Is the function sending each set to its corresponding Hartogs number injective?
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Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.
As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.
I find it difficult to imagine the Hartogs numbers of infinite sets.
My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?
Thank you in advance for your help!
elementary-set-theory ordinals
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
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up vote
0
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Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.
As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.
I find it difficult to imagine the Hartogs numbers of infinite sets.
My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?
Thank you in advance for your help!
elementary-set-theory ordinals
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 at 7:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.
As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.
I find it difficult to imagine the Hartogs numbers of infinite sets.
My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?
Thank you in advance for your help!
elementary-set-theory ordinals
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
Let $V$ be the class of all sets, $text{Ord}$ be the class of all ordinals, $text{InOrd}$ be the class of all initial ordinals, and $f:V to text{Ord}$ be the function that sends each set to its corresponding Hartogs number.
As a result, $f(a)=a+1$ for all $a in Bbb N$ and thus $f restriction Bbb N$ is injective.
I find it difficult to imagine the Hartogs numbers of infinite sets.
My question: Are $f,frestriction text{Ord},frestriction text{InOrd}$ injective?
Thank you in advance for your help!
elementary-set-theory ordinals
elementary-set-theory ordinals
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
edited Dec 5 at 3:16
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
asked Dec 5 at 3:09
Le Anh Dung
1,0081421
1,0081421
What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 at 7:35
add a comment |
What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 at 7:35
What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 at 7:35
What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 at 7:35
add a comment |
2 Answers
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2
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The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?
The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:
Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.
We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.
But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.
Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.
answered Dec 5 at 3:24
Noah Schweber
120k10146278
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
add a comment |
up vote
0
down vote
Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.
answered Dec 5 at 3:25
Gödel
1,393319
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?
The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:
Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.
We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.
But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.
Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.
answered Dec 5 at 3:24
Noah Schweber
120k10146278
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
add a comment |
up vote
2
down vote
accepted
The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?
The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:
Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.
We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.
But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.
Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.
answered Dec 5 at 3:24
Noah Schweber
120k10146278
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?
The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:
Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.
We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.
But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.
Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.
answered Dec 5 at 3:24
Noah Schweber
120k10146278
The answers to $1$ and $2$ are no: for example, it's easy to show that $f(omega)=f(omega+1)=omega_1$ (where $omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?
The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:
Suppose $kappa<lambda$ are infinite cardinals and $f(kappa)=f(lambda)$.
We clearly have $lambda<f(lambda)$ (since there's a bijection between $lambda$ and $lambda+1$), so we get $lambda<f(kappa)$.
But then there must be a surjection from $kappa$ to $lambda$. So $f$ is injective on the initial ordinals.
Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.
answered Dec 5 at 3:24
Noah Schweber
120k10146278
answered Dec 5 at 3:24
Noah Schweber
120k10146278
answered Dec 5 at 3:24
Noah Schweber
120k10146278
answered Dec 5 at 3:24
Noah Schweber
120k10146278
120k10146278
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
add a comment |
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
Thank you so much for your detailed answer!
– Le Anh Dung
Dec 5 at 16:25
add a comment |
up vote
0
down vote
Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.
answered Dec 5 at 3:25
Gödel
1,393319
add a comment |
up vote
0
down vote
Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.
answered Dec 5 at 3:25
Gödel
1,393319
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.
answered Dec 5 at 3:25
Gödel
1,393319
Note that $f(omega)=f(omega+1)=omega_1$ and $omeganeqomega+1$.
answered Dec 5 at 3:25
Gödel
1,393319
answered Dec 5 at 3:25
Gödel
1,393319
answered Dec 5 at 3:25
Gödel
1,393319
answered Dec 5 at 3:25
Gödel
1,393319
1,393319
add a comment |
add a comment |
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What is the reason for downvoting my question? Did I do anything wrong?
– Le Anh Dung
Dec 5 at 7:35