finite total orders. So it is essential with some asymmetry in the assumptions in order to get the rich theory of ordinals, even though it is formally irrelevant whether the asymmetry is to one side or the other.
answered Jul 14 at 16:06
Henning Makholm
226k16291520
edited Jul 14 at 16:15
answered Jul 14 at 16:06
Henning Makholm
226k16291520
answered Jul 14 at 16:06
Henning Makholm
226k16291520
answered Jul 14 at 16:06
Henning Makholm
226k16291520
226k16291520
3
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
1
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
add a comment |Â
3
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
1
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
3
3
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
+1. To sketch (for the OP) a proof of the claim in the last paragraph, supposing $A$ is an infinite linear order we define $R_A=ain A: exists$ infinitely many $bin A$ with $b<_Aa$, and can now argue that either $R_A$ has no least element, or - taking $alpha$ to be the least element of $A$ - the set $bin A: b<_Aalpha$ has no greatest element, since otherwise that element would be $alpha$'s immediate predecessor and the immediate predecessor of any element of $R_A$ (if such a thing exists) is also in $R_A$.
– Noah Schweber
Jul 14 at 16:42
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
There is a famous German quote, usually attributed to Hilbert, that stresses the point that logic is blind to any naming convention: "Man muss jederzeit an Stelle von Punkt, Gerade, Ebene auch Tisch, Bank, Bierseidel sagen können"
– Stefan Mesken
Jul 15 at 11:27
1
1
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
@StefanMesken: Ein Tisch ist ein Tisch.
– Henning Makholm
Jul 15 at 11:40
add a comment |Â
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3
Having a least element is not a particularly interesting property. The fact that any non-empty subset has a least element is noteworthy.
– Saucy O'Path
Jul 14 at 14:57
1
There's nothing preventing you from flipping everything upside down and defining a reverse notion of well ordering in terms of all nonempty subsets having a greatest element, other than common sense and convention: the set of natural numbers is, almost by definition, a more natural exemplar of well ordering than the set of negative integers. (If you do want to define a reverse notion, perhaps it should be call "ill" ordering.)
– Barry Cipra
Jul 14 at 15:17