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Is a set containing itself already a paradox?
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This is inspired by Russel's paradox stating there is not set of all sets. It uses the presupposition that set can contain itself. However, this already seems paradoxical.
Suppose a set A = . Then the smallest set containing it will be . Let's call it B. Since != , we can say that A != B. Namely, we can say that does not belong to . Therefore there is no set containing itself. According to this Russel's paradox itself is indefinable, since any set will not contain itself.
Of course, maybe in naive set theory it was presupposed that a set belongs to itself, but then we should say = and, for example, a, a = a.
Therefore, I can't understand how is Russel's paradox really a paradox if a set containing itself isn't. Am I missing something? Is this issue recognized and if so how is it dealt with?
philosophy-of-mathematics paradox bertrand-russell set-theory
asked Sep 4 at 8:11
rus9384
7291116
 |Â
show 10 more comments
up vote
2
down vote
favorite
This is inspired by Russel's paradox stating there is not set of all sets. It uses the presupposition that set can contain itself. However, this already seems paradoxical.
Suppose a set A = . Then the smallest set containing it will be . Let's call it B. Since != , we can say that A != B. Namely, we can say that does not belong to . Therefore there is no set containing itself. According to this Russel's paradox itself is indefinable, since any set will not contain itself.
Of course, maybe in naive set theory it was presupposed that a set belongs to itself, but then we should say = and, for example, a, a = a.
Therefore, I can't understand how is Russel's paradox really a paradox if a set containing itself isn't. Am I missing something? Is this issue recognized and if so how is it dealt with?
philosophy-of-mathematics paradox bertrand-russell set-theory
asked Sep 4 at 8:11
rus9384
7291116
1
If you want to model sets which may contain themselves, you can't write them down in curly brackets notation, think more of directed, pointed graphs where the set containing only itself and nothing else would be expressed by a single node with a loop.
– fweth
Sep 4 at 8:34
2
The issue you're gonna run into is that even if you just say "I'm using a naive set theory", when it comes to certain questions like this, and like the continuum hypothesis, you are going to have to eventually be explicit about what assumptions you're making (what axioms you adopt) and you'll be able to see whether or not you get a contradiction from them. This is analogous to why Cantor couldn't figure out if CH was true, his naive theory was too vague.
– Not_Here
Sep 4 at 9:51
1
The same goes for "can a set be a member of itself?" It depends on what axioms you are assuming. There are some systems where that is totally fine, there are some where it is explicitly not fine. Just assuming a naive set theory is not enough information to figure out whether or not it is fine, the same exact issue CH has.
– Not_Here
Sep 4 at 11:35
1
L is a model of ZFC. L is a model of ZF and ZF+V=L, which L is also a model of, implies the axiom of choice, so L is a model of ZFC, idk what "ZFC differs from L" means. That means within the model L, the axioms of ZFC are true and so is CH. Like, of course it differs, one is a set of axioms and one is a model, but when did I saw they were the same thing, or even type of thing?
– Not_Here
Sep 4 at 11:52
4
" doesn't contain itself, therefore no set contains itself" ?? Not sure I follow
– BallpointBen
Sep 4 at 14:25
 |Â
show 10 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is inspired by Russel's paradox stating there is not set of all sets. It uses the presupposition that set can contain itself. However, this already seems paradoxical.
Suppose a set A = . Then the smallest set containing it will be . Let's call it B. Since != , we can say that A != B. Namely, we can say that does not belong to . Therefore there is no set containing itself. According to this Russel's paradox itself is indefinable, since any set will not contain itself.
Of course, maybe in naive set theory it was presupposed that a set belongs to itself, but then we should say = and, for example, a, a = a.
Therefore, I can't understand how is Russel's paradox really a paradox if a set containing itself isn't. Am I missing something? Is this issue recognized and if so how is it dealt with?
philosophy-of-mathematics paradox bertrand-russell set-theory
asked Sep 4 at 8:11
rus9384
7291116
This is inspired by Russel's paradox stating there is not set of all sets. It uses the presupposition that set can contain itself. However, this already seems paradoxical.
Suppose a set A = . Then the smallest set containing it will be . Let's call it B. Since != , we can say that A != B. Namely, we can say that does not belong to . Therefore there is no set containing itself. According to this Russel's paradox itself is indefinable, since any set will not contain itself.
