Question about ZF set theory and the way of defining a set











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I have some questions about ZF set theory. I was reading some courses about the construction of $mathbb{N}$, $mathbb{Z}$, $mathbb{Q}$ and $mathbb{R}$ and something disturbes me a bit.



I saw things like :
$$mathbb{N} = {n mid n in I text{ for every inductive set } I},$$
or
$$mathbb{Q} = left{frac{p}{q} mathrel{}middle|mathrel{} p in mathbb{Z}, q in mathbb{N}^{ast}right},$$
and I was wondering if it is really correct to write these sets in such a way.



If I'm not wrong, in ZF, if we consider a set $A$, we can always construct a set $B$ such that :
$$B = {x in A mid phi(x)} , $$
where $phi$ is a formula in ZF language (I'm not really a specialist in logic, but I get the idea of "formula") and it is called the "axiom schema of specification". But we cannot construct set of the form :
$${x mid phi(x)}$$
with this axiom schema right (otherwise, we can find things like Russell's paradox) ?



Now, if we take for example the way I wrote $mathbb{N}$ above, there are two possibilities for me :




  1. This way of writing is not correct. In that case, how should we write in particular the set $mathbb{N}$ ? I read some things about von Neumann's construction of $mathbb{N}$ and about Peano's axioms, but I didn't see exactly a way to "write succinctly" $mathbb{N}$...


  2. This way of writing is correct. In that case, which axiom of ZF theory is used to write for example $mathbb{N}$ or $mathbb{Q}$ in that way ?



Thank you for your help.










share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    I have some questions about ZF set theory. I was reading some courses about the construction of $mathbb{N}$, $mathbb{Z}$, $mathbb{Q}$ and $mathbb{R}$ and something disturbes me a bit.



    I saw things like :
    $$mathbb{N} = {n mid n in I text{ for every inductive set } I},$$
    or
    $$mathbb{Q} = left{frac{p}{q} mathrel{}middle|mathrel{} p in mathbb{Z}, q in mathbb{N}^{ast}right},$$
    and I was wondering if it is really correct to write these sets in such a way.



    If I'm not wrong, in ZF, if we consider a set $A$, we can always construct a set $B$ such that :
    $$B = {x in A mid phi(x)} , $$
    where $phi$ is a formula in ZF language (I'm not really a specialist in logic, but I get the idea of "formula") and it is called the "axiom schema of specification". But we cannot construct set of the form :
    $${x mid phi(x)}$$
    with this axiom schema right (otherwise, we can find things like Russell's paradox) ?



    Now, if we take for example the way I wrote $mathbb{N}$ above, there are two possibilities for me :




    1. This way of writing is not correct. In that case, how should we write in particular the set $mathbb{N}$ ? I read some things about von Neumann's construction of $mathbb{N}$ and about Peano's axioms, but I didn't see exactly a way to "write succinctly" $mathbb{N}$...


    2. This way of writing is correct. In that case, which axiom of ZF theory is used to write for example $mathbb{N}$ or $mathbb{Q}$ in that way ?



    Thank you for your help.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have some questions about ZF set theory. I was reading some courses about the construction of $mathbb{N}$, $mathbb{Z}$, $mathbb{Q}$ and $mathbb{R}$ and something disturbes me a bit.



      I saw things like :
      $$mathbb{N} = {n mid n in I text{ for every inductive set } I},$$
      or
      $$mathbb{Q} = left{frac{p}{q} mathrel{}middle|mathrel{} p in mathbb{Z}, q in mathbb{N}^{ast}right},$$
      and I was wondering if it is really correct to write these sets in such a way.



      If I'm not wrong, in ZF, if we consider a set $A$, we can always construct a set $B$ such that :
      $$B = {x in A mid phi(x)} , $$
      where $phi$ is a formula in ZF language (I'm not really a specialist in logic, but I get the idea of "formula") and it is called the "axiom schema of specification". But we cannot construct set of the form :
      $${x mid phi(x)}$$
      with this axiom schema right (otherwise, we can find things like Russell's paradox) ?



      Now, if we take for example the way I wrote $mathbb{N}$ above, there are two possibilities for me :




      1. This way of writing is not correct. In that case, how should we write in particular the set $mathbb{N}$ ? I read some things about von Neumann's construction of $mathbb{N}$ and about Peano's axioms, but I didn't see exactly a way to "write succinctly" $mathbb{N}$...


      2. This way of writing is correct. In that case, which axiom of ZF theory is used to write for example $mathbb{N}$ or $mathbb{Q}$ in that way ?



      Thank you for your help.










      share|cite|improve this question















      I have some questions about ZF set theory. I was reading some courses about the construction of $mathbb{N}$, $mathbb{Z}$, $mathbb{Q}$ and $mathbb{R}$ and something disturbes me a bit.



