Isn't the construction of successors of numbers up to infinity the same as what the axiom of infinity says?
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In the book of "Näive Set Theory", the author writes first 3 numbers as a set of previous numbers building by this successors of previous numbers and then writes "etc." which he clarifies that from what has been said it doesn't follow the construction of successors continues ad infinitum.
Then, he introduces the axiom of infinity which says there exists a set that contains $0$ and successor of each of its elements.
Isn't it the same thing as building successors up to infinity? What am I missing and why I can't I see the difference between two concepts?
elementary-set-theory
$endgroup$
|
show 4 more comments
$begingroup$
In the book of "Näive Set Theory", the author writes first 3 numbers as a set of previous numbers building by this successors of previous numbers and then writes "etc." which he clarifies that from what has been said it doesn't follow the construction of successors continues ad infinitum.
Then, he introduces the axiom of infinity which says there exists a set that contains $0$ and successor of each of its elements.
Isn't it the same thing as building successors up to infinity? What am I missing and why I can't I see the difference between two concepts?
elementary-set-theory
$endgroup$
3
$begingroup$
The fact that we have a "procedure" to manufacture a set "named" $n$ representing the natural number $n$ does not mean that we have also the set of all $n$.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:01
$begingroup$
A set exists when : either (i) we have an axiom asserting its existence, or (ii) we can prove its existence from axioms or already proven theorems.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:02
$begingroup$
@MauroALLEGRANZA, this is beautiful. Can't I just say $A = { x : text{x can be manufactured} }$? Then I don't need the axiom. What is the problem in this way of thinking?
$endgroup$
– Turkhan Badalov
Dec 30 '18 at 11:06
1
$begingroup$
Russell's paradox and many others...See Halmos, page 6-7.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:17
1
$begingroup$
NO. From the Ax of Spec : "for a set to exist I just need to make sure its elements exist as elements of an already existsing set."
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:37
|
show 4 more comments
$begingroup$
In the book of "Näive Set Theory", the author writes first 3 numbers as a set of previous numbers building by this successors of previous numbers and then writes "etc." which he clarifies that from what has been said it doesn't follow the construction of successors continues ad infinitum.
Then, he introduces the axiom of infinity which says there exists a set that contains $0$ and successor of each of its elements.
Isn't it the same thing as building successors up to infinity? What am I missing and why I can't I see the difference between two concepts?
elementary-set-theory
$endgroup$
In the book of "Näive Set Theory", the author writes first 3 numbers as a set of previous numbers building by this successors of previous numbers and then writes "etc." which he clarifies that from what has been said it doesn't follow the construction of successors continues ad infinitum.
Then, he introduces the axiom of infinity which says there exists a set that contains $0$ and successor of each of its elements.
Isn't it the same thing as building successors up to infinity? What am I missing and why I can't I see the difference between two concepts?
elementary-set-theory
elementary-set-theory
edited Dec 30 '18 at 13:25
Andrés E. Caicedo
65.1k8158246
65.1k8158246
asked Dec 30 '18 at 10:55
Turkhan BadalovTurkhan Badalov
459312
459312
3
$begingroup$
The fact that we have a "procedure" to manufacture a set "named" $n$ representing the natural number $n$ does not mean that we have also the set of all $n$.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:01
$begingroup$
A set exists when : either (i) we have an axiom asserting its existence, or (ii) we can prove its existence from axioms or already proven theorems.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:02
$begingroup$
@MauroALLEGRANZA, this is beautiful. Can't I just say $A = { x : text{x can be manufactured} }$? Then I don't need the axiom. What is the problem in this way of thinking?
$endgroup$
– Turkhan Badalov
Dec 30 '18 at 11:06
1
$begingroup$
Russell's paradox and many others...See Halmos, page 6-7.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:17
1
$begingroup$
NO. From the Ax of Spec : "for a set to exist I just need to make sure its elements exist as elements of an already existsing set."