Of course, maybe in naive set theory it was presupposed that a set belongs to itself, but then we should say = and, for example, a, a = a.
Therefore, I can't understand how is Russel's paradox really a paradox if a set containing itself isn't. Am I missing something? Is this issue recognized and if so how is it dealt with?
philosophy-of-mathematics paradox bertrand-russell set-theory
asked Sep 4 at 8:11
rus9384
7291116
edited Sep 4 at 8:22
asked Sep 4 at 8:11
rus9384
7291116
asked Sep 4 at 8:11
rus9384
7291116
asked Sep 4 at 8:11
rus9384
7291116
7291116
1
If you want to model sets which may contain themselves, you can't write them down in curly brackets notation, think more of directed, pointed graphs where the set containing only itself and nothing else would be expressed by a single node with a loop.
– fweth
Sep 4 at 8:34
2
The issue you're gonna run into is that even if you just say "I'm using a naive set theory", when it comes to certain questions like this, and like the continuum hypothesis, you are going to have to eventually be explicit about what assumptions you're making (what axioms you adopt) and you'll be able to see whether or not you get a contradiction from them. This is analogous to why Cantor couldn't figure out if CH was true, his naive theory was too vague.
– Not_Here
Sep 4 at 9:51
1
The same goes for "can a set be a member of itself?" It depends on what axioms you are assuming. There are some systems where that is totally fine, there are some where it is explicitly not fine. Just assuming a naive set theory is not enough information to figure out whether or not it is fine, the same exact issue CH has.
– Not_Here
Sep 4 at 11:35
1
L is a model of ZFC. L is a model of ZF and ZF+V=L, which L is also a model of, implies the axiom of choice, so L is a model of ZFC, idk what "ZFC differs from L" means. That means within the model L, the axioms of ZFC are true and so is CH. Like, of course it differs, one is a set of axioms and one is a model, but when did I saw they were the same thing, or even type of thing?
– Not_Here
Sep 4 at 11:52
4
" doesn't contain itself, therefore no set contains itself" ?? Not sure I follow
– BallpointBen
Sep 4 at 14:25
 |Â
show 10 more comments
1
If you want to model sets which may contain themselves, you can't write them down in curly brackets notation, think more of directed, pointed graphs where the set containing only itself and nothing else would be expressed by a single node with a loop.
– fweth
Sep 4 at 8:34
2
The issue you're gonna run into is that even if you just say "I'm using a naive set theory", when it comes to certain questions like this, and like the continuum hypothesis, you are going to have to eventually be explicit about what assumptions you're making (what axioms you adopt) and you'll be able to see whether or not you get a contradiction from them. This is analogous to why Cantor couldn't figure out if CH was true, his naive theory was too vague.
– Not_Here
Sep 4 at 9:51
1
The same goes for "can a set be a member of itself?" It depends on what axioms you are assuming. There are some systems where that is totally fine, there are some where it is explicitly not fine. Just assuming a naive set theory is not enough information to figure out whether or not it is fine, the same exact issue CH has.
– Not_Here
Sep 4 at 11:35
1
L is a model of ZFC. L is a model of ZF and ZF+V=L, which L is also a model of, implies the axiom of choice, so L is a model of ZFC, idk what "ZFC differs from L" means. That means within the model L, the axioms of ZFC are true and so is CH. Like, of course it differs, one is a set of axioms and one is a model, but when did I saw they were the same thing, or even type of thing?
– Not_Here
Sep 4 at 11:52
4
" doesn't contain itself, therefore no set contains itself" ?? Not sure I follow
– BallpointBen
Sep 4 at 14:25
1
1
If you want to model sets which may contain themselves, you can't write them down in curly brackets notation, think more of directed, pointed graphs where the set containing only itself and nothing else would be expressed by a single node with a loop.
– fweth
Sep 4 at 8:34
If you want to model sets which may contain themselves, you can't write them down in curly brackets notation, think more of directed, pointed graphs where the set containing only itself and nothing else would be expressed by a single node with a loop.