      I saw things like :
      $$mathbb{N} = {n mid n in I text{ for every inductive set } I},$$
      or
      $$mathbb{Q} = left{frac{p}{q} mathrel{}middle|mathrel{} p in mathbb{Z}, q in mathbb{N}^{ast}right},$$
      and I was wondering if it is really correct to write these sets in such a way.



      If I'm not wrong, in ZF, if we consider a set $A$, we can always construct a set $B$ such that :
      $$B = {x in A mid phi(x)} , $$
      where $phi$ is a formula in ZF language (I'm not really a specialist in logic, but I get the idea of "formula") and it is called the "axiom schema of specification". But we cannot construct set of the form :
      $${x mid phi(x)}$$
      with this axiom schema right (otherwise, we can find things like Russell's paradox) ?



      Now, if we take for example the way I wrote $mathbb{N}$ above, there are two possibilities for me :




      1. This way of writing is not correct. In that case, how should we write in particular the set $mathbb{N}$ ? I read some things about von Neumann's construction of $mathbb{N}$ and about Peano's axioms, but I didn't see exactly a way to "write succinctly" $mathbb{N}$...


      2. This way of writing is correct. In that case, which axiom of ZF theory is used to write for example $mathbb{N}$ or $mathbb{Q}$ in that way ?



      Thank you for your help.







      elementary-set-theory






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      edited Nov 21 at 7:58









      Asaf Karagila

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      asked Nov 20 at 15:53









      deeppinkwater

      618




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          The axioms of ZF only talk about abstraction of the form ${xin X:|:varphi}$, true; but that doesn't mean we can't write the same set another way. In particular, ${x:|:varphi}$ is still a perfectly acceptable way to write something when you know it's a set. It's worth noting, since writers and instructors don't always make this clear, that ${x:|:xin Xwedgevarphi}$ is exactly the same thing as ${xin X:|:varphi}$, so nothing specifically rides on whether the membership claim is in that first space. The important thing is that ${x:|:varphi}$ is a set exactly when there's an $X$ such that $xin Xwedgevarphiiffvarphi$, which is exactly what's happening in the two examples you give.



          In the case of $mathbb{N}$ we are given by the ZF axioms that there is an inductive set $I$; the elements that are in every inductive set will also all be in $I$, so there's no difference in the between the extension of "$x$ is a member of every inductive set" and "$xin I$ and $x$ is a member of every inductive set."



          In the case of $mathbb{Q}$, as it's written above it's a little bit question-begging. We might write it that way if we're working in $mathbb{R}$ and defining the rational numbers in $mathbb{R}$ by the naturals and integers in $mathbb{R}$; in which case because we've taken two sets of reals and applied an operation defined only on the reals, under which the reals are closed, "$x$ is a quotient of an integer by a non-zero natural" and "$xin mathbb{R}$ and $x$ is a quotient of an integer by a non-zero natural" once again refer to exactly the same collection.



          You are right to observe that in a fully formal proof, you would probably prove existence via first using that ${x:|:xin Xwedgevarphi}$ is an instance of separation, and then proving that $xin Xwedgevarphiiffvarphi$ to justify later use of ${x:|:varphi}$; but in practice it's usually obvious (or presumed obvious) how to supply the appropriate $X$ if we really need to, so we skip it and just write the more natural form.






          share|cite|improve this answer





















          • Okay, I understand now. Thank you !
            – deeppinkwater
            Nov 21 at 7:40












          • Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
            – deeppinkwater
            Nov 21 at 7:48










          • @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
            – Malice Vidrine
            Nov 21 at 9:58










          • Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
            – deeppinkwater
            Nov 21 at 10:45












          • No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
            – Malice Vidrine
            Nov 21 at 10:51













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          The axioms of ZF only talk about abstraction of the form ${xin X:|:varphi}$, true; but that doesn't mean we can't write the same set another way. In particular, ${x:|:varphi}$ is still a perfectly acceptable way to write something when you know it's a set. It's worth noting, since writers and instructors don't always make this clear, that ${x:|:xin Xwedgevarphi}$ is exactly the same thing as ${xin X:|:varphi}$, so nothing specifically rides on whether the membership claim is in that first space. The important thing is that ${x:|:varphi}$ is a set exactly when there's an $X$ such that $xin Xwedgevarphiiffvarphi$, which is exactly what's happening in the two examples you give.



          In the case of $mathbb{N}$ we are given by the ZF axioms that there is an inductive set $I$; the elements that are in every inductive set will also all be in $I$, so there's no difference in the between the extension of "$x$ is a member of every inductive set" and "$xin I$ and $x$ is a member of every inductive set."