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:37
|
show 4 more comments
3
$begingroup$
The fact that we have a "procedure" to manufacture a set "named" $n$ representing the natural number $n$ does not mean that we have also the set of all $n$.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:01
$begingroup$
A set exists when : either (i) we have an axiom asserting its existence, or (ii) we can prove its existence from axioms or already proven theorems.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:02
$begingroup$
@MauroALLEGRANZA, this is beautiful. Can't I just say $A = { x : text{x can be manufactured} }$? Then I don't need the axiom. What is the problem in this way of thinking?
$endgroup$
– Turkhan Badalov
Dec 30 '18 at 11:06
1
$begingroup$
Russell's paradox and many others...See Halmos, page 6-7.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:17
1
$begingroup$
NO. From the Ax of Spec : "for a set to exist I just need to make sure its elements exist as elements of an already existsing set."
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:37
3
3
$begingroup$
The fact that we have a "procedure" to manufacture a set "named" $n$ representing the natural number $n$ does not mean that we have also the set of all $n$.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:01
$begingroup$
The fact that we have a "procedure" to manufacture a set "named" $n$ representing the natural number $n$ does not mean that we have also the set of all $n$.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:01
$begingroup$
A set exists when : either (i) we have an axiom asserting its existence, or (ii) we can prove its existence from axioms or already proven theorems.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:02
$begingroup$
A set exists when : either (i) we have an axiom asserting its existence, or (ii) we can prove its existence from axioms or already proven theorems.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:02
$begingroup$
@MauroALLEGRANZA, this is beautiful. Can't I just say $A = { x : text{x can be manufactured} }$? Then I don't need the axiom. What is the problem in this way of thinking?
$endgroup$
– Turkhan Badalov
Dec 30 '18 at 11:06
$begingroup$
@MauroALLEGRANZA, this is beautiful. Can't I just say $A = { x : text{x can be manufactured} }$? Then I don't need the axiom. What is the problem in this way of thinking?
$endgroup$
– Turkhan Badalov
Dec 30 '18 at 11:06
1
1
$begingroup$
Russell's paradox and many others...See Halmos, page 6-7.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:17
$begingroup$
Russell's paradox and many others...See Halmos, page 6-7.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:17
1
1
$begingroup$
NO. From the Ax of Spec : "for a set to exist I just need to make sure its elements exist as elements of an already existsing set."
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:37
$begingroup$
NO. From the Ax of Spec : "for a set to exist I just need to make sure its elements exist as elements of an already existsing set."
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:37
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
No. Actually, the existence of arbitrarily many successors and the content of the axiom of infinity are two different things.
- "Arbitrarily many successors" means just that - that the successors $S(n)$, $S(S(n))$, $S(S(s(n)))$, ... exist.
- "Axiom of infinity" means that not only do they exist, but they can be collected into a set, which we call $mathbb{N}$.
It is entirely possible that the first can hold true but not the second. In fact, in full ZFC, there is a sort of "super" version of the natural numbers called the ordinal numbers, all of which "exist", yet their collection into a set does not. Actually, the sense of "non-existence" here is in a sense worse than the case for $mathbb{N}$ because here it actually leads to a contradiction if we try to assert this set exists, whereas in the case of $mathbb{N}$ we can freely assert, without contradiction to the other axioms, either its existence or its non-existence. But the point is more to illustrate how all things you might consider as "members" of a set can exist and yet the set itself not, and moreover that the non-existence of the latter does not say anything about the non-existence of the former.
$endgroup$
add a comment |
$begingroup$
Naively speaking, you might expect that every collection you can define exists. But existence in standard set theoretic contexts means being a set.
Russell, with his eponymous paradox, showed that not every collection is a set. The reaction to that was to formulate some axioms and state that things you can derive to exist from these axioms will be sets, and things you can prove to not be sets, are not sets.
Naively speaking there is no reason for the collection of successors or collection of natural numbers to be sets, even if individually we can prove each natural number exists. Other than "we really want them to be" anyway. So this was formalized into the axiom of infinity that states that this is indeed a set.
You could argue that this is not justified to do just that. But in effect this is necessary for modern mathematics, and we know that this axiom gives us significant power, so other than giving it some philosophical justification, there is no way to derive it naively.
