Mixing
Why do we use von Neumann ordinals and not Zermelo ordinals?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
Why do we use von Neumann ordinals,
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
edited 35 mins ago
Peter Mortensen
539310
asked 9 hours ago
Maicake
312
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.
add a comment |Â
up vote
5
down vote
favorite
Why do we use von Neumann ordinals,
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
edited 35 mins ago
Peter Mortensen
539310
asked 9 hours ago
Maicake
312
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.
2
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
9 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Why do we use von Neumann ordinals,
$$ 0 = emptyset $$
$$ n+1 = n cup n $$
and not Zermelo ordinals?
$$ 0 = emptyset $$
$$ n+1 = n $$
set-theory
edited 35 mins ago
Peter Mortensen
539310
asked 9 hours ago
Maicake
312
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 do 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
edited 35 mins ago
Peter Mortensen
539310
asked 9 hours ago
Maicake
312
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.
edited 35 mins ago
Peter Mortensen
539310
asked 9 hours ago
Maicake
312
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.
edited 35 mins ago
Peter Mortensen
539310
edited 35 mins ago
Peter Mortensen
539310
edited 35 mins ago
Peter Mortensen
539310
539310
asked 9 hours ago
Maicake
312
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 9 hours ago
Maicake
312
asked 9 hours ago
Maicake
312
312
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.
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.
Check out our Code of Conduct.
2
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
9 hours ago
add a comment |Â
2
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
9 hours ago
2
2
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
9 hours ago
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
9 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
18
down vote
accepted
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 9 hours ago
Mees de Vries
16k12654
13
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
2
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
add a comment |Â
up vote
2
down vote
A simple motivation for the Von Neumann style is to write it in this way:
$$
n := m<n.
$$
I.e., the ordinal $n$ is the set of all the ordinals up to $n-1$. This is, I'd say, the idea behind the Von Neumann ordinals, though obviously it isn't quite a proper definition ($m$ from what base set? What is $<$?).
Alternative way to express it:
$$beginalign
0 &:=
\ n+1 &:= 0 ldots n
endalign$$
answered 6 hours ago
leftaroundabout
3,5511528
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
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 9 hours ago
Mees de Vries
16k12654
13
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
2
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
add a comment |Â
up vote
18
down vote
accepted
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 9 hours ago
Mees de Vries
16k12654
13
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
2
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
add a comment |Â
up vote
18
down vote
accepted
up vote
18
down vote
accepted
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 9 hours ago
Mees de Vries
16k12654
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 9 hours ago
Mees de Vries
16k12654
answered 9 hours ago
Mees de Vries
16k12654
answered 9 hours ago
Mees de Vries
16k12654
answered 9 hours ago
Mees de Vries
16k12654
16k12654
13
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
2
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
add a comment |Â
13
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
2
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
13
13
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental.
– Asaf Karagila♦
9 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals?
– Maicake
8 hours ago
2
2
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
@Maicake $1 = emptyset not in emptyset = 3$ and $1 = emptyset not subseteq emptyset = 3$.
– Stefan Mesken
7 hours ago
add a comment |Â
up vote
2
down vote
A simple motivation for the Von Neumann style is to write it in this way:
$$
n := m<n.
$$
I.e., the ordinal $n$ is the set of all the ordinals up to $n-1$. This is, I'd say, the idea behind the Von Neumann ordinals, though obviously it isn't quite a proper definition ($m$ from what base set? What is $<$?).
Alternative way to express it:
$$beginalign
0 &:=
\ n+1 &:= 0 ldots n
endalign$$
answered 6 hours ago
leftaroundabout
3,5511528
add a comment |Â
up vote
2
down vote
A simple motivation for the Von Neumann style is to write it in this way:
$$
n := m<n.
$$
I.e., the ordinal $n$ is the set of all the ordinals up to $n-1$. This is, I'd say, the idea behind the Von Neumann ordinals, though obviously it isn't quite a proper definition ($m$ from what base set? What is $<$?).
Alternative way to express it:
$$beginalign
0 &:=
\ n+1 &:= 0 ldots n
endalign$$
answered 6 hours ago
leftaroundabout
3,5511528
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A simple motivation for the Von Neumann style is to write it in this way:
$$
n := m<n.
$$
I.e., the ordinal $n$ is the set of all the ordinals up to $n-1$. This is, I'd say, the idea behind the Von Neumann ordinals, though obviously it isn't quite a proper definition ($m$ from what base set? What is $<$?).
Alternative way to express it:
$$beginalign
0 &:=
\ n+1 &:= 0 ldots n
endalign$$
answered 6 hours ago
leftaroundabout
3,5511528
A simple motivation for the Von Neumann style is to write it in this way:
$$
n := m<n.
$$
I.e., the ordinal $n$ is the set of all the ordinals up to $n-1$. This is, I'd say, the idea behind the Von Neumann ordinals, though obviously it isn't quite a proper definition ($m$ from what base set? What is $<$?).
Alternative way to express it:
$$beginalign
0 &:=
\ n+1 &:= 0 ldots n
endalign$$
answered 6 hours ago
leftaroundabout
3,5511528
answered 6 hours ago
leftaroundabout
3,5511528
answered 6 hours ago
leftaroundabout
3,5511528
answered 6 hours ago
leftaroundabout
3,5511528
3,5511528
add a comment |Â
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.
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2975907%2fwhy-do-we-use-von-neumann-ordinals-and-not-zermelo-ordinals%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
math.stackexchange.com/questions/85672/…
– Asaf Karagila♦
9 hours ago