          In the case of $mathbb{Q}$, as it's written above it's a little bit question-begging. We might write it that way if we're working in $mathbb{R}$ and defining the rational numbers in $mathbb{R}$ by the naturals and integers in $mathbb{R}$; in which case because we've taken two sets of reals and applied an operation defined only on the reals, under which the reals are closed, "$x$ is a quotient of an integer by a non-zero natural" and "$xin mathbb{R}$ and $x$ is a quotient of an integer by a non-zero natural" once again refer to exactly the same collection.



          You are right to observe that in a fully formal proof, you would probably prove existence via first using that ${x:|:xin Xwedgevarphi}$ is an instance of separation, and then proving that $xin Xwedgevarphiiffvarphi$ to justify later use of ${x:|:varphi}$; but in practice it's usually obvious (or presumed obvious) how to supply the appropriate $X$ if we really need to, so we skip it and just write the more natural form.






          share|cite|improve this answer





















          • Okay, I understand now. Thank you !
            – deeppinkwater
            Nov 21 at 7:40












          • Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
            – deeppinkwater
            Nov 21 at 7:48










          • @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
            – Malice Vidrine
            Nov 21 at 9:58










          • Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
            – deeppinkwater
            Nov 21 at 10:45












          • No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
            – Malice Vidrine
            Nov 21 at 10:51

















          up vote
          1
          down vote



          accepted










          The axioms of ZF only talk about abstraction of the form ${xin X:|:varphi}$, true; but that doesn't mean we can't write the same set another way. In particular, ${x:|:varphi}$ is still a perfectly acceptable way to write something when you know it's a set. It's worth noting, since writers and instructors don't always make this clear, that ${x:|:xin Xwedgevarphi}$ is exactly the same thing as ${xin X:|:varphi}$, so nothing specifically rides on whether the membership claim is in that first space. The important thing is that ${x:|:varphi}$ is a set exactly when there's an $X$ such that $xin Xwedgevarphiiffvarphi$, which is exactly what's happening in the two examples you give.



          In the case of $mathbb{N}$ we are given by the ZF axioms that there is an inductive set $I$; the elements that are in every inductive set will also all be in $I$, so there's no difference in the between the extension of "$x$ is a member of every inductive set" and "$xin I$ and $x$ is a member of every inductive set."



          In the case of $mathbb{Q}$, as it's written above it's a little bit question-begging. We might write it that way if we're working in $mathbb{R}$ and defining the rational numbers in $mathbb{R}$ by the naturals and integers in $mathbb{R}$; in which case because we've taken two sets of reals and applied an operation defined only on the reals, under which the reals are closed, "$x$ is a quotient of an integer by a non-zero natural" and "$xin mathbb{R}$ and $x$ is a quotient of an integer by a non-zero natural" once again refer to exactly the same collection.



          You are right to observe that in a fully formal proof, you would probably prove existence via first using that ${x:|:xin Xwedgevarphi}$ is an instance of separation, and then proving that $xin Xwedgevarphiiffvarphi$ to justify later use of ${x:|:varphi}$; but in practice it's usually obvious (or presumed obvious) how to supply the appropriate $X$ if we really need to, so we skip it and just write the more natural form.






          share|cite|improve this answer





















          • Okay, I understand now. Thank you !
            – deeppinkwater
            Nov 21 at 7:40












          • Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
            – deeppinkwater
            Nov 21 at 7:48










          • @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
            – Malice Vidrine
            Nov 21 at 9:58










          • Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
            – deeppinkwater
            Nov 21 at 10:45












          • No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
            – Malice Vidrine
            Nov 21 at 10:51















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The axioms of ZF only talk about abstraction of the form ${xin X:|:varphi}$, true; but that doesn't mean we can't write the same set another way. In particular, ${x:|:varphi}$ is still a perfectly acceptable way to write something when you know it's a set. It's worth noting, since writers and instructors don't always make this clear, that ${x:|:xin Xwedgevarphi}$ is exactly the same thing as ${xin X:|:varphi}$, so nothing specifically rides on whether the membership claim is in that first space. The important thing is that ${x:|:varphi}$ is a set exactly when there's an $X$ such that $xin Xwedgevarphiiffvarphi$, which is exactly what's happening in the two examples you give.



          In the case of $mathbb{N}$ we are given by the ZF axioms that there is an inductive set $I$; the elements that are in every inductive set will also all be in $I$, so there's no difference in the between the extension of "$x$ is a member of every inductive set" and "$xin I$ and $x$ is a member of every inductive set."



          In the case of $mathbb{Q}$, as it's written above it's a little bit question-begging. We might write it that way if we're working in $mathbb{R}$ and defining the rational numbers in $mathbb{R}$ by the naturals and integers in $mathbb{R}$; in which case because we've taken two sets of reals and applied an operation defined only on the reals, under which the reals are closed, "$x$ is a quotient of an integer by a non-zero natural" and "$xin mathbb{R}$ and $x$ is a quotient of an integer by a non-zero natural" once again refer to exactly the same collection.



