What ZFC axiom guarantees the existence of a domain set and a range set?












2












$begingroup$


Suppose we are working with the Kuratowski definition of ordered pairs, and that a relation is a set of ordered pairs. What axiom guarantees that $dom(R) = { x : exists y langle x, y rangle in R}$ and $ran(R) = {y : exists x langle x, y rangle in R}$ are sets? I think that the axiom of replacement should be enough, but I am not so sure.






set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am trying to imagine encountering a relation where I do not know the set of departure and the set of destination. en.wikipedia.org/wiki/Binary_relation#Formal_definition But thought provoking question.
    $endgroup$
    – CopyPasteIt
    Dec 19 '18 at 23:37








  • 1




    $begingroup$
    @CopyPasteIt Maybe the relation $n mathrel{R} kappa$ defined by $n in mathbb{N}$, $kappa$ is a (von Neumann) cardinal, and there exist exactly $n$ infinite cardinals $in kappa$? If you can find the range of this $R$, you can define $aleph_omega$ as the sup of the cardinals in this range...
    $endgroup$
    – Daniel Schepler
    Dec 19 '18 at 23:42
















2












$begingroup$


Suppose we are working with the Kuratowski definition of ordered pairs, and that a relation is a set of ordered pairs. What axiom guarantees that $dom(R) = { x : exists y langle x, y rangle in R}$ and $ran(R) = {y : exists x langle x, y rangle in R}$ are sets? I think that the axiom of replacement should be enough, but I am not so sure.






set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am trying to imagine encountering a relation where I do not know the set of departure and the set of destination. en.wikipedia.org/wiki/Binary_relation#Formal_definition But thought provoking question.
    $endgroup$
    – CopyPasteIt
    Dec 19 '18 at 23:37








  • 1




    $begingroup$
    @CopyPasteIt Maybe the relation $n mathrel{R} kappa$ defined by $n in mathbb{N}$, $kappa$ is a (von Neumann) cardinal, and there exist exactly $n$ infinite cardinals $in kappa$? If you can find the range of this $R$, you can define $aleph_omega$ as the sup of the cardinals in this range...
    $endgroup$
    – Daniel Schepler
    Dec 19 '18 at 23:42














2












2








2





$begingroup$


Suppose we are working with the Kuratowski definition of ordered pairs, and that a relation is a set of ordered pairs. What axiom guarantees that $dom(R) = { x : exists y langle x, y rangle in R}$ and $ran(R) = {y : exists x langle x, y rangle in R}$ are sets? I think that the axiom of replacement should be enough, but I am not so sure.






set-theory







share|cite|improve this question











$endgroup$




Suppose we are working with the Kuratowski definition of ordered pairs, and that a relation is a set of ordered pairs. What axiom guarantees that $dom(R) = { x : exists y langle x, y rangle in R}$ and $ran(R) = {y : exists x langle x, y rangle in R}$ are sets? I think that the axiom of replacement should be enough, but I am not so sure.






set-theory




set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 17:55









Andrés E. Caicedo

65.9k8160252




65.9k8160252










asked Dec 19 '18 at 23:25









Nuntractatuses AmávelNuntractatuses Amável

618112




618112












  • $begingroup$
    I am trying to imagine encountering a relation where I do not know the set of departure and the set of destination. en.wikipedia.org/wiki/Binary_relation#Formal_definition But thought provoking question.
    $endgroup$
    – CopyPasteIt
    Dec 19 '18 at 23:37








  • 1




    $begingroup$
    @CopyPasteIt Maybe the relation $n mathrel{R} kappa$ defined by $n in mathbb{N}$, $kappa$ is a (von Neumann) cardinal, and there exist exactly $n$ infinite cardinals $in kappa$? If you can find the range of this $R$, you can define $aleph_omega$ as the sup of the cardinals in this range...
    $endgroup$
    – Daniel Schepler
    Dec 19 '18 at 23:42


















  • $begingroup$
    I am trying to imagine encountering a relation where I do not know the set of departure and the set of destination. en.wikipedia.org/wiki/Binary_relation#Formal_definition But thought provoking question.
    $endgroup$
    – CopyPasteIt
    Dec 19 '18 at 23:37








  • 1




    $begingroup$
    @CopyPasteIt Maybe the relation $n mathrel{R} kappa$ defined by $n in mathbb{N}$, $kappa$ is a (von Neumann) cardinal, and there exist exactly $n$ infinite cardinals $in kappa$? If you can find the range of this $R$, you can define $aleph_omega$ as the sup of the cardinals in this range...
    $endgroup$
    – Daniel Schepler
    Dec 19 '18 at 23:42
















