Mixing
A possible mistake in the statement “The set of all rationals $Bbb Q$ is the closure of $emptyset$ in $(Bbb...
0
$begingroup$
In Chapter 5. Operations and Structures from textbook Introduction to Set Theory by Hrbacek and Jech, the authors define closure as follows
Then the authors mention that
The set of all rationals $Bbb Q$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
I think this statement is possibly wrong. I think it should be
$emptyset$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
Could you please verify my observation? Thank you for your help!
elementary-set-theory
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
add a comment |
0
$begingroup$
In Chapter 5. Operations and Structures from textbook Introduction to Set Theory by Hrbacek and Jech, the authors define closure as follows
Then the authors mention that
The set of all rationals $Bbb Q$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
I think this statement is possibly wrong. I think it should be
$emptyset$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
Could you please verify my observation? Thank you for your help!
elementary-set-theory
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
2
$begingroup$
"...all constants of A belong to B". If one believes this phrase, one should believe that 0 and 1 belong to any closed set of the structure R. In particular, the emptyset is not closed, hence it cannot be the closure of the emptyset.
$endgroup$
– BUI Quang-Tu
Jan 6 at 2:10
$begingroup$
Hi @BUIQuang-Tu, I don't understand whether the statement all constant of $mathfrak{A}$ belongs to $B$ is a condition in the definition of closed, or it is just a corollary follows from that definition.
$endgroup$
– Le Anh Dung
Jan 6 at 3:05
$begingroup$
By definition of closed which is "A set $Bsubseteq A$ is called closed if the result of applying any operation to elements of $B$ is again in $B$, i.e., if for all $jle n-1$ and for all $a_0,cdots, a_{f_{j-1}}in B, F_j(a_0,cdots, a_{f_{j-1}})in B$ provided that it is defined", I can not infer how all constants of $mathfrak{A}$ belong to $B$. For reference, in the textbook, the authors identify $0$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,0)}$ and constant $1$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,1)}$. Please shed some lights!
$endgroup$
– Le Anh Dung
Jan 6 at 3:24
add a comment |
0
0
0
2
$begingroup$
In Chapter 5. Operations and Structures from textbook Introduction to Set Theory by Hrbacek and Jech, the authors define closure as follows
Then the authors mention that
The set of all rationals $Bbb Q$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
I think this statement is possibly wrong. I think it should be
$emptyset$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
Could you please verify my observation? Thank you for your help!
elementary-set-theory
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
In Chapter 5. Operations and Structures from textbook Introduction to Set Theory by Hrbacek and Jech, the authors define closure as follows
Then the authors mention that
The set of all rationals $Bbb Q$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
I think this statement is possibly wrong. I think it should be
$emptyset$ is the closure of $emptyset$ in $(Bbb R,0,1,+,-,times,div)$.
Could you please verify my observation? Thank you for your help!
elementary-set-theory
elementary-set-theory
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
asked Jan 6 at 2:04
Le Anh DungLe Anh Dung
1,0291521
1,0291521
2
$begingroup$
"...all constants of A belong to B". If one believes this phrase, one should believe that 0 and 1 belong to any closed set of the structure R. In particular, the emptyset is not closed, hence it cannot be the closure of the emptyset.
$endgroup$
– BUI Quang-Tu
Jan 6 at 2:10
$begingroup$
Hi @BUIQuang-Tu, I don't understand whether the statement all constant of $mathfrak{A}$ belongs to $B$ is a condition in the definition of closed, or it is just a corollary follows from that definition.
$endgroup$
– Le Anh Dung
Jan 6 at 3:05
$begingroup$
By definition of closed which is "A set $Bsubseteq A$ is called closed if the result of applying any operation to elements of $B$ is again in $B$, i.e., if for all $jle n-1$ and for all $a_0,cdots, a_{f_{j-1}}in B, F_j(a_0,cdots, a_{f_{j-1}})in B$ provided that it is defined", I can not infer how all constants of $mathfrak{A}$ belong to $B$. For reference, in the textbook, the authors identify $0$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,0)}$ and constant $1$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,1)}$. Please shed some lights!
