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Example of Set which possesses well ordering property other than Integers
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During Studying Elementary Number theory I had encountered in property called as well ordering property which tell every nonempty set of natural number has least element.
I had interested in is such property hold for any other set.If not then How to prove that. As I had checked one question A well-order on a uncountable set but I had not understood.
Any Help will be appreciated.
real-analysis number-theory well-orders
asked Jul 31 at 11:39
SRJ
1,014317
add a comment |Â
up vote
3
down vote
favorite
During Studying Elementary Number theory I had encountered in property called as well ordering property which tell every nonempty set of natural number has least element.
I had interested in is such property hold for any other set.If not then How to prove that. As I had checked one question A well-order on a uncountable set but I had not understood.
Any Help will be appreciated.
real-analysis number-theory well-orders
asked Jul 31 at 11:39
SRJ
1,014317
1
All proper subsets of the naturals are also well ordered. If you take the set $A=mathbbNcupx$ for $xnotin mathbbN$, and define $n<x$ for all $ninmathbbN$. Then $A$ is well ordered too. In both of these examples, the resulting sets are not order isomorphic to the natural numbers. For uncountable sets, since you can't just exhibit it, you have to prove its existence from rather bold axioms or postulate it.
– spiralstotheleft
Jul 31 at 11:44
1
The property of being well-ordered crucially depends on the fact that there is an "initial" natural number : $0$. With the "standard" ordering : $le$ neither the integers nor the reals are well-ordered.
– Mauro ALLEGRANZA
Jul 31 at 11:51
As mentioned in the question you referenced, there is an assertion called the well-ordering principle which states that every non-empty set can be well ordered. This turns out to be equivalent(given some conditions) to the axiom of choice, an important assertion in axiomatic set theory. So depending on the set theoretic foundations you might choose, the answer to your question can vary.
– zzuussee
Jul 31 at 11:53
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
During Studying Elementary Number theory I had encountered in property called as well ordering property which tell every nonempty set of natural number has least element.
I had interested in is such property hold for any other set.If not then How to prove that. As I had checked one question A well-order on a uncountable set but I had not understood.
Any Help will be appreciated.
real-analysis number-theory well-orders
asked Jul 31 at 11:39
SRJ
1,014317
During Studying Elementary Number theory I had encountered in property called as well ordering property which tell every nonempty set of natural number has least element.
I had interested in is such property hold for any other set.If not then How to prove that. As I had checked one question A well-order on a uncountable set but I had not understood.
Any Help will be appreciated.
real-analysis number-theory well-orders
asked Jul 31 at 11:39
SRJ
1,014317
asked Jul 31 at 11:39
SRJ
1,014317
asked Jul 31 at 11:39
SRJ
1,014317
asked Jul 31 at 11:39
SRJ
1,014317
1,014317
1
All proper subsets of the naturals are also well ordered. If you take the set $A=mathbbNcupx$ for $xnotin mathbbN$, and define $n<x$ for all $ninmathbbN$. Then $A$ is well ordered too. In both of these examples, the resulting sets are not order isomorphic to the natural numbers. For uncountable sets, since you can't just exhibit it, you have to prove its existence from rather bold axioms or postulate it.
– spiralstotheleft
Jul 31 at 11:44
1
The property of being well-ordered crucially depends on the fact that there is an "initial" natural number : $0$. With the "standard" ordering : $le$ neither the integers nor the reals are well-ordered.
– Mauro ALLEGRANZA
Jul 31 at 11:51
As mentioned in the question you referenced, there is an assertion called the well-ordering principle which states that every non-empty set can be well ordered. This turns out to be equivalent(given some conditions) to the axiom of choice, an important assertion in axiomatic set theory. So depending on the set theoretic foundations you might choose, the answer to your question can vary.
– zzuussee
Jul 31 at 11:53
add a comment |Â
1
All proper subsets of the naturals are also well ordered. If you take the set $A=mathbbNcupx$ for $xnotin mathbbN$, and define $n<x$ for all $ninmathbbN$. Then $A$ is well ordered too. In both of these examples, the resulting sets are not order isomorphic to the natural numbers. For uncountable sets, since you can't just exhibit it, you have to prove its existence from rather bold axioms or postulate it.
– spiralstotheleft
Jul 31 at 11:44
1
The property of being well-ordered crucially depends on the fact that there is an "initial" natural number : $0$. With the "standard" ordering : $le$ neither the integers nor the reals are well-ordered.
– Mauro ALLEGRANZA
Jul 31 at 11:51
As mentioned in the question you referenced, there is an assertion called the well-ordering principle which states that every non-empty set can be well ordered. This turns out to be equivalent(given some conditions) to the axiom of choice, an important assertion in axiomatic set theory. So depending on the set theoretic foundations you might choose, the answer to your question can vary.