– fweth
Sep 4 at 8:34
2
2
The issue you're gonna run into is that even if you just say "I'm using a naive set theory", when it comes to certain questions like this, and like the continuum hypothesis, you are going to have to eventually be explicit about what assumptions you're making (what axioms you adopt) and you'll be able to see whether or not you get a contradiction from them. This is analogous to why Cantor couldn't figure out if CH was true, his naive theory was too vague.
– Not_Here
Sep 4 at 9:51
The issue you're gonna run into is that even if you just say "I'm using a naive set theory", when it comes to certain questions like this, and like the continuum hypothesis, you are going to have to eventually be explicit about what assumptions you're making (what axioms you adopt) and you'll be able to see whether or not you get a contradiction from them. This is analogous to why Cantor couldn't figure out if CH was true, his naive theory was too vague.
– Not_Here
Sep 4 at 9:51
1
1
The same goes for "can a set be a member of itself?" It depends on what axioms you are assuming. There are some systems where that is totally fine, there are some where it is explicitly not fine. Just assuming a naive set theory is not enough information to figure out whether or not it is fine, the same exact issue CH has.
– Not_Here
Sep 4 at 11:35
The same goes for "can a set be a member of itself?" It depends on what axioms you are assuming. There are some systems where that is totally fine, there are some where it is explicitly not fine. Just assuming a naive set theory is not enough information to figure out whether or not it is fine, the same exact issue CH has.
– Not_Here
Sep 4 at 11:35
1
1
L is a model of ZFC. L is a model of ZF and ZF+V=L, which L is also a model of, implies the axiom of choice, so L is a model of ZFC, idk what "ZFC differs from L" means. That means within the model L, the axioms of ZFC are true and so is CH. Like, of course it differs, one is a set of axioms and one is a model, but when did I saw they were the same thing, or even type of thing?
– Not_Here
Sep 4 at 11:52
L is a model of ZFC. L is a model of ZF and ZF+V=L, which L is also a model of, implies the axiom of choice, so L is a model of ZFC, idk what "ZFC differs from L" means. That means within the model L, the axioms of ZFC are true and so is CH. Like, of course it differs, one is a set of axioms and one is a model, but when did I saw they were the same thing, or even type of thing?
– Not_Here
Sep 4 at 11:52
4
4
" doesn't contain itself, therefore no set contains itself" ?? Not sure I follow
– BallpointBen
Sep 4 at 14:25
" doesn't contain itself, therefore no set contains itself" ?? Not sure I follow
– BallpointBen
Sep 4 at 14:25
 |Â
show 10 more comments
2 Answers
2
active
oldest
votes
up vote
11
down vote
accepted
Russell's paradox arises within naïve set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself. Hence the paradox.
The "root" of the paradox is the so-called unrestriceted Comprehension Principle of naïve set theory:
for every property Æ(x) expressible in the language, there is the set x: Æ(x) of all and only those objects satisfying that property.
The paradox arises considering as Æ(x) the property “~(x ∈ x)â€Â.
Zermelo's solution to the paradox is built on the replacement of Comprehension principle with the Axiom schema of specification:
in order to assert the existence of the set B satisfying the property Æ(x), we have to "separate" it from an already existsing set A.
How does Zermelo's theory avoid Russell’s paradox?
Assuming the existence of V – the whole universe of sets – and let Æ be x ∉ x, a contradiction again appears to arise. But in this case, all the contradiction shows is that V is not a set.
All the contradiction shows is that “V†is an empty name, i.e., that it has no reference, that it does not exist.
In conclusion, the non-existence of the set of all sets is a feature of Zermelo-Fraenkel Set Theory.
There are Alternative Axiomatic Set Theories, like Quine's New Foundations, where the universal set V exists.
The impossibility of "circular" sets, like e.g. x ∈ x, is due to the Axiom of regularity.
The axiom implies that no set is an element of itself, and that there is no infinite sequence a(n) such that a(i+1) is an element of a(i).
Ther are other axiomatic set theories that are non-well-founded, i.e. that allow sets to contain themselves and otherwise violate the rule of well-foundedness.
Conclusion : a set containing itself is not per se a paradox. Paradoxes can arise in cooperation with other basic assumptions regarding sets.
The universal set is an example of "set containing itself" and its conception is quite natural.
