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Why are used Von Neumann ordinals and not Zermelo ordinals?
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Why we use Von Neumann ordinals
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
asked 27 mins ago
Maicake
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up vote
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Why we use Von Neumann ordinals
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
asked 27 mins ago
Maicake
61
New contributor
Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
12 mins ago
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
Why we use Von Neumann ordinals
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
asked 27 mins ago
Maicake
61
New contributor
Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Why we use Von Neumann ordinals
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
set-theory
asked 27 mins ago
Maicake
61
New contributor
Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 27 mins ago
Maicake
61
New contributor
Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 27 mins ago
Maicake
61
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Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 27 mins ago
Maicake
61
asked 27 mins ago
Maicake
61
61
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Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Maicake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
12 mins ago
add a comment |Â
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
12 mins ago
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
12 mins ago
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
12 mins ago
add a comment |Â
1 Answer
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There is no real, deep, fundamental reason. You can find a bijection between the set of Von Neumann naturals and Zermelo naturals, so anything you can do with the one set you can do with the other.
However, Von Neumann naturals are more convenient in practice for a lot of reasons. For one, the element we call $n$ also has exactly $n$ elements. That means that we can use the actual set $n$ as a cardinality, defining "the set $A$ has $n$ elements" to mean that there is a bijection between $A$ and $n$. For another, a convenient way to define the ordinals is to say that they are the transitive sets which are linearly ordered by $in$. Then the Von Neumann naturals are precisely the finite ordinals, which is a natural and important way to think about the finite ordinals.
answered 22 mins ago
Mees de Vries
15.9k12553
1
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There is no real, deep, fundamental reason. You can find a bijection between the set of Von Neumann naturals and Zermelo naturals, so anything you can do with the one set you can do with the other.
However, Von Neumann naturals are more convenient in practice for a lot of reasons. For one, the element we call $n$ also has exactly $n$ elements. That means that we can use the actual set $n$ as a cardinality, defining "the set $A$ has $n$ elements" to mean that there is a bijection between $A$ and $n$. For another, a convenient way to define the ordinals is to say that they are the transitive sets which are linearly ordered by $in$. Then the Von Neumann naturals are precisely the finite ordinals, which is a natural and important way to think about the finite ordinals.
answered 22 mins ago
Mees de Vries
15.9k12553
1
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
add a comment |Â
up vote
4
down vote
There is no real, deep, fundamental reason. You can find a bijection between the set of Von Neumann naturals and Zermelo naturals, so anything you can do with the one set you can do with the other.
However, Von Neumann naturals are more convenient in practice for a lot of reasons. For one, the element we call $n$ also has exactly $n$ elements. That means that we can use the actual set $n$ as a cardinality, defining "the set $A$ has $n$ elements" to mean that there is a bijection between $A$ and $n$. For another, a convenient way to define the ordinals is to say that they are the transitive sets which are linearly ordered by $in$. Then the Von Neumann naturals are precisely the finite ordinals, which is a natural and important way to think about the finite ordinals.
answered 22 mins ago
Mees de Vries
15.9k12553
1
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There is no real, deep, fundamental reason. You can find a bijection between the set of Von Neumann naturals and Zermelo naturals, so anything you can do with the one set you can do with the other.
However, Von Neumann naturals are more convenient in practice for a lot of reasons. For one, the element we call $n$ also has exactly $n$ elements. That means that we can use the actual set $n$ as a cardinality, defining "the set $A$ has $n$ elements" to mean that there is a bijection between $A$ and $n$. For another, a convenient way to define the ordinals is to say that they are the transitive sets which are linearly ordered by $in$. Then the Von Neumann naturals are precisely the finite ordinals, which is a natural and important way to think about the finite ordinals.
answered 22 mins ago
Mees de Vries
15.9k12553
There is no real, deep, fundamental reason. You can find a bijection between the set of Von Neumann naturals and Zermelo naturals, so anything you can do with the one set you can do with the other.
However, Von Neumann naturals are more convenient in practice for a lot of reasons. For one, the element we call $n$ also has exactly $n$ elements. That means that we can use the actual set $n$ as a cardinality, defining "the set $A$ has $n$ elements" to mean that there is a bijection between $A$ and $n$. For another, a convenient way to define the ordinals is to say that they are the transitive sets which are linearly ordered by $in$. Then the Von Neumann naturals are precisely the finite ordinals, which is a natural and important way to think about the finite ordinals.
answered 22 mins ago
Mees de Vries
15.9k12553
answered 22 mins ago
Mees de Vries
15.9k12553
answered 22 mins ago
Mees de Vries
15.9k12553
answered 22 mins ago
Mees de Vries
15.9k12553
15.9k12553
1
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
add a comment |Â
1
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
1
1
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
8 mins ago
add a comment |Â
Maicake is a new contributor. Be nice, and check out our Code of Conduct.
Maicake is a new contributor. Be nice, and check out our Code of Conduct.
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math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
12 mins ago