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Is there conflict between the law of excluded middle and “no set is its own member�
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And if yes, then how can it be resolved?
As far as I know in standard set theory it's true that "no set is its own member".
Also in standard logic the law of excluded middle is true, either $A$ or $lnot A$.
Now let's consider simple sentence "Entity $X$ is either a real number or not a real number".
This can be restated in terms of set theory as "Element $x$ either belongs to set $mathbbR$ or
to set non-$mathbbR$".
Looks like a tautology, doesn't it? But I will show you that it's not because there is one
thing that isn't a member of either set. Namely, it's set "non-$mathbbR$". This can't belong to
set $mathbbR$ because it isn't a real number. This also can't belong to set non-$mathbbR$ (i.e. it can't belong
to itself) because of "no set is its own member". Thus "Element $x$ either belongs to set $mathbbR$ or to set non-$mathbbR$" isn't a tautology.
elementary-set-theory logic propositional-calculus
edited 40 mins ago
Taroccoesbrocco
3,78951433
asked 59 mins ago
user161005
1138
 |Â
show 5 more comments
up vote
4
down vote
favorite
And if yes, then how can it be resolved?
As far as I know in standard set theory it's true that "no set is its own member".
Also in standard logic the law of excluded middle is true, either $A$ or $lnot A$.
Now let's consider simple sentence "Entity $X$ is either a real number or not a real number".
This can be restated in terms of set theory as "Element $x$ either belongs to set $mathbbR$ or
to set non-$mathbbR$".
Looks like a tautology, doesn't it? But I will show you that it's not because there is one
thing that isn't a member of either set. Namely, it's set "non-$mathbbR$". This can't belong to
set $mathbbR$ because it isn't a real number. This also can't belong to set non-$mathbbR$ (i.e. it can't belong
to itself) because of "no set is its own member". Thus "Element $x$ either belongs to set $mathbbR$ or to set non-$mathbbR$" isn't a tautology.
elementary-set-theory logic propositional-calculus
edited 40 mins ago
Taroccoesbrocco
3,78951433
asked 59 mins ago
user161005
1138
2
I suspect the answer is something like "there is no such thing as the set non-$mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things.
– littleO
44 mins ago
That a set cannot be "its own member"is due in $mathsf ZFC$ to the Axiom of regularity : "The axiom implies that no set is an element of itself, ". We may have theories without it, and this does not mean leave exluded middle. See Non-well-founded set theory.
– Mauro ALLEGRANZA
39 mins ago
Your example absolutely does not work. The proper translation is "$xin mathbb R$ or $neg xin mathbb R$". What you wrote literally makes no sense.
– Git Gud
37 mins ago
See How can I prove that set membership is irreflexive ?
– Mauro ALLEGRANZA
36 mins ago
1
@AsafKaragila Does it mean that there are no absolute complements in ZFC?
– user161005
12 mins ago
 |Â
show 5 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
And if yes, then how can it be resolved?
As far as I know in standard set theory it's true that "no set is its own member".
Also in standard logic the law of excluded middle is true, either $A$ or $lnot A$.
Now let's consider simple sentence "Entity $X$ is either a real number or not a real number".
This can be restated in terms of set theory as "Element $x$ either belongs to set $mathbbR$ or
to set non-$mathbbR$".
Looks like a tautology, doesn't it? But I will show you that it's not because there is one
thing that isn't a member of either set. Namely, it's set "non-$mathbbR$". This can't belong to
set $mathbbR$ because it isn't a real number. This also can't belong to set non-$mathbbR$ (i.e. it can't belong
to itself) because of "no set is its own member". Thus "Element $x$ either belongs to set $mathbbR$ or to set non-$mathbbR$" isn't a tautology.
elementary-set-theory logic propositional-calculus
edited 40 mins ago
Taroccoesbrocco
3,78951433
asked 59 mins ago
user161005
1138
And if yes, then how can it be resolved?
As far as I know in standard set theory it's true that "no set is its own member".
Also in standard logic the law of excluded middle is true, either $A$ or $lnot A$.
Now let's consider simple sentence "Entity $X$ is either a real number or not a real number".
This can be restated in terms of set theory as "Element $x$ either belongs to set $mathbbR$ or
to set non-$mathbbR$".