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
1
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
add a comment |Â
up vote
0
down vote
For any binary relation R (not just "is an element of"), it is trivial to prove, using the ordinary rules of logic, that:
~Ex: Ay: [yRx <=> ~yRy]
(Assume the contrary and obtain a contradiction in 1 or 2 lines.)
So, the root of the problem was not sets being elements of themselves, whatever that may mean. The problem was that early attempts to axiomatize set theory allowed you to prove that, contrary to the above mentioned rules of logic:
Ex: Ay: [y in x <=> ~ y in y]
This "bug" was fixed by various means in subsequent versions of the axioms of set theory (or logic in some cases) that would, in one way or another, disallow this theorem.
answered Sep 5 at 14:53
Dan Christensen
73038
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Russell's paradox arises within naïve set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself. Hence the paradox.
The "root" of the paradox is the so-called unrestriceted Comprehension Principle of naïve set theory:
for every property Æ(x) expressible in the language, there is the set x: Æ(x) of all and only those objects satisfying that property.
The paradox arises considering as Æ(x) the property “~(x ∈ x)â€Â.
Zermelo's solution to the paradox is built on the replacement of Comprehension principle with the Axiom schema of specification:
in order to assert the existence of the set B satisfying the property Æ(x), we have to "separate" it from an already existsing set A.
How does Zermelo's theory avoid Russell’s paradox?
Assuming the existence of V – the whole universe of sets – and let Æ be x ∉ x, a contradiction again appears to arise. But in this case, all the contradiction shows is that V is not a set.
All the contradiction shows is that “V†is an empty name, i.e., that it has no reference, that it does not exist.
In conclusion, the non-existence of the set of all sets is a feature of Zermelo-Fraenkel Set Theory.
There are Alternative Axiomatic Set Theories, like Quine's New Foundations, where the universal set V exists.
The impossibility of "circular" sets, like e.g. x ∈ x, is due to the Axiom of regularity.
The axiom implies that no set is an element of itself, and that there is no infinite sequence a(n) such that a(i+1) is an element of a(i).
Ther are other axiomatic set theories that are non-well-founded, i.e. that allow sets to contain themselves and otherwise violate the rule of well-foundedness.
Conclusion : a set containing itself is not per se a paradox. Paradoxes can arise in cooperation with other basic assumptions regarding sets.
The universal set is an example of "set containing itself" and its conception is quite natural.
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
1
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
add a comment |Â
up vote
11
down vote
accepted
Russell's paradox arises within naïve set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself. Hence the paradox.
The "root" of the paradox is the so-called unrestriceted Comprehension Principle of naïve set theory:
for every property Æ(x) expressible in the language, there is the set x: Æ(x) of all and only those objects satisfying that property.
The paradox arises considering as Æ(x) the property “~(x ∈ x)â€Â.
Zermelo's solution to the paradox is built on the replacement of Comprehension principle with the Axiom schema of specification:
in order to assert the existence of the set B satisfying the property Æ(x), we have to "separate" it from an already existsing set A.
How does Zermelo's theory avoid Russell’s paradox?
Assuming the existence of V – the whole universe of sets – and let Æ be x ∉ x, a contradiction again appears to arise. But in this case, all the contradiction shows is that V is not a set.
All the contradiction shows is that “V†is an empty name, i.e., that it has no reference, that it does not exist.
In conclusion, the non-existence of the set of all sets is a feature of Zermelo-Fraenkel Set Theory.
There are Alternative Axiomatic Set Theories, like Quine's New Foundations, where the universal set V exists.
The impossibility of "circular" sets, like e.g. x ∈ x, is due to the Axiom of regularity.
The axiom implies that no set is an element of itself, and that there is no infinite sequence a(n) such that a(i+1) is an element of a(i).
Ther are other axiomatic set theories that are non-well-founded, i.e. that allow sets to contain themselves and otherwise violate the rule of well-foundedness.
Conclusion : a set containing itself is not per se a paradox. Paradoxes can arise in cooperation with other basic assumptions regarding sets.
The universal set is an example of "set containing itself" and its conception is quite natural.
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
1
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Russell's paradox arises within naïve set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself. Hence the paradox.