$endgroup$
– Le Anh Dung
Jan 6 at 3:24
add a comment |
2
$begingroup$
"...all constants of A belong to B". If one believes this phrase, one should believe that 0 and 1 belong to any closed set of the structure R. In particular, the emptyset is not closed, hence it cannot be the closure of the emptyset.
$endgroup$
– BUI Quang-Tu
Jan 6 at 2:10
$begingroup$
Hi @BUIQuang-Tu, I don't understand whether the statement all constant of $mathfrak{A}$ belongs to $B$ is a condition in the definition of closed, or it is just a corollary follows from that definition.
$endgroup$
– Le Anh Dung
Jan 6 at 3:05
$begingroup$
By definition of closed which is "A set $Bsubseteq A$ is called closed if the result of applying any operation to elements of $B$ is again in $B$, i.e., if for all $jle n-1$ and for all $a_0,cdots, a_{f_{j-1}}in B, F_j(a_0,cdots, a_{f_{j-1}})in B$ provided that it is defined", I can not infer how all constants of $mathfrak{A}$ belong to $B$. For reference, in the textbook, the authors identify $0$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,0)}$ and constant $1$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,1)}$. Please shed some lights!
$endgroup$
– Le Anh Dung
Jan 6 at 3:24
2
2
$begingroup$
"...all constants of A belong to B". If one believes this phrase, one should believe that 0 and 1 belong to any closed set of the structure R. In particular, the emptyset is not closed, hence it cannot be the closure of the emptyset.
$endgroup$
– BUI Quang-Tu
Jan 6 at 2:10
$begingroup$
"...all constants of A belong to B". If one believes this phrase, one should believe that 0 and 1 belong to any closed set of the structure R. In particular, the emptyset is not closed, hence it cannot be the closure of the emptyset.
$endgroup$
– BUI Quang-Tu
Jan 6 at 2:10
$begingroup$
Hi @BUIQuang-Tu, I don't understand whether the statement all constant of $mathfrak{A}$ belongs to $B$ is a condition in the definition of closed, or it is just a corollary follows from that definition.
$endgroup$
– Le Anh Dung
Jan 6 at 3:05
$begingroup$
Hi @BUIQuang-Tu, I don't understand whether the statement all constant of $mathfrak{A}$ belongs to $B$ is a condition in the definition of closed, or it is just a corollary follows from that definition.
$endgroup$
– Le Anh Dung
Jan 6 at 3:05
$begingroup$
By definition of closed which is "A set $Bsubseteq A$ is called closed if the result of applying any operation to elements of $B$ is again in $B$, i.e., if for all $jle n-1$ and for all $a_0,cdots, a_{f_{j-1}}in B, F_j(a_0,cdots, a_{f_{j-1}})in B$ provided that it is defined", I can not infer how all constants of $mathfrak{A}$ belong to $B$. For reference, in the textbook, the authors identify $0$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,0)}$ and constant $1$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,1)}$. Please shed some lights!
$endgroup$
– Le Anh Dung
Jan 6 at 3:24
$begingroup$
By definition of closed which is "A set $Bsubseteq A$ is called closed if the result of applying any operation to elements of $B$ is again in $B$, i.e., if for all $jle n-1$ and for all $a_0,cdots, a_{f_{j-1}}in B, F_j(a_0,cdots, a_{f_{j-1}})in B$ provided that it is defined", I can not infer how all constants of $mathfrak{A}$ belong to $B$. For reference, in the textbook, the authors identify $0$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,0)}$ and constant $1$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,1)}$. Please shed some lights!
$endgroup$
– Le Anh Dung
Jan 6 at 3:24
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
Note that the definition of "closed" forces $0,1$ to be in the closure of $emptyset$. Once you have those, then by repeating $+$, you get all of $mathbb{N}$, etc.
In summary, $mathbb{Q}$ is the correct answer.
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
$endgroup$
1
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
1
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
1
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
1
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
1
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
|
show 7 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
Note that the definition of "closed" forces $0,1$ to be in the closure of $emptyset$. Once you have those, then by repeating $+$, you get all of $mathbb{N}$, etc.
