Mixing
Is this a typo in the hint for theorem “The complete ordered field is unique up to isomorphism”
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In Chapter 10. Sets of Real Numbers from textbook Introduction to Set Theory by Hrbacek and Jech, the authors mention:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be the structure of real numbers. Then the closure of ${0,1}$ under $+$ and $cdot$ in $mathfrak{R}$ is $Bbb N$. As such, I guess $mathfrak{C}$ should be isomorphic to $mathfrak{N}=langle Bbb N,<,+,cdot,0,1 rangle$, rather than to $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$ as mentioned by the authors.
Please verify my observation! Thank you so much!
elementary-set-theory real-numbers
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
|
show 4 more comments
$begingroup$
In Chapter 10. Sets of Real Numbers from textbook Introduction to Set Theory by Hrbacek and Jech, the authors mention:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be the structure of real numbers. Then the closure of ${0,1}$ under $+$ and $cdot$ in $mathfrak{R}$ is $Bbb N$. As such, I guess $mathfrak{C}$ should be isomorphic to $mathfrak{N}=langle Bbb N,<,+,cdot,0,1 rangle$, rather than to $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$ as mentioned by the authors.
Please verify my observation! Thank you so much!
elementary-set-theory real-numbers
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
$begingroup$
Perhaps the book makes the (nonstandard) definition that the closure of a set under some group operations also must be closed under their corresponding inverses?
$endgroup$
– Micah
Jan 6 at 4:05
$begingroup$
Hi @Micah, I have posted a related question here in which i also quoted the authors' definition of closure. Please have a look for reference!
$endgroup$
– Le Anh Dung
Jan 6 at 4:08
3
$begingroup$
You want an ordered field. The natural numbers are not a field. You have to include multiplicative inverses so you get the rationals.
$endgroup$
– John Douma
Jan 6 at 4:10
$begingroup$
You're right. The author was so immersed in fields he forgot to clarify that the closure is a field closure. If it were a group or ring closure it'd be Z instead of N.
$endgroup$
– William Elliot
Jan 6 at 4:57
1
$begingroup$
No. The author did not make a mistake. You are trying to calculate the smallest ordered field that contains $0$ and $1$. The division sign is not part of the language of fields.
$endgroup$
– John Douma
Jan 6 at 7:03
|
show 4 more comments
$begingroup$
In Chapter 10. Sets of Real Numbers from textbook Introduction to Set Theory by Hrbacek and Jech, the authors mention:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be the structure of real numbers. Then the closure of ${0,1}$ under $+$ and $cdot$ in $mathfrak{R}$ is $Bbb N$. As such, I guess $mathfrak{C}$ should be isomorphic to $mathfrak{N}=langle Bbb N,<,+,cdot,0,1 rangle$, rather than to $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$ as mentioned by the authors.
Please verify my observation! Thank you so much!
elementary-set-theory real-numbers
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
In Chapter 10. Sets of Real Numbers from textbook Introduction to Set Theory by Hrbacek and Jech, the authors mention:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ be the structure of real numbers. Then the closure of ${0,1}$ under $+$ and $cdot$ in $mathfrak{R}$ is $Bbb N$. As such, I guess $mathfrak{C}$ should be isomorphic to $mathfrak{N}=langle Bbb N,<,+,cdot,0,1 rangle$, rather than to $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$ as mentioned by the authors.
Please verify my observation! Thank you so much!
elementary-set-theory real-numbers
elementary-set-theory real-numbers
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 3:56
Le Anh DungLe Anh Dung
1,0291521
1,0291521
$begingroup$
Perhaps the book makes the (nonstandard) definition that the closure of a set under some group operations also must be closed under their corresponding inverses?
$endgroup$
– Micah
Jan 6 at 4:05
$begingroup$
Hi @Micah, I have posted a related question here in which i also quoted the authors' definition of closure. Please have a look for reference!
$endgroup$
– Le Anh Dung
Jan 6 at 4:08
3
$begingroup$
You want an ordered field. The natural numbers are not a field. You have to include multiplicative inverses so you get the rationals.
$endgroup$
– John Douma
Jan 6 at 4:10
$begingroup$
You're right. The author was so immersed in fields he forgot to clarify that the closure is a field closure. If it were a group or ring closure it'd be Z instead of N.