The "root" of the paradox is the so-called unrestriceted Comprehension Principle of naïve set theory:
for every property Æ(x) expressible in the language, there is the set x: Æ(x) of all and only those objects satisfying that property.
The paradox arises considering as Æ(x) the property “~(x ∈ x)â€Â.
Zermelo's solution to the paradox is built on the replacement of Comprehension principle with the Axiom schema of specification:
in order to assert the existence of the set B satisfying the property Æ(x), we have to "separate" it from an already existsing set A.
How does Zermelo's theory avoid Russell’s paradox?
Assuming the existence of V – the whole universe of sets – and let Æ be x ∉ x, a contradiction again appears to arise. But in this case, all the contradiction shows is that V is not a set.
All the contradiction shows is that “V†is an empty name, i.e., that it has no reference, that it does not exist.
In conclusion, the non-existence of the set of all sets is a feature of Zermelo-Fraenkel Set Theory.
There are Alternative Axiomatic Set Theories, like Quine's New Foundations, where the universal set V exists.
The impossibility of "circular" sets, like e.g. x ∈ x, is due to the Axiom of regularity.
The axiom implies that no set is an element of itself, and that there is no infinite sequence a(n) such that a(i+1) is an element of a(i).
Ther are other axiomatic set theories that are non-well-founded, i.e. that allow sets to contain themselves and otherwise violate the rule of well-foundedness.
Conclusion : a set containing itself is not per se a paradox. Paradoxes can arise in cooperation with other basic assumptions regarding sets.
The universal set is an example of "set containing itself" and its conception is quite natural.
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
Russell's paradox arises within naïve set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself. Hence the paradox.
The "root" of the paradox is the so-called unrestriceted Comprehension Principle of naïve set theory:
for every property Æ(x) expressible in the language, there is the set x: Æ(x) of all and only those objects satisfying that property.
The paradox arises considering as Æ(x) the property “~(x ∈ x)â€Â.
Zermelo's solution to the paradox is built on the replacement of Comprehension principle with the Axiom schema of specification:
in order to assert the existence of the set B satisfying the property Æ(x), we have to "separate" it from an already existsing set A.
How does Zermelo's theory avoid Russell’s paradox?
Assuming the existence of V – the whole universe of sets – and let Æ be x ∉ x, a contradiction again appears to arise. But in this case, all the contradiction shows is that V is not a set.
All the contradiction shows is that “V†is an empty name, i.e., that it has no reference, that it does not exist.
In conclusion, the non-existence of the set of all sets is a feature of Zermelo-Fraenkel Set Theory.
There are Alternative Axiomatic Set Theories, like Quine's New Foundations, where the universal set V exists.
The impossibility of "circular" sets, like e.g. x ∈ x, is due to the Axiom of regularity.
The axiom implies that no set is an element of itself, and that there is no infinite sequence a(n) such that a(i+1) is an element of a(i).
Ther are other axiomatic set theories that are non-well-founded, i.e. that allow sets to contain themselves and otherwise violate the rule of well-foundedness.
Conclusion : a set containing itself is not per se a paradox. Paradoxes can arise in cooperation with other basic assumptions regarding sets.
The universal set is an example of "set containing itself" and its conception is quite natural.
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
edited Sep 4 at 16:23
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
answered Sep 4 at 8:42
Mauro ALLEGRANZA
25.1k21756
25.1k21756
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
1
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
add a comment |Â
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
1
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
Does well-foundedness rule make sense in multisets? I assumed some kind of atom (which is not a multiset) and multisets which only differ by their size (number of that atom occurences). Then I assume any multiset contains itself, but still can't be an atom.
– rus9384
Sep 4 at 8:51
1
1
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
I assume you meant to write ~(x ∈ x), not ~(x = x). The set x: ~(x = x) is just the empty set, and there's nothing paradoxical about it.
– Ilmari Karonen
Sep 4 at 15:28
add a comment |Â
up vote
0
down vote
For any binary relation R (not just "is an element of"), it is trivial to prove, using the ordinary rules of logic, that:
~Ex: Ay: [yRx <=> ~yRy]
(Assume the contrary and obtain a contradiction in 1 or 2 lines.)
