Does set theory allow for infinite ascending membership chains? Is every set itself a member of another set?
0
$begingroup$
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
65.7k8160250
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
$endgroup$
add a comment |
0
$begingroup$
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
65.7k8160250
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
$endgroup$
$begingroup$
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
$begingroup$
@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...
$endgroup$
– Matt D
Dec 26 '18 at 18:17
$begingroup$
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
$endgroup$
– Matt D
Dec 26 '18 at 18:28
add a comment |
0
0
0
$begingroup$
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
65.7k8160250
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
$endgroup$
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
65.7k8160250
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
65.7k8160250
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
65.7k8160250
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
65.7k8160250
edited Dec 26 '18 at 18:05
Andrés E. Caicedo
65.7k8160250
65.7k8160250
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
asked Dec 26 '18 at 18:01
Matt DMatt D
7319
7319
$begingroup$
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
$begingroup$
@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...
$endgroup$
– Matt D
Dec 26 '18 at 18:17
$begingroup$
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
$endgroup$
– Matt D
Dec 26 '18 at 18:28
add a comment |
$begingroup$
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
$begingroup$
@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...
$endgroup$
– Matt D
Dec 26 '18 at 18:17
$begingroup$
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
$endgroup$
– Matt D
Dec 26 '18 at 18:28
$begingroup$
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
$begingroup$
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
$begingroup$
@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...
$endgroup$
– Matt D
Dec 26 '18 at 18:17
$begingroup$
@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...
$endgroup$
– Matt D
Dec 26 '18 at 18:17
$begingroup$
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
$endgroup$
– Matt D
Dec 26 '18 at 18:28
$begingroup$
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
$endgroup$
– Matt D
Dec 26 '18 at 18:28
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
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 EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
1
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053157%2fdoes-set-theory-allow-for-infinite-ascending-membership-chains-is-every-set-its%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
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 EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
1
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
5
$begingroup$
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 EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
1
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
5
5
5
$begingroup$
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 EitzenHagen von Eitzen
282k23272507
$endgroup$
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 EitzenHagen von Eitzen
282k23272507
answered Dec 26 '18 at 18:04
Hagen von EitzenHagen von Eitzen
282k23272507
answered Dec 26 '18 at 18:04
Hagen von EitzenHagen von Eitzen
282k23272507
answered Dec 26 '18 at 18:04
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
1
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
1
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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.
$endgroup$
– Matt D
Dec 26 '18 at 18:06
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
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}}} --> ...)
$endgroup$
– Matt D
Dec 26 '18 at 18:08
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
$begingroup$
@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).
$endgroup$
– Arturo Magidin
Dec 26 '18 at 18:17
1
1
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
$begingroup$
@MattD Nope. $V_omega$ - the set of hereditarily finite sets - satisfies all the ZFC axioms except Infinity.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:37
1
1
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 26 '18 at 18:59
|
show 5 more comments
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053157%2fdoes-set-theory-allow-for-infinite-ascending-membership-chains-is-every-set-its%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
See Zermelo ordinals : "Each natural number is then equal to the set containing just the natural number preceding it."
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 18:14
$begingroup$
@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...
$endgroup$
– Matt D
Dec 26 '18 at 18:17
$begingroup$
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
$endgroup$
– Matt D
Dec 26 '18 at 18:28