Does set theory allow for infinite ascending membership chains? Is every set itself a member of another set?
0
I was thinking about a variant of Russell's paradox: Is it possible to specify a set of all objects that are themselves not members of any set?
Membership in this set would be clearly contradictory. However, there's another option: perhaps this set is empty. But if that's the case, then that would imply that every set is itself a member of another set. In other words: an infinite ascending membership chain.
Does set theory allow for this? Or is my definition -- a set of all objects that are themselves not members of any set -- disallowed by one of the axioms of modern set theory?
set-theory
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:01
Matt D
7819
add a comment |
0
I was thinking about a variant of Russell's paradox: Is it possible to specify a set of all objects that are themselves not members of any set?
Membership in this set would be clearly contradictory. However, there's another option: perhaps this set is empty. But if that's the case, then that would imply that every set is itself a member of another set. In other words: an infinite ascending membership chain.
Does set theory allow for this? Or is my definition -- a set of all objects that are themselves not members of any set -- disallowed by one of the axioms of modern set theory?
set-theory
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:01
Matt D
7819
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
@MauroALLEGRANZA Thanks! That makes sense... so my original definition would translate to: "a number that has no greater number", perhaps? It would seem that the infinite ascending membership chain is nothing more than the cardinality of the natural numbers encoded as Zermelo ordinals...
– Matt D
Dec 26 '18 at 18:17
Reading up on this, it seems like my question might be best answered by the Axiom of Infinity: en.wikipedia.org/wiki/Axiom_of_infinity
– Matt D
Dec 26 '18 at 18:28
add a comment |
0
0
0
I was thinking about a variant of Russell's paradox: Is it possible to specify a set of all objects that are themselves not members of any set?
Membership in this set would be clearly contradictory. However, there's another option: perhaps this set is empty. But if that's the case, then that would imply that every set is itself a member of another set. In other words: an infinite ascending membership chain.
Does set theory allow for this? Or is my definition -- a set of all objects that are themselves not members of any set -- disallowed by one of the axioms of modern set theory?
set-theory
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:01
Matt D
7819
I was thinking about a variant of Russell's paradox: Is it possible to specify a set of all objects that are themselves not members of any set?
Membership in this set would be clearly contradictory. However, there's another option: perhaps this set is empty. But if that's the case, then that would imply that every set is itself a member of another set. In other words: an infinite ascending membership chain.
Does set theory allow for this? Or is my definition -- a set of all objects that are themselves not members of any set -- disallowed by one of the axioms of modern set theory?
set-theory
set-theory
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:01
Matt D
7819
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
asked Dec 26 '18 at 18:01
Matt D
7819
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
64.7k8158246
64.7k8158246
asked Dec 26 '18 at 18:01
Matt D
7819
asked Dec 26 '18 at 18:01
Matt D
7819
asked Dec 26 '18 at 18:01
Matt D
7819
7819
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
@MauroALLEGRANZA Thanks! That makes sense... so my original definition would translate to: "a number that has no greater number", perhaps? It would seem that the infinite ascending membership chain is nothing more than the cardinality of the natural numbers encoded as Zermelo ordinals...
– Matt D
Dec 26 '18 at 18:17
Reading up on this, it seems like my question might be best answered by the Axiom of Infinity: en.wikipedia.org/wiki/Axiom_of_infinity
– Matt D
Dec 26 '18 at 18:28
add a comment |
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
@MauroALLEGRANZA Thanks! That makes sense... so my original definition would translate to: "a number that has no greater number", perhaps? It would seem that the infinite ascending membership chain is nothing more than the cardinality of the natural numbers encoded as Zermelo ordinals...
– Matt D
Dec 26 '18 at 18:17
Reading up on this, it seems like my question might be best answered by the Axiom of Infinity: en.wikipedia.org/wiki/Axiom_of_infinity
– Matt D
Dec 26 '18 at 18:28
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
@MauroALLEGRANZA Thanks! That makes sense... so my original definition would translate to: "a number that has no greater number", perhaps? It would seem that the infinite ascending membership chain is nothing more than the cardinality of the natural numbers encoded as Zermelo ordinals...
– Matt D
Dec 26 '18 at 18:17
@MauroALLEGRANZA Thanks! That makes sense... so my original definition would translate to: "a number that has no greater number", perhaps? It would seem that the infinite ascending membership chain is nothing more than the cardinality of the natural numbers encoded as Zermelo ordinals...
– Matt D
Dec 26 '18 at 18:17
Reading up on this, it seems like my question might be best answered by the Axiom of Infinity: en.wikipedia.org/wiki/Axiom_of_infinity
– Matt D
Dec 26 '18 at 18:28
Reading up on this, it seems like my question might be best answered by the Axiom of Infinity: en.wikipedia.org/wiki/Axiom_of_infinity
– Matt D
Dec 26 '18 at 18:28
add a comment |
1 Answer
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Note that for every set $x$, $$xin{x},$$ where the latter set is postulated to exist per the Pairing Axiom
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
1
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
1
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
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Note that for every set $x$, $$xin{x},$$ where the latter set is postulated to exist per the Pairing Axiom
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
1
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
1
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
5
Note that for every set $x$, $$xin{x},$$ where the latter set is postulated to exist per the Pairing Axiom
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
1
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
1
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
5
5
5
Note that for every set $x$, $$xin{x},$$ where the latter set is postulated to exist per the Pairing Axiom
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
Note that for every set $x$, $$xin{x},$$ where the latter set is postulated to exist per the Pairing Axiom
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
answered Dec 26 '18 at 18:04
Hagen von Eitzen
276k21269496
276k21269496
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
1
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
1
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
1
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
1
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
That makes sense, thanks! I need to wait 10 minutes to accept an answer, but this seems right: you're saying that every object is a member of its own singleton set.
– Matt D
Dec 26 '18 at 18:06
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
Wait, would this mean that the singleton set, as a distinct object, itself has a containing singleton set? Would that imply an infinite ascending chain? (e.g. A --> {A} --> {{A}} --> {{{A}}} --> ...)
– Matt D
Dec 26 '18 at 18:08
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
@MattD: Yes, it would; the fact that they are distinct would follow from Foundation, but if you accept an anti-foundation postulate that allows $Ain A$, you would have ${A}=A$, and you would also get an infinite ascending chain (though each set in the chain is equal).
– Arturo Magidin
Dec 26 '18 at 18:17
1
1
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
– Noah Schweber
Dec 26 '18 at 18:37
1
1
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
@MattD $V_omega$ is an infinite set, but remember: it doesn't contain itself as an element! So $V_omega$ doesn't think that there is an infinite set: from $V_omega$'s perspective, $V_omega$ itself is a proper class (namely, the class of all sets). This can be pretty confusing at first; to keep everything clear you'll want to learn the basics of model theory, which makes notions like "from $V_omega$'s perspective" precise.
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
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See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
@MauroALLEGRANZA Thanks! That makes sense... so my original definition would translate to: "a number that has no greater number", perhaps? It would seem that the infinite ascending membership chain is nothing more than the cardinality of the natural numbers encoded as Zermelo ordinals...
– Matt D
Dec 26 '18 at 18:17
Reading up on this, it seems like my question might be best answered by the Axiom of Infinity: en.wikipedia.org/wiki/Axiom_of_infinity
– Matt D
Dec 26 '18 at 18:28