In summary, $mathbb{Q}$ is the correct answer.
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
$endgroup$
1
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
1
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
1
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
1
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
1
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
|
show 7 more comments
5
$begingroup$
Note that the definition of "closed" forces $0,1$ to be in the closure of $emptyset$. Once you have those, then by repeating $+$, you get all of $mathbb{N}$, etc.
In summary, $mathbb{Q}$ is the correct answer.
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
$endgroup$
1
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
1
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
1
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
1
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
1
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
|
show 7 more comments
5
5
5
$begingroup$
Note that the definition of "closed" forces $0,1$ to be in the closure of $emptyset$. Once you have those, then by repeating $+$, you get all of $mathbb{N}$, etc.
In summary, $mathbb{Q}$ is the correct answer.
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
$endgroup$
Note that the definition of "closed" forces $0,1$ to be in the closure of $emptyset$. Once you have those, then by repeating $+$, you get all of $mathbb{N}$, etc.
In summary, $mathbb{Q}$ is the correct answer.
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
edited Jan 6 at 2:16
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
answered Jan 6 at 2:08
vadim123vadim123
75.9k897189
75.9k897189
1
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
1
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
1
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
1
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
1
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
|
show 7 more comments
1
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
1
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
1
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
1
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
1
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
1
1
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
$begingroup$
A constant is the result of a 0-argument operation.
$endgroup$
– Ted
Jan 6 at 4:01
1
1
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
$begingroup$
@LeAnhDung, all constants belonging to the closure is part of the definition, as quoted by you. Whether this part of the definition is redundant (and can be inferred from the rest of the definition) or not is immaterial. Your claim that $emptyset$ is closed is incorrect.
$endgroup$
– vadim123
Jan 6 at 4:13
1
1
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
$begingroup$
@LeAnhDung The domain of an $n$-ary operation on $A$ is the set of all $n$-tuples on $A$. If $B subset A$, then $B$ is closed under this operation if for every $n$-tuple all of whose elements belong $B$, applying this operation gives an element of $B$. A constant $c$ is the 0-ary operation which produces the result $c$. If $B = emptyset$, then it is not closed under this 0-ary operation because all elements of the 0-tuple belong to $B$ vacuously (since the 0-tuple has no elements) but the result $c$ does not belong to $B$.
$endgroup$
– Ted
Jan 6 at 4:49
1
1
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
$begingroup$
Closure says that if a particular tuple in the domain of the operation has all its elements belonging to $B$, then the result of applying the operation to that tuple is also in $B$. The fact that $emptyset not in emptyset$ is not relevant here, because $emptyset$ is not an element of any tuple in the domain of the 0-ary operation: $emptyset$ is the 0-tuple, which has no elements.
$endgroup$
– Ted
Jan 6 at 5:02
1
1
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
$begingroup$
@LeAnhDung Yes, that is exactly correct!
$endgroup$
– Ted
Jan 6 at 6:02
|
show 7 more comments
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"...all constants of A belong to B". If one believes this phrase, one should believe that 0 and 1 belong to any closed set of the structure R. In particular, the emptyset is not closed, hence it cannot be the closure of the emptyset.
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– BUI Quang-Tu
Jan 6 at 2:10
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Hi @BUIQuang-Tu, I don't understand whether the statement all constant of $mathfrak{A}$ belongs to $B$ is a condition in the definition of closed, or it is just a corollary follows from that definition.
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– Le Anh Dung
Jan 6 at 3:05
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By definition of closed which is "A set $Bsubseteq A$ is called closed if the result of applying any operation to elements of $B$ is again in $B$, i.e., if for all $jle n-1$ and for all $a_0,cdots, a_{f_{j-1}}in B, F_j(a_0,cdots, a_{f_{j-1}})in B$ provided that it is defined", I can not infer how all constants of $mathfrak{A}$ belong to $B$. For reference, in the textbook, the authors identify $0$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,0)}$ and constant $1$ in $(Bbb R,0,1,+,-,times,div)$ with operation ${(emptyset,1)}$. Please shed some lights!
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– Le Anh Dung
Jan 6 at 3:24