Need help proving that the set of natural numbers is transitive.
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The problem reads: Prove that $mathbb{N}$ is transitive.
With the definition of a transitive set that im using is $forall x in X$, $x subset X$.
I cant help but feel that i'm missing something. Any help/advice is greatly appreciated. This is what i have so far:
Let $A$ be the set of all transitive natural numbers
$n in A$, iff $n in mathbb{N}$, $x subset n$, $forall x in n$
$ 0 in A $
Suppose $n in A$, consider $n^+ = n cup$ {n}
if $x in n^+$, then $x in n$ or $x=n$
If $x in n$, then $x subset n subset n^+$
or
If $x=n$, then $x subset n^+$
It follows that every element of $n^+$ is a subset of $n^+$
$n^+ in A$
By induction $A=mathbb{N}$.
elementary-set-theory
asked 6 hours ago
noobisko
353
add a comment |
up vote
0
down vote
favorite
The problem reads: Prove that $mathbb{N}$ is transitive.
With the definition of a transitive set that im using is $forall x in X$, $x subset X$.
I cant help but feel that i'm missing something. Any help/advice is greatly appreciated. This is what i have so far:
Let $A$ be the set of all transitive natural numbers
$n in A$, iff $n in mathbb{N}$, $x subset n$, $forall x in n$
$ 0 in A $
Suppose $n in A$, consider $n^+ = n cup$ {n}
if $x in n^+$, then $x in n$ or $x=n$
If $x in n$, then $x subset n subset n^+$
or
If $x=n$, then $x subset n^+$
It follows that every element of $n^+$ is a subset of $n^+$
$n^+ in A$
By induction $A=mathbb{N}$.
elementary-set-theory
asked 6 hours ago
noobisko
353
1
What is your definition of $mathbb{N}$? (No, seriously; to do this problem, you need to know both what "transitive" means, and also exactly what "$mathbb{N}$" is). And what is the definition of $n$?
– Arturo Magidin
6 hours ago
Note that assuming everything you did were correct, and you have proven that "the set of all transitive natural numbers" is equal to $mathbb{N}$, then what you have proven is that each natural number is transitive. But that is not what you are trying to prove. You are trying to prove that the set $mathbb{N}$ is transitive, not that each element of the set is transitive.
– Arturo Magidin
6 hours ago
Well I'm using the peano axioms that define the natural numbers ( $mathbb{N}$). And a transitive set has the property that it includes "$subset$" everything that it contains "$in$". I hope that helps.
– noobisko
6 hours ago
1
The Peano axioms don't define $mathbb{N}$ inside of set theory; if you have a set, then the set must be constructed in some way, within your set theory. Once you construct it, then the Peano axioms become theorems about your constructed set. However, what you are trying to prove here is a set-theoretic property of the set you have constructed, so you cannot derive it from the Peano axioms; the Peano axioms assume you already have a set and a function, and do not care about what the set actually is. So you cannot just be "using the Peano axioms".
– Arturo Magidin
6 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The problem reads: Prove that $mathbb{N}$ is transitive.
With the definition of a transitive set that im using is $forall x in X$, $x subset X$.
I cant help but feel that i'm missing something. Any help/advice is greatly appreciated. This is what i have so far:
Let $A$ be the set of all transitive natural numbers
$n in A$, iff $n in mathbb{N}$, $x subset n$, $forall x in n$
$ 0 in A $
Suppose $n in A$, consider $n^+ = n cup$ {n}
if $x in n^+$, then $x in n$ or $x=n$
If $x in n$, then $x subset n subset n^+$
or
If $x=n$, then $x subset n^+$
It follows that every element of $n^+$ is a subset of $n^+$
$n^+ in A$
By induction $A=mathbb{N}$.
elementary-set-theory
asked 6 hours ago
noobisko
353
The problem reads: Prove that $mathbb{N}$ is transitive.
With the definition of a transitive set that im using is $forall x in X$, $x subset X$.
I cant help but feel that i'm missing something. Any help/advice is greatly appreciated. This is what i have so far:
Let $A$ be the set of all transitive natural numbers
$n in A$, iff $n in mathbb{N}$, $x subset n$, $forall x in n$
$ 0 in A $
Suppose $n in A$, consider $n^+ = n cup$ {n}
if $x in n^+$, then $x in n$ or $x=n$
If $x in n$, then $x subset n subset n^+$
or
If $x=n$, then $x subset n^+$
It follows that every element of $n^+$ is a subset of $n^+$
$n^+ in A$
By induction $A=mathbb{N}$.
elementary-set-theory
elementary-set-theory
asked 6 hours ago
noobisko
353
asked 6 hours ago
noobisko
353
asked 6 hours ago
noobisko
353
asked 6 hours ago
noobisko
353
asked 6 hours ago
noobisko
353
353
1
What is your definition of $mathbb{N}$? (No, seriously; to do this problem, you need to know both what "transitive" means, and also exactly what "$mathbb{N}$" is). And what is the definition of $n$?
– Arturo Magidin
6 hours ago
Note that assuming everything you did were correct, and you have proven that "the set of all transitive natural numbers" is equal to $mathbb{N}$, then what you have proven is that each natural number is transitive. But that is not what you are trying to prove. You are trying to prove that the set $mathbb{N}$ is transitive, not that each element of the set is transitive.
– Arturo Magidin
6 hours ago
Well I'm using the peano axioms that define the natural numbers ( $mathbb{N}$). And a transitive set has the property that it includes "$subset$" everything that it contains "$in$". I hope that helps.