Finally, there are many many ways to formulate this axiom. We can require different kinds of sets to exist, and from them to prove the existence of the set of all natural numbers.
$endgroup$
add a comment |
$begingroup$
The OP is reading an informal presentation of ZFC axiomatic set theory. The author knows that the axiom of infinity is necessary in a formal development of set theory, so puts his readers 'on notice' when stating that 'it doesn't follow the construction of successors continues ad infinitum'.
Now, if the OP wants to ignore this 'ZFC backdrop' they can accept on faith the following quote from Georg Cantor:
A set is a gathering together into a whole of definite, distinct
objects of our perception or of our thought—which are called elements
of the set.
They can then 'manufacture sets' to their hearts content, but if they get too carried away they will have difficulties communicating with $21^{st}$ century mathematicians.
$endgroup$
add a comment |
Your Answer
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3 Answers
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active
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3 Answers
3
active
oldest
votes
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$begingroup$
No. Actually, the existence of arbitrarily many successors and the content of the axiom of infinity are two different things.
- "Arbitrarily many successors" means just that - that the successors $S(n)$, $S(S(n))$, $S(S(s(n)))$, ... exist.
- "Axiom of infinity" means that not only do they exist, but they can be collected into a set, which we call $mathbb{N}$.
It is entirely possible that the first can hold true but not the second. In fact, in full ZFC, there is a sort of "super" version of the natural numbers called the ordinal numbers, all of which "exist", yet their collection into a set does not. Actually, the sense of "non-existence" here is in a sense worse than the case for $mathbb{N}$ because here it actually leads to a contradiction if we try to assert this set exists, whereas in the case of $mathbb{N}$ we can freely assert, without contradiction to the other axioms, either its existence or its non-existence. But the point is more to illustrate how all things you might consider as "members" of a set can exist and yet the set itself not, and moreover that the non-existence of the latter does not say anything about the non-existence of the former.
$endgroup$
add a comment |
$begingroup$
No. Actually, the existence of arbitrarily many successors and the content of the axiom of infinity are two different things.
- "Arbitrarily many successors" means just that - that the successors $S(n)$, $S(S(n))$, $S(S(s(n)))$, ... exist.
- "Axiom of infinity" means that not only do they exist, but they can be collected into a set, which we call $mathbb{N}$.
It is entirely possible that the first can hold true but not the second. In fact, in full ZFC, there is a sort of "super" version of the natural numbers called the ordinal numbers, all of which "exist", yet their collection into a set does not. Actually, the sense of "non-existence" here is in a sense worse than the case for $mathbb{N}$ because here it actually leads to a contradiction if we try to assert this set exists, whereas in the case of $mathbb{N}$ we can freely assert, without contradiction to the other axioms, either its existence or its non-existence. But the point is more to illustrate how all things you might consider as "members" of a set can exist and yet the set itself not, and moreover that the non-existence of the latter does not say anything about the non-existence of the former.
$endgroup$
add a comment |
$begingroup$
No. Actually, the existence of arbitrarily many successors and the content of the axiom of infinity are two different things.
- "Arbitrarily many successors" means just that - that the successors $S(n)$, $S(S(n))$, $S(S(s(n)))$, ... exist.
- "Axiom of infinity" means that not only do they exist, but they can be collected into a set, which we call $mathbb{N}$.
It is entirely possible that the first can hold true but not the second. In fact, in full ZFC, there is a sort of "super" version of the natural numbers called the ordinal numbers, all of which "exist", yet their collection into a set does not. Actually, the sense of "non-existence" here is in a sense worse than the case for $mathbb{N}$ because here it actually leads to a contradiction if we try to assert this set exists, whereas in the case of $mathbb{N}$ we can freely assert, without contradiction to the other axioms, either its existence or its non-existence. But the point is more to illustrate how all things you might consider as "members" of a set can exist and yet the set itself not, and moreover that the non-existence of the latter does not say anything about the non-existence of the former.
$endgroup$
No. Actually, the existence of arbitrarily many successors and the content of the axiom of infinity are two different things.