So, the root of the problem was not sets being elements of themselves, whatever that may mean. The problem was that early attempts to axiomatize set theory allowed you to prove that, contrary to the above mentioned rules of logic:
Ex: Ay: [y in x <=> ~ y in y]
This "bug" was fixed by various means in subsequent versions of the axioms of set theory (or logic in some cases) that would, in one way or another, disallow this theorem.
answered Sep 5 at 14:53
Dan Christensen
73038
add a comment |Â
up vote
0
down vote
For any binary relation R (not just "is an element of"), it is trivial to prove, using the ordinary rules of logic, that:
~Ex: Ay: [yRx <=> ~yRy]
(Assume the contrary and obtain a contradiction in 1 or 2 lines.)
So, the root of the problem was not sets being elements of themselves, whatever that may mean. The problem was that early attempts to axiomatize set theory allowed you to prove that, contrary to the above mentioned rules of logic:
Ex: Ay: [y in x <=> ~ y in y]
This "bug" was fixed by various means in subsequent versions of the axioms of set theory (or logic in some cases) that would, in one way or another, disallow this theorem.
answered Sep 5 at 14:53
Dan Christensen
73038
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For any binary relation R (not just "is an element of"), it is trivial to prove, using the ordinary rules of logic, that:
~Ex: Ay: [yRx <=> ~yRy]
(Assume the contrary and obtain a contradiction in 1 or 2 lines.)
So, the root of the problem was not sets being elements of themselves, whatever that may mean. The problem was that early attempts to axiomatize set theory allowed you to prove that, contrary to the above mentioned rules of logic:
Ex: Ay: [y in x <=> ~ y in y]
This "bug" was fixed by various means in subsequent versions of the axioms of set theory (or logic in some cases) that would, in one way or another, disallow this theorem.
answered Sep 5 at 14:53
Dan Christensen
73038
For any binary relation R (not just "is an element of"), it is trivial to prove, using the ordinary rules of logic, that:
~Ex: Ay: [yRx <=> ~yRy]
(Assume the contrary and obtain a contradiction in 1 or 2 lines.)
So, the root of the problem was not sets being elements of themselves, whatever that may mean. The problem was that early attempts to axiomatize set theory allowed you to prove that, contrary to the above mentioned rules of logic:
Ex: Ay: [y in x <=> ~ y in y]
This "bug" was fixed by various means in subsequent versions of the axioms of set theory (or logic in some cases) that would, in one way or another, disallow this theorem.
answered Sep 5 at 14:53
Dan Christensen
73038
edited Sep 7 at 13:37
answered Sep 5 at 14:53
Dan Christensen
73038
answered Sep 5 at 14:53
Dan Christensen
73038
answered Sep 5 at 14:53
Dan Christensen
73038
73038
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1
If you want to model sets which may contain themselves, you can't write them down in curly brackets notation, think more of directed, pointed graphs where the set containing only itself and nothing else would be expressed by a single node with a loop.
– fweth
Sep 4 at 8:34
2
The issue you're gonna run into is that even if you just say "I'm using a naive set theory", when it comes to certain questions like this, and like the continuum hypothesis, you are going to have to eventually be explicit about what assumptions you're making (what axioms you adopt) and you'll be able to see whether or not you get a contradiction from them. This is analogous to why Cantor couldn't figure out if CH was true, his naive theory was too vague.
– Not_Here
Sep 4 at 9:51
1
The same goes for "can a set be a member of itself?" It depends on what axioms you are assuming. There are some systems where that is totally fine, there are some where it is explicitly not fine. Just assuming a naive set theory is not enough information to figure out whether or not it is fine, the same exact issue CH has.
– Not_Here
Sep 4 at 11:35
1
L is a model of ZFC. L is a model of ZF and ZF+V=L, which L is also a model of, implies the axiom of choice, so L is a model of ZFC, idk what "ZFC differs from L" means. That means within the model L, the axioms of ZFC are true and so is CH. Like, of course it differs, one is a set of axioms and one is a model, but when did I saw they were the same thing, or even type of thing?
– Not_Here
Sep 4 at 11:52
4
" doesn't contain itself, therefore no set contains itself" ?? Not sure I follow
– BallpointBen
Sep 4 at 14:25