– zzuussee
Jul 31 at 11:53
1
1
All proper subsets of the naturals are also well ordered. If you take the set $A=mathbbNcupx$ for $xnotin mathbbN$, and define $n<x$ for all $ninmathbbN$. Then $A$ is well ordered too. In both of these examples, the resulting sets are not order isomorphic to the natural numbers. For uncountable sets, since you can't just exhibit it, you have to prove its existence from rather bold axioms or postulate it.
– spiralstotheleft
Jul 31 at 11:44
All proper subsets of the naturals are also well ordered. If you take the set $A=mathbbNcupx$ for $xnotin mathbbN$, and define $n<x$ for all $ninmathbbN$. Then $A$ is well ordered too. In both of these examples, the resulting sets are not order isomorphic to the natural numbers. For uncountable sets, since you can't just exhibit it, you have to prove its existence from rather bold axioms or postulate it.
– spiralstotheleft
Jul 31 at 11:44
1
1
The property of being well-ordered crucially depends on the fact that there is an "initial" natural number : $0$. With the "standard" ordering : $le$ neither the integers nor the reals are well-ordered.
– Mauro ALLEGRANZA
Jul 31 at 11:51
The property of being well-ordered crucially depends on the fact that there is an "initial" natural number : $0$. With the "standard" ordering : $le$ neither the integers nor the reals are well-ordered.
– Mauro ALLEGRANZA
Jul 31 at 11:51
As mentioned in the question you referenced, there is an assertion called the well-ordering principle which states that every non-empty set can be well ordered. This turns out to be equivalent(given some conditions) to the axiom of choice, an important assertion in axiomatic set theory. So depending on the set theoretic foundations you might choose, the answer to your question can vary.
– zzuussee
Jul 31 at 11:53
As mentioned in the question you referenced, there is an assertion called the well-ordering principle which states that every non-empty set can be well ordered. This turns out to be equivalent(given some conditions) to the axiom of choice, an important assertion in axiomatic set theory. So depending on the set theoretic foundations you might choose, the answer to your question can vary.
– zzuussee
Jul 31 at 11:53
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
What you call "well ordering property" is probably the well ordering principle and is formulated explicitly for the set of positive integers.
"I had interested in is such property hold for any other set"
It can be projected, but not purely on sets. For that you need sets that are equipped with an ordering.
The axiom of choice is in the context of ZF equivalent with the statement that every set can be equipped with such an ordering, which is a so-called well-ordering.
edited Jul 31 at 12:23
Bernard
110k635102
answered Jul 31 at 11:52
drhab
85.8k540118
add a comment |Â
up vote
1
down vote
Here's another example to think about. The underlying set is the natural numbers, but we redefine the ordering: if $m$ and $n$ are both odd or both even then $m "<" n$ just when $m < n$ in the usual sense, but every even number is (by definition) $"<"$ than every odd number. The ordering looks like this
$$
0,2,4, ldots 1, 3, 5, ldots
$$
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
add a comment |Â
up vote
0
down vote
Let $N$ be a countable set. This means that there's a bijection $bcolonmathbbNlongrightarrow N$. Define this order relation on $N$: $aleqslant a'$ if and only if $b^-1(a)leqslant b^-1(a')$. Then $(N,leqslant)$ is a well-ordered set.
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
What you call "well ordering property" is probably the well ordering principle and is formulated explicitly for the set of positive integers.
"I had interested in is such property hold for any other set"
It can be projected, but not purely on sets. For that you need sets that are equipped with an ordering.
The axiom of choice is in the context of ZF equivalent with the statement that every set can be equipped with such an ordering, which is a so-called well-ordering.
edited Jul 31 at 12:23
Bernard
110k635102
answered Jul 31 at 11:52
drhab
85.8k540118
add a comment |Â
up vote
2
down vote
What you call "well ordering property" is probably the well ordering principle and is formulated explicitly for the set of positive integers.
"I had interested in is such property hold for any other set"
It can be projected, but not purely on sets. For that you need sets that are equipped with an ordering.
The axiom of choice is in the context of ZF equivalent with the statement that every set can be equipped with such an ordering, which is a so-called well-ordering.
edited Jul 31 at 12:23
Bernard
110k635102
answered Jul 31 at 11:52
drhab
85.8k540118
add a comment |Â
up vote
2
down vote
up vote
2
down vote
What you call "well ordering property" is probably the well ordering principle and is formulated explicitly for the set of positive integers.
"I had interested in is such property hold for any other set"
It can be projected, but not purely on sets. For that you need sets that are equipped with an ordering.