          You are right to observe that in a fully formal proof, you would probably prove existence via first using that ${x:|:xin Xwedgevarphi}$ is an instance of separation, and then proving that $xin Xwedgevarphiiffvarphi$ to justify later use of ${x:|:varphi}$; but in practice it's usually obvious (or presumed obvious) how to supply the appropriate $X$ if we really need to, so we skip it and just write the more natural form.






          share|cite|improve this answer












          The axioms of ZF only talk about abstraction of the form ${xin X:|:varphi}$, true; but that doesn't mean we can't write the same set another way. In particular, ${x:|:varphi}$ is still a perfectly acceptable way to write something when you know it's a set. It's worth noting, since writers and instructors don't always make this clear, that ${x:|:xin Xwedgevarphi}$ is exactly the same thing as ${xin X:|:varphi}$, so nothing specifically rides on whether the membership claim is in that first space. The important thing is that ${x:|:varphi}$ is a set exactly when there's an $X$ such that $xin Xwedgevarphiiffvarphi$, which is exactly what's happening in the two examples you give.



          In the case of $mathbb{N}$ we are given by the ZF axioms that there is an inductive set $I$; the elements that are in every inductive set will also all be in $I$, so there's no difference in the between the extension of "$x$ is a member of every inductive set" and "$xin I$ and $x$ is a member of every inductive set."



          In the case of $mathbb{Q}$, as it's written above it's a little bit question-begging. We might write it that way if we're working in $mathbb{R}$ and defining the rational numbers in $mathbb{R}$ by the naturals and integers in $mathbb{R}$; in which case because we've taken two sets of reals and applied an operation defined only on the reals, under which the reals are closed, "$x$ is a quotient of an integer by a non-zero natural" and "$xin mathbb{R}$ and $x$ is a quotient of an integer by a non-zero natural" once again refer to exactly the same collection.



          You are right to observe that in a fully formal proof, you would probably prove existence via first using that ${x:|:xin Xwedgevarphi}$ is an instance of separation, and then proving that $xin Xwedgevarphiiffvarphi$ to justify later use of ${x:|:varphi}$; but in practice it's usually obvious (or presumed obvious) how to supply the appropriate $X$ if we really need to, so we skip it and just write the more natural form.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 20:52









          Malice Vidrine

          5,92121122




          5,92121122












          • Okay, I understand now. Thank you !
            – deeppinkwater
            Nov 21 at 7:40












          • Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
            – deeppinkwater
            Nov 21 at 7:48










          • @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
            – Malice Vidrine
            Nov 21 at 9:58










          • Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
            – deeppinkwater
            Nov 21 at 10:45












          • No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
            – Malice Vidrine
            Nov 21 at 10:51




















          • Okay, I understand now. Thank you !
            – deeppinkwater
            Nov 21 at 7:40












          • Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
            – deeppinkwater
            Nov 21 at 7:48










          • @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
            – Malice Vidrine
            Nov 21 at 9:58










          • Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
            – deeppinkwater
            Nov 21 at 10:45












          • No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
            – Malice Vidrine
            Nov 21 at 10:51


















          Okay, I understand now. Thank you !
          – deeppinkwater
          Nov 21 at 7:40






          Okay, I understand now. Thank you !
          – deeppinkwater
          Nov 21 at 7:40














          Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
          – deeppinkwater
          Nov 21 at 7:48




          Just a little thing, is there a link between “axiom schema of replacement” in ZF ?
          – deeppinkwater
          Nov 21 at 7:48












          @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
          – Malice Vidrine
          Nov 21 at 9:58




          @deeppinkwater - A link between the axiom schema of replacement and what? I'm not sure I understand the question.
          – Malice Vidrine
          Nov 21 at 9:58












          Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
          – deeppinkwater
          Nov 21 at 10:45






          Sorry, between the axiom schema of replacement and the fact that you said : $x in X wedge varphi Leftrightarrow varphi$ ? When I ask if "there is a link", it is in the sense that I'm not sure to have really understand this axiom, but I have the feeling that there is something similar.
          – deeppinkwater
          Nov 21 at 10:45














          No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
          – Malice Vidrine
          Nov 21 at 10:51






          No, this doesn't necessarily involve replacement. It's because "${x:|:varphi}$ exists" is short for (the universal closure of) "$exists yforall x(xin yLeftrightarrow varphi)$." If a sentence of this form is a theorem, and $forall x(varphiLeftrightarrowpsi)$ is a theorem, then one can infer "${x:|:psi}$ exists," too. It's just a matter of the logic.
          – Malice Vidrine
          Nov 21 at 10:51




















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