$begingroup$
I am trying to imagine encountering a relation where I do not know the set of departure and the set of destination. en.wikipedia.org/wiki/Binary_relation#Formal_definition But thought provoking question.
$endgroup$
– CopyPasteIt
Dec 19 '18 at 23:37






$begingroup$
I am trying to imagine encountering a relation where I do not know the set of departure and the set of destination. en.wikipedia.org/wiki/Binary_relation#Formal_definition But thought provoking question.
$endgroup$
– CopyPasteIt
Dec 19 '18 at 23:37






1




1




$begingroup$
@CopyPasteIt Maybe the relation $n mathrel{R} kappa$ defined by $n in mathbb{N}$, $kappa$ is a (von Neumann) cardinal, and there exist exactly $n$ infinite cardinals $in kappa$? If you can find the range of this $R$, you can define $aleph_omega$ as the sup of the cardinals in this range...
$endgroup$
– Daniel Schepler
Dec 19 '18 at 23:42




$begingroup$
@CopyPasteIt Maybe the relation $n mathrel{R} kappa$ defined by $n in mathbb{N}$, $kappa$ is a (von Neumann) cardinal, and there exist exactly $n$ infinite cardinals $in kappa$? If you can find the range of this $R$, you can define $aleph_omega$ as the sup of the cardinals in this range...
$endgroup$
– Daniel Schepler
Dec 19 '18 at 23:42










2 Answers
2






active

oldest

votes


















5












$begingroup$

Instead of replacement, you could use the axioms of union and of separation (unless of course you treat separation as a consequence of replacement rather than an axiom). With Kuratowski's definition of ordered pairs, if $R$ is a relation then $bigcupbigcup R$ is the union of the domain and range of $R$,from which you can extract the domain and the range by separation.



The reason for pointing out this alternative approach is that it works even in Zermelo set theory, which is weaker than Zermelo-Fraenkel set theory.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 0:21










  • $begingroup$
    @AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
    $endgroup$
    – Andreas Blass
    Dec 20 '18 at 0:28



















3












$begingroup$

You're right that that'll work (though as another answer mentions, it's not the only usable approach). Given a relation $R$, replacement maps any $(x,,y)in R$ to $x$, giving the domain. The range follows similarly. All you need is a definition of ordered pairs that implies their coordinates are uniquely defined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:34












  • $begingroup$
    I think I made myself clearer now, sorry for the nonsensical comment.
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:38










  • $begingroup$
    Ah, I understand now. Yes, that'll do.
    $endgroup$
    – J.G.
    Dec 19 '18 at 23:43












Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Instead of replacement, you could use the axioms of union and of separation (unless of course you treat separation as a consequence of replacement rather than an axiom). With Kuratowski's definition of ordered pairs, if $R$ is a relation then $bigcupbigcup R$ is the union of the domain and range of $R$,from which you can extract the domain and the range by separation.



The reason for pointing out this alternative approach is that it works even in Zermelo set theory, which is weaker than Zermelo-Fraenkel set theory.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 0:21










  • $begingroup$
    @AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
    $endgroup$
    – Andreas Blass
    Dec 20 '18 at 0:28
















5












$begingroup$

Instead of replacement, you could use the axioms of union and of separation (unless of course you treat separation as a consequence of replacement rather than an axiom). With Kuratowski's definition of ordered pairs, if $R$ is a relation then $bigcupbigcup R$ is the union of the domain and range of $R$,from which you can extract the domain and the range by separation.



The reason for pointing out this alternative approach is that it works even in Zermelo set theory, which is weaker than Zermelo-Fraenkel set theory.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 0:21










  • $begingroup$
    @AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
    $endgroup$
    – Andreas Blass
    Dec 20 '18 at 0:28














5












5








5





$begingroup$

Instead of replacement, you could use the axioms of union and of separation (unless of course you treat separation as a consequence of replacement rather than an axiom). With Kuratowski's definition of ordered pairs, if $R$ is a relation then $bigcupbigcup R$ is the union of the domain and range of $R$,from which you can extract the domain and the range by separation.



The reason for pointing out this alternative approach is that it works even in Zermelo set theory, which is weaker than Zermelo-Fraenkel set theory.






share|cite|improve this answer









$endgroup$



Instead of replacement, you could use the axioms of union and of separation (unless of course you treat separation as a consequence of replacement rather than an axiom). With Kuratowski's definition of ordered pairs, if $R$ is a relation then $bigcupbigcup R$ is the union of the domain and range of $R$,from which you can extract the domain and the range by separation.