The axiom of choice is in the context of ZF equivalent with the statement that every set can be equipped with such an ordering, which is a so-called well-ordering.
edited Jul 31 at 12:23
Bernard
110k635102
answered Jul 31 at 11:52
drhab
85.8k540118
What you call "well ordering property" is probably the well ordering principle and is formulated explicitly for the set of positive integers.
"I had interested in is such property hold for any other set"
It can be projected, but not purely on sets. For that you need sets that are equipped with an ordering.
The axiom of choice is in the context of ZF equivalent with the statement that every set can be equipped with such an ordering, which is a so-called well-ordering.
edited Jul 31 at 12:23
Bernard
110k635102
answered Jul 31 at 11:52
drhab
85.8k540118
edited Jul 31 at 12:23
Bernard
110k635102
edited Jul 31 at 12:23
Bernard
110k635102
edited Jul 31 at 12:23
Bernard
110k635102
110k635102
answered Jul 31 at 11:52
drhab
85.8k540118
answered Jul 31 at 11:52
drhab
85.8k540118
answered Jul 31 at 11:52
drhab
85.8k540118
85.8k540118
add a comment |Â
add a comment |Â
up vote
1
down vote
Here's another example to think about. The underlying set is the natural numbers, but we redefine the ordering: if $m$ and $n$ are both odd or both even then $m "<" n$ just when $m < n$ in the usual sense, but every even number is (by definition) $"<"$ than every odd number. The ordering looks like this
$$
0,2,4, ldots 1, 3, 5, ldots
$$
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
add a comment |Â
up vote
1
down vote
Here's another example to think about. The underlying set is the natural numbers, but we redefine the ordering: if $m$ and $n$ are both odd or both even then $m "<" n$ just when $m < n$ in the usual sense, but every even number is (by definition) $"<"$ than every odd number. The ordering looks like this
$$
0,2,4, ldots 1, 3, 5, ldots
$$
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here's another example to think about. The underlying set is the natural numbers, but we redefine the ordering: if $m$ and $n$ are both odd or both even then $m "<" n$ just when $m < n$ in the usual sense, but every even number is (by definition) $"<"$ than every odd number. The ordering looks like this
$$
0,2,4, ldots 1, 3, 5, ldots
$$
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
Here's another example to think about. The underlying set is the natural numbers, but we redefine the ordering: if $m$ and $n$ are both odd or both even then $m "<" n$ just when $m < n$ in the usual sense, but every even number is (by definition) $"<"$ than every odd number. The ordering looks like this
$$
0,2,4, ldots 1, 3, 5, ldots
$$
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
answered Jul 31 at 12:29
Ethan Bolker
35.7k54199
35.7k54199
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $N$ be a countable set. This means that there's a bijection $bcolonmathbbNlongrightarrow N$. Define this order relation on $N$: $aleqslant a'$ if and only if $b^-1(a)leqslant b^-1(a')$. Then $(N,leqslant)$ is a well-ordered set.
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
add a comment |Â
up vote
0
down vote
Let $N$ be a countable set. This means that there's a bijection $bcolonmathbbNlongrightarrow N$. Define this order relation on $N$: $aleqslant a'$ if and only if $b^-1(a)leqslant b^-1(a')$. Then $(N,leqslant)$ is a well-ordered set.
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $N$ be a countable set. This means that there's a bijection $bcolonmathbbNlongrightarrow N$. Define this order relation on $N$: $aleqslant a'$ if and only if $b^-1(a)leqslant b^-1(a')$. Then $(N,leqslant)$ is a well-ordered set.
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
Let $N$ be a countable set. This means that there's a bijection $bcolonmathbbNlongrightarrow N$. Define this order relation on $N$: $aleqslant a'$ if and only if $b^-1(a)leqslant b^-1(a')$. Then $(N,leqslant)$ is a well-ordered set.
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
answered Jul 31 at 11:47
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
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1
All proper subsets of the naturals are also well ordered. If you take the set $A=mathbbNcupx$ for $xnotin mathbbN$, and define $n<x$ for all $ninmathbbN$. Then $A$ is well ordered too. In both of these examples, the resulting sets are not order isomorphic to the natural numbers. For uncountable sets, since you can't just exhibit it, you have to prove its existence from rather bold axioms or postulate it.
– spiralstotheleft
Jul 31 at 11:44
1
The property of being well-ordered crucially depends on the fact that there is an "initial" natural number : $0$. With the "standard" ordering : $le$ neither the integers nor the reals are well-ordered.
– Mauro ALLEGRANZA
Jul 31 at 11:51
As mentioned in the question you referenced, there is an assertion called the well-ordering principle which states that every non-empty set can be well ordered. This turns out to be equivalent(given some conditions) to the axiom of choice, an important assertion in axiomatic set theory. So depending on the set theoretic foundations you might choose, the answer to your question can vary.
– zzuussee
Jul 31 at 11:53