Looks like a tautology, doesn't it? But I will show you that it's not because there is one
thing that isn't a member of either set. Namely, it's set "non-$mathbbR$". This can't belong to
set $mathbbR$ because it isn't a real number. This also can't belong to set non-$mathbbR$ (i.e. it can't belong
to itself) because of "no set is its own member". Thus "Element $x$ either belongs to set $mathbbR$ or to set non-$mathbbR$" isn't a tautology.
elementary-set-theory logic propositional-calculus
elementary-set-theory logic propositional-calculus
edited 40 mins ago
Taroccoesbrocco
3,78951433
asked 59 mins ago
user161005
1138
edited 40 mins ago
Taroccoesbrocco
3,78951433
asked 59 mins ago
user161005
1138
edited 40 mins ago
Taroccoesbrocco
3,78951433
edited 40 mins ago
Taroccoesbrocco
3,78951433
edited 40 mins ago
Taroccoesbrocco
3,78951433
3,78951433
asked 59 mins ago
user161005
1138
asked 59 mins ago
user161005
1138
asked 59 mins ago
user161005
1138
1138
2
I suspect the answer is something like "there is no such thing as the set non-$mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things.
– littleO
44 mins ago
That a set cannot be "its own member"is due in $mathsf ZFC$ to the Axiom of regularity : "The axiom implies that no set is an element of itself, ". We may have theories without it, and this does not mean leave exluded middle. See Non-well-founded set theory.
– Mauro ALLEGRANZA
39 mins ago
Your example absolutely does not work. The proper translation is "$xin mathbb R$ or $neg xin mathbb R$". What you wrote literally makes no sense.
– Git Gud
37 mins ago
See How can I prove that set membership is irreflexive ?
– Mauro ALLEGRANZA
36 mins ago
1
@AsafKaragila Does it mean that there are no absolute complements in ZFC?
– user161005
12 mins ago
 |Â
show 5 more comments
2
I suspect the answer is something like "there is no such thing as the set non-$mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things.
– littleO
44 mins ago
That a set cannot be "its own member"is due in $mathsf ZFC$ to the Axiom of regularity : "The axiom implies that no set is an element of itself, ". We may have theories without it, and this does not mean leave exluded middle. See Non-well-founded set theory.
– Mauro ALLEGRANZA
39 mins ago
Your example absolutely does not work. The proper translation is "$xin mathbb R$ or $neg xin mathbb R$". What you wrote literally makes no sense.
– Git Gud
37 mins ago
See How can I prove that set membership is irreflexive ?
– Mauro ALLEGRANZA
36 mins ago
1
@AsafKaragila Does it mean that there are no absolute complements in ZFC?
– user161005
12 mins ago
2
2
I suspect the answer is something like "there is no such thing as the set non-$mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things.
– littleO
44 mins ago
I suspect the answer is something like "there is no such thing as the set non-$mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things.
– littleO
44 mins ago
That a set cannot be "its own member"is due in $mathsf ZFC$ to the Axiom of regularity : "The axiom implies that no set is an element of itself, ". We may have theories without it, and this does not mean leave exluded middle. See Non-well-founded set theory.
– Mauro ALLEGRANZA
39 mins ago
That a set cannot be "its own member"is due in $mathsf ZFC$ to the Axiom of regularity : "The axiom implies that no set is an element of itself, ". We may have theories without it, and this does not mean leave exluded middle. See Non-well-founded set theory.
– Mauro ALLEGRANZA
39 mins ago
Your example absolutely does not work. The proper translation is "$xin mathbb R$ or $neg xin mathbb R$". What you wrote literally makes no sense.
– Git Gud
37 mins ago
Your example absolutely does not work. The proper translation is "$xin mathbb R$ or $neg xin mathbb R$". What you wrote literally makes no sense.
– Git Gud
37 mins ago
See How can I prove that set membership is irreflexive ?
– Mauro ALLEGRANZA
36 mins ago
See How can I prove that set membership is irreflexive ?
– Mauro ALLEGRANZA
36 mins ago
1
1
@AsafKaragila Does it mean that there are no absolute complements in ZFC?