The reason for pointing out this alternative approach is that it works even in Zermelo set theory, which is weaker than Zermelo-Fraenkel set theory.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 23:59









Andreas BlassAndreas Blass

50.5k452109




50.5k452109












  • $begingroup$
    One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 0:21










  • $begingroup$
    @AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
    $endgroup$
    – Andreas Blass
    Dec 20 '18 at 0:28


















  • $begingroup$
    One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
    $endgroup$
    – Asaf Karagila♦
    Dec 20 '18 at 0:21










  • $begingroup$
    @AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
    $endgroup$
    – Andreas Blass
    Dec 20 '18 at 0:28
















$begingroup$
One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 0:21




$begingroup$
One might wonder why we need Replacement, then. But we need it because it guarantees that definable functions with a domain are in fact functions in the universe (as opposed to proper classes).
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 0:21












$begingroup$
@AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
$endgroup$
– Andreas Blass
Dec 20 '18 at 0:28




$begingroup$
@AsafKaragila Right. The proof in Zermelo set theory that the domain and range of a relation are sets depends on the relation's being a set to begin with. (As far as I know, the original motivation for introducing the axiom(s) of replacement was that Zerrmelo set theory can't prove the existence of $beth_omega$.)
$endgroup$
– Andreas Blass
Dec 20 '18 at 0:28











3












$begingroup$

You're right that that'll work (though as another answer mentions, it's not the only usable approach). Given a relation $R$, replacement maps any $(x,,y)in R$ to $x$, giving the domain. The range follows similarly. All you need is a definition of ordered pairs that implies their coordinates are uniquely defined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:34












  • $begingroup$
    I think I made myself clearer now, sorry for the nonsensical comment.
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:38










  • $begingroup$
    Ah, I understand now. Yes, that'll do.
    $endgroup$
    – J.G.
    Dec 19 '18 at 23:43
















3












$begingroup$

You're right that that'll work (though as another answer mentions, it's not the only usable approach). Given a relation $R$, replacement maps any $(x,,y)in R$ to $x$, giving the domain. The range follows similarly. All you need is a definition of ordered pairs that implies their coordinates are uniquely defined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:34












  • $begingroup$
    I think I made myself clearer now, sorry for the nonsensical comment.
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:38










  • $begingroup$
    Ah, I understand now. Yes, that'll do.
    $endgroup$
    – J.G.
    Dec 19 '18 at 23:43














3












3








3





$begingroup$

You're right that that'll work (though as another answer mentions, it's not the only usable approach). Given a relation $R$, replacement maps any $(x,,y)in R$ to $x$, giving the domain. The range follows similarly. All you need is a definition of ordered pairs that implies their coordinates are uniquely defined.






share|cite|improve this answer











$endgroup$



You're right that that'll work (though as another answer mentions, it's not the only usable approach). Given a relation $R$, replacement maps any $(x,,y)in R$ to $x$, giving the domain. The range follows similarly. All you need is a definition of ordered pairs that implies their coordinates are uniquely defined.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 9:27

























answered Dec 19 '18 at 23:32









J.G.J.G.

32.6k23250




32.6k23250












  • $begingroup$
    Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:34












  • $begingroup$
    I think I made myself clearer now, sorry for the nonsensical comment.
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:38










  • $begingroup$
    Ah, I understand now. Yes, that'll do.
    $endgroup$
    – J.G.
    Dec 19 '18 at 23:43


















  • $begingroup$
    Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:34












  • $begingroup$
    I think I made myself clearer now, sorry for the nonsensical comment.
    $endgroup$
    – Nuntractatuses Amável
    Dec 19 '18 at 23:38










  • $begingroup$
    Ah, I understand now. Yes, that'll do.
    $endgroup$
    – J.G.
    Dec 19 '18 at 23:43
















$begingroup$
Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
$endgroup$
– Nuntractatuses Amável
Dec 19 '18 at 23:34






$begingroup$
Would this formula $exists c(x = langle y, c rangle)$ be enough to legitimate the definition of domain, with the replacement axiom (take an instance of the axiom, with this formula as $phi$)?
$endgroup$
– Nuntractatuses Amável
Dec 19 '18 at 23:34














$begingroup$
I think I made myself clearer now, sorry for the nonsensical comment.
$endgroup$
– Nuntractatuses Amável
Dec 19 '18 at 23:38




$begingroup$
I think I made myself clearer now, sorry for the nonsensical comment.
$endgroup$
– Nuntractatuses Amável
Dec 19 '18 at 23:38












$begingroup$
Ah, I understand now. Yes, that'll do.
$endgroup$
– J.G.
Dec 19 '18 at 23:43




$begingroup$
Ah, I understand now. Yes, that'll do.
$endgroup$
– J.G.
Dec 19 '18 at 23:43


















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