- "Arbitrarily many successors" means just that - that the successors $S(n)$, $S(S(n))$, $S(S(s(n)))$, ... exist.
- "Axiom of infinity" means that not only do they exist, but they can be collected into a set, which we call $mathbb{N}$.
It is entirely possible that the first can hold true but not the second. In fact, in full ZFC, there is a sort of "super" version of the natural numbers called the ordinal numbers, all of which "exist", yet their collection into a set does not. Actually, the sense of "non-existence" here is in a sense worse than the case for $mathbb{N}$ because here it actually leads to a contradiction if we try to assert this set exists, whereas in the case of $mathbb{N}$ we can freely assert, without contradiction to the other axioms, either its existence or its non-existence. But the point is more to illustrate how all things you might consider as "members" of a set can exist and yet the set itself not, and moreover that the non-existence of the latter does not say anything about the non-existence of the former.
answered Dec 30 '18 at 14:59
The_SympathizerThe_Sympathizer
7,4852245
7,4852245
add a comment |
add a comment |
$begingroup$
Naively speaking, you might expect that every collection you can define exists. But existence in standard set theoretic contexts means being a set.
Russell, with his eponymous paradox, showed that not every collection is a set. The reaction to that was to formulate some axioms and state that things you can derive to exist from these axioms will be sets, and things you can prove to not be sets, are not sets.
Naively speaking there is no reason for the collection of successors or collection of natural numbers to be sets, even if individually we can prove each natural number exists. Other than "we really want them to be" anyway. So this was formalized into the axiom of infinity that states that this is indeed a set.
You could argue that this is not justified to do just that. But in effect this is necessary for modern mathematics, and we know that this axiom gives us significant power, so other than giving it some philosophical justification, there is no way to derive it naively.
Finally, there are many many ways to formulate this axiom. We can require different kinds of sets to exist, and from them to prove the existence of the set of all natural numbers.
$endgroup$
add a comment |
$begingroup$
Naively speaking, you might expect that every collection you can define exists. But existence in standard set theoretic contexts means being a set.
Russell, with his eponymous paradox, showed that not every collection is a set. The reaction to that was to formulate some axioms and state that things you can derive to exist from these axioms will be sets, and things you can prove to not be sets, are not sets.
Naively speaking there is no reason for the collection of successors or collection of natural numbers to be sets, even if individually we can prove each natural number exists. Other than "we really want them to be" anyway. So this was formalized into the axiom of infinity that states that this is indeed a set.
You could argue that this is not justified to do just that. But in effect this is necessary for modern mathematics, and we know that this axiom gives us significant power, so other than giving it some philosophical justification, there is no way to derive it naively.
Finally, there are many many ways to formulate this axiom. We can require different kinds of sets to exist, and from them to prove the existence of the set of all natural numbers.
$endgroup$
add a comment |
$begingroup$
Naively speaking, you might expect that every collection you can define exists. But existence in standard set theoretic contexts means being a set.
Russell, with his eponymous paradox, showed that not every collection is a set. The reaction to that was to formulate some axioms and state that things you can derive to exist from these axioms will be sets, and things you can prove to not be sets, are not sets.
Naively speaking there is no reason for the collection of successors or collection of natural numbers to be sets, even if individually we can prove each natural number exists. Other than "we really want them to be" anyway. So this was formalized into the axiom of infinity that states that this is indeed a set.
You could argue that this is not justified to do just that. But in effect this is necessary for modern mathematics, and we know that this axiom gives us significant power, so other than giving it some philosophical justification, there is no way to derive it naively.
Finally, there are many many ways to formulate this axiom. We can require different kinds of sets to exist, and from them to prove the existence of the set of all natural numbers.
$endgroup$
Naively speaking, you might expect that every collection you can define exists. But existence in standard set theoretic contexts means being a set.
Russell, with his eponymous paradox, showed that not every collection is a set. The reaction to that was to formulate some axioms and state that things you can derive to exist from these axioms will be sets, and things you can prove to not be sets, are not sets.
Naively speaking there is no reason for the collection of successors or collection of natural numbers to be sets, even if individually we can prove each natural number exists. Other than "we really want them to be" anyway. So this was formalized into the axiom of infinity that states that this is indeed a set.