– user161005
12 mins ago
@AsafKaragila Does it mean that there are no absolute complements in ZFC?
– user161005
12 mins ago
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
This is similar to Russell's paradox.
The usual approach is to cast blame on the principle of unrestricted comprehension — the hypothesis that for any property $P$ of sets, there is a set of everything satisfying $P$.
ZFC replaces this with the more modest hypothesis that if you're additionally given a set $S$, then you can form the set of everything in $S$ satisfying $P$.
In particular, in the usual formulation of modern set theory, the thing you call "non-R" is not a set. (c.f. "proper class")
answered 47 mins ago
Hurkyl
109k9113254
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is similar to Russell's paradox.
The usual approach is to cast blame on the principle of unrestricted comprehension — the hypothesis that for any property $P$ of sets, there is a set of everything satisfying $P$.
ZFC replaces this with the more modest hypothesis that if you're additionally given a set $S$, then you can form the set of everything in $S$ satisfying $P$.
In particular, in the usual formulation of modern set theory, the thing you call "non-R" is not a set. (c.f. "proper class")
answered 47 mins ago
Hurkyl
109k9113254
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
add a comment |Â
up vote
3
down vote
This is similar to Russell's paradox.
The usual approach is to cast blame on the principle of unrestricted comprehension — the hypothesis that for any property $P$ of sets, there is a set of everything satisfying $P$.
ZFC replaces this with the more modest hypothesis that if you're additionally given a set $S$, then you can form the set of everything in $S$ satisfying $P$.
In particular, in the usual formulation of modern set theory, the thing you call "non-R" is not a set. (c.f. "proper class")
answered 47 mins ago
Hurkyl
109k9113254
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is similar to Russell's paradox.
The usual approach is to cast blame on the principle of unrestricted comprehension — the hypothesis that for any property $P$ of sets, there is a set of everything satisfying $P$.
ZFC replaces this with the more modest hypothesis that if you're additionally given a set $S$, then you can form the set of everything in $S$ satisfying $P$.
In particular, in the usual formulation of modern set theory, the thing you call "non-R" is not a set. (c.f. "proper class")
answered 47 mins ago
Hurkyl
109k9113254
This is similar to Russell's paradox.
The usual approach is to cast blame on the principle of unrestricted comprehension — the hypothesis that for any property $P$ of sets, there is a set of everything satisfying $P$.
ZFC replaces this with the more modest hypothesis that if you're additionally given a set $S$, then you can form the set of everything in $S$ satisfying $P$.
In particular, in the usual formulation of modern set theory, the thing you call "non-R" is not a set. (c.f. "proper class")
answered 47 mins ago
Hurkyl
109k9113254
answered 47 mins ago
Hurkyl
109k9113254
answered 47 mins ago
Hurkyl
109k9113254
answered 47 mins ago
Hurkyl
109k9113254
109k9113254
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
add a comment |Â
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
Wait a minute, isn't "non-R set" the same thing as the absolute complement of set R?
– user161005
13 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
@user161005: Yes, if I understand what you mean by the term. Thus the class "R" and the class "non-R" cannot both be sets.
– Hurkyl
11 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
Does it mean that there are no absolute complements in ZFC?
– user161005
8 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
@user161005 Exactly.
– M. Winter
2 mins ago
add a comment |Â
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2
I suspect the answer is something like "there is no such thing as the set non-$mathbb R$". A similar idea is that there is no "set of all things", because it would have to contain itself. Trying to introduce a set of all things that are not real numbers is similar to trying to introduce a set of all things.
– littleO
44 mins ago
That a set cannot be "its own member"is due in $mathsf ZFC$ to the Axiom of regularity : "The axiom implies that no set is an element of itself, ". We may have theories without it, and this does not mean leave exluded middle. See Non-well-founded set theory.
– Mauro ALLEGRANZA
39 mins ago
Your example absolutely does not work. The proper translation is "$xin mathbb R$ or $neg xin mathbb R$". What you wrote literally makes no sense.
– Git Gud
37 mins ago
See How can I prove that set membership is irreflexive ?
– Mauro ALLEGRANZA
36 mins ago
1
@AsafKaragila Does it mean that there are no absolute complements in ZFC?
– user161005
12 mins ago