– noobisko
6 hours ago
1
The Peano axioms don't define $mathbb{N}$ inside of set theory; if you have a set, then the set must be constructed in some way, within your set theory. Once you construct it, then the Peano axioms become theorems about your constructed set. However, what you are trying to prove here is a set-theoretic property of the set you have constructed, so you cannot derive it from the Peano axioms; the Peano axioms assume you already have a set and a function, and do not care about what the set actually is. So you cannot just be "using the Peano axioms".
– Arturo Magidin
6 hours ago
add a comment |
1
What is your definition of $mathbb{N}$? (No, seriously; to do this problem, you need to know both what "transitive" means, and also exactly what "$mathbb{N}$" is). And what is the definition of $n$?
– Arturo Magidin
6 hours ago
Note that assuming everything you did were correct, and you have proven that "the set of all transitive natural numbers" is equal to $mathbb{N}$, then what you have proven is that each natural number is transitive. But that is not what you are trying to prove. You are trying to prove that the set $mathbb{N}$ is transitive, not that each element of the set is transitive.
– Arturo Magidin
6 hours ago
Well I'm using the peano axioms that define the natural numbers ( $mathbb{N}$). And a transitive set has the property that it includes "$subset$" everything that it contains "$in$". I hope that helps.
– noobisko
6 hours ago
1
The Peano axioms don't define $mathbb{N}$ inside of set theory; if you have a set, then the set must be constructed in some way, within your set theory. Once you construct it, then the Peano axioms become theorems about your constructed set. However, what you are trying to prove here is a set-theoretic property of the set you have constructed, so you cannot derive it from the Peano axioms; the Peano axioms assume you already have a set and a function, and do not care about what the set actually is. So you cannot just be "using the Peano axioms".
– Arturo Magidin
6 hours ago
1
1
What is your definition of $mathbb{N}$? (No, seriously; to do this problem, you need to know both what "transitive" means, and also exactly what "$mathbb{N}$" is). And what is the definition of $n$?
– Arturo Magidin
6 hours ago
What is your definition of $mathbb{N}$? (No, seriously; to do this problem, you need to know both what "transitive" means, and also exactly what "$mathbb{N}$" is). And what is the definition of $n$?
– Arturo Magidin
6 hours ago
Note that assuming everything you did were correct, and you have proven that "the set of all transitive natural numbers" is equal to $mathbb{N}$, then what you have proven is that each natural number is transitive. But that is not what you are trying to prove. You are trying to prove that the set $mathbb{N}$ is transitive, not that each element of the set is transitive.
– Arturo Magidin
6 hours ago
Note that assuming everything you did were correct, and you have proven that "the set of all transitive natural numbers" is equal to $mathbb{N}$, then what you have proven is that each natural number is transitive. But that is not what you are trying to prove. You are trying to prove that the set $mathbb{N}$ is transitive, not that each element of the set is transitive.
– Arturo Magidin
6 hours ago
Well I'm using the peano axioms that define the natural numbers ( $mathbb{N}$). And a transitive set has the property that it includes "$subset$" everything that it contains "$in$". I hope that helps.
– noobisko
6 hours ago
Well I'm using the peano axioms that define the natural numbers ( $mathbb{N}$). And a transitive set has the property that it includes "$subset$" everything that it contains "$in$". I hope that helps.
– noobisko
6 hours ago
1
1
The Peano axioms don't define $mathbb{N}$ inside of set theory; if you have a set, then the set must be constructed in some way, within your set theory. Once you construct it, then the Peano axioms become theorems about your constructed set. However, what you are trying to prove here is a set-theoretic property of the set you have constructed, so you cannot derive it from the Peano axioms; the Peano axioms assume you already have a set and a function, and do not care about what the set actually is. So you cannot just be "using the Peano axioms".
– Arturo Magidin
6 hours ago
The Peano axioms don't define $mathbb{N}$ inside of set theory; if you have a set, then the set must be constructed in some way, within your set theory. Once you construct it, then the Peano axioms become theorems about your constructed set. However, what you are trying to prove here is a set-theoretic property of the set you have constructed, so you cannot derive it from the Peano axioms; the Peano axioms assume you already have a set and a function, and do not care about what the set actually is. So you cannot just be "using the Peano axioms".
– Arturo Magidin
6 hours ago
add a comment |
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What is your definition of $mathbb{N}$? (No, seriously; to do this problem, you need to know both what "transitive" means, and also exactly what "$mathbb{N}$" is). And what is the definition of $n$?
– Arturo Magidin
6 hours ago
Note that assuming everything you did were correct, and you have proven that "the set of all transitive natural numbers" is equal to $mathbb{N}$, then what you have proven is that each natural number is transitive. But that is not what you are trying to prove. You are trying to prove that the set $mathbb{N}$ is transitive, not that each element of the set is transitive.
– Arturo Magidin
6 hours ago
Well I'm using the peano axioms that define the natural numbers ( $mathbb{N}$). And a transitive set has the property that it includes "$subset$" everything that it contains "$in$". I hope that helps.
– noobisko
6 hours ago
1
The Peano axioms don't define $mathbb{N}$ inside of set theory; if you have a set, then the set must be constructed in some way, within your set theory. Once you construct it, then the Peano axioms become theorems about your constructed set. However, what you are trying to prove here is a set-theoretic property of the set you have constructed, so you cannot derive it from the Peano axioms; the Peano axioms assume you already have a set and a function, and do not care about what the set actually is. So you cannot just be "using the Peano axioms".
– Arturo Magidin
6 hours ago