You could argue that this is not justified to do just that. But in effect this is necessary for modern mathematics, and we know that this axiom gives us significant power, so other than giving it some philosophical justification, there is no way to derive it naively.
Finally, there are many many ways to formulate this axiom. We can require different kinds of sets to exist, and from them to prove the existence of the set of all natural numbers.
answered Dec 30 '18 at 11:41
Asaf Karagila♦Asaf Karagila
302k32427758
302k32427758
add a comment |
add a comment |
$begingroup$
The OP is reading an informal presentation of ZFC axiomatic set theory. The author knows that the axiom of infinity is necessary in a formal development of set theory, so puts his readers 'on notice' when stating that 'it doesn't follow the construction of successors continues ad infinitum'.
Now, if the OP wants to ignore this 'ZFC backdrop' they can accept on faith the following quote from Georg Cantor:
A set is a gathering together into a whole of definite, distinct
objects of our perception or of our thought—which are called elements
of the set.
They can then 'manufacture sets' to their hearts content, but if they get too carried away they will have difficulties communicating with $21^{st}$ century mathematicians.
$endgroup$
add a comment |
$begingroup$
The OP is reading an informal presentation of ZFC axiomatic set theory. The author knows that the axiom of infinity is necessary in a formal development of set theory, so puts his readers 'on notice' when stating that 'it doesn't follow the construction of successors continues ad infinitum'.
Now, if the OP wants to ignore this 'ZFC backdrop' they can accept on faith the following quote from Georg Cantor:
A set is a gathering together into a whole of definite, distinct
objects of our perception or of our thought—which are called elements
of the set.
They can then 'manufacture sets' to their hearts content, but if they get too carried away they will have difficulties communicating with $21^{st}$ century mathematicians.
$endgroup$
add a comment |
$begingroup$
The OP is reading an informal presentation of ZFC axiomatic set theory. The author knows that the axiom of infinity is necessary in a formal development of set theory, so puts his readers 'on notice' when stating that 'it doesn't follow the construction of successors continues ad infinitum'.
Now, if the OP wants to ignore this 'ZFC backdrop' they can accept on faith the following quote from Georg Cantor:
A set is a gathering together into a whole of definite, distinct
objects of our perception or of our thought—which are called elements
of the set.
They can then 'manufacture sets' to their hearts content, but if they get too carried away they will have difficulties communicating with $21^{st}$ century mathematicians.
$endgroup$
The OP is reading an informal presentation of ZFC axiomatic set theory. The author knows that the axiom of infinity is necessary in a formal development of set theory, so puts his readers 'on notice' when stating that 'it doesn't follow the construction of successors continues ad infinitum'.
Now, if the OP wants to ignore this 'ZFC backdrop' they can accept on faith the following quote from Georg Cantor:
A set is a gathering together into a whole of definite, distinct
objects of our perception or of our thought—which are called elements
of the set.
They can then 'manufacture sets' to their hearts content, but if they get too carried away they will have difficulties communicating with $21^{st}$ century mathematicians.
answered Dec 30 '18 at 12:14
CopyPasteItCopyPasteIt
4,1061628
4,1061628
add a comment |
add a comment |
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$begingroup$
The fact that we have a "procedure" to manufacture a set "named" $n$ representing the natural number $n$ does not mean that we have also the set of all $n$.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:01
$begingroup$
A set exists when : either (i) we have an axiom asserting its existence, or (ii) we can prove its existence from axioms or already proven theorems.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:02
$begingroup$
@MauroALLEGRANZA, this is beautiful. Can't I just say $A = { x : text{x can be manufactured} }$? Then I don't need the axiom. What is the problem in this way of thinking?
$endgroup$
– Turkhan Badalov
Dec 30 '18 at 11:06
1
$begingroup$
Russell's paradox and many others...See Halmos, page 6-7.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:17
1
$begingroup$
NO. From the Ax of Spec : "for a set to exist I just need to make sure its elements exist as elements of an already existsing set."
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 11:37