$endgroup$
– William Elliot
Jan 6 at 4:57
1
$begingroup$
No. The author did not make a mistake. You are trying to calculate the smallest ordered field that contains $0$ and $1$. The division sign is not part of the language of fields.
$endgroup$
– John Douma
Jan 6 at 7:03
|
show 4 more comments
$begingroup$
Perhaps the book makes the (nonstandard) definition that the closure of a set under some group operations also must be closed under their corresponding inverses?
$endgroup$
– Micah
Jan 6 at 4:05
$begingroup$
Hi @Micah, I have posted a related question here in which i also quoted the authors' definition of closure. Please have a look for reference!
$endgroup$
– Le Anh Dung
Jan 6 at 4:08
3
$begingroup$
You want an ordered field. The natural numbers are not a field. You have to include multiplicative inverses so you get the rationals.
$endgroup$
– John Douma
Jan 6 at 4:10
$begingroup$
You're right. The author was so immersed in fields he forgot to clarify that the closure is a field closure. If it were a group or ring closure it'd be Z instead of N.
$endgroup$
– William Elliot
Jan 6 at 4:57
1
$begingroup$
No. The author did not make a mistake. You are trying to calculate the smallest ordered field that contains $0$ and $1$. The division sign is not part of the language of fields.
$endgroup$
– John Douma
Jan 6 at 7:03
$begingroup$
Perhaps the book makes the (nonstandard) definition that the closure of a set under some group operations also must be closed under their corresponding inverses?
$endgroup$
– Micah
Jan 6 at 4:05
$begingroup$
Perhaps the book makes the (nonstandard) definition that the closure of a set under some group operations also must be closed under their corresponding inverses?
$endgroup$
– Micah
Jan 6 at 4:05
$begingroup$
Hi @Micah, I have posted a related question here in which i also quoted the authors' definition of closure. Please have a look for reference!
$endgroup$
– Le Anh Dung
Jan 6 at 4:08
$begingroup$
Hi @Micah, I have posted a related question here in which i also quoted the authors' definition of closure. Please have a look for reference!
$endgroup$
– Le Anh Dung
Jan 6 at 4:08
3
3
$begingroup$
You want an ordered field. The natural numbers are not a field. You have to include multiplicative inverses so you get the rationals.
$endgroup$
– John Douma
Jan 6 at 4:10
$begingroup$
You want an ordered field. The natural numbers are not a field. You have to include multiplicative inverses so you get the rationals.
$endgroup$
– John Douma
Jan 6 at 4:10
$begingroup$
You're right. The author was so immersed in fields he forgot to clarify that the closure is a field closure. If it were a group or ring closure it'd be Z instead of N.
$endgroup$
– William Elliot
Jan 6 at 4:57
$begingroup$
You're right. The author was so immersed in fields he forgot to clarify that the closure is a field closure. If it were a group or ring closure it'd be Z instead of N.
$endgroup$
– William Elliot
Jan 6 at 4:57
1
1
$begingroup$
No. The author did not make a mistake. You are trying to calculate the smallest ordered field that contains $0$ and $1$. The division sign is not part of the language of fields.
$endgroup$
– John Douma
Jan 6 at 7:03
$begingroup$
No. The author did not make a mistake. You are trying to calculate the smallest ordered field that contains $0$ and $1$. The division sign is not part of the language of fields.
$endgroup$
– John Douma
Jan 6 at 7:03
|
show 4 more comments
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$begingroup$
Perhaps the book makes the (nonstandard) definition that the closure of a set under some group operations also must be closed under their corresponding inverses?
$endgroup$
– Micah
Jan 6 at 4:05
$begingroup$
Hi @Micah, I have posted a related question here in which i also quoted the authors' definition of closure. Please have a look for reference!
$endgroup$
– Le Anh Dung
Jan 6 at 4:08
3
$begingroup$
You want an ordered field. The natural numbers are not a field. You have to include multiplicative inverses so you get the rationals.
$endgroup$
– John Douma
Jan 6 at 4:10
$begingroup$
You're right. The author was so immersed in fields he forgot to clarify that the closure is a field closure. If it were a group or ring closure it'd be Z instead of N.
$endgroup$
– William Elliot
Jan 6 at 4:57
1
$begingroup$
No. The author did not make a mistake. You are trying to calculate the smallest ordered field that contains $0$ and $1$. The division sign is not part of the language of fields.
$endgroup$
– John Douma
Jan 6 at 7:03