Why can't we just add “nothing else is a set” as an axiom?












5












$begingroup$


The axioms of ZF define what a set is by:





  1. $omega$ is a set

  2. If $x$ and $y$ are sets, then ${x, y}$ is a set

  3. If $x$ is a set, then $bigcup x$ is a set

  4. If $x$ is a set, then $mathcal{P}(x)$ is a set

  5. If $x, y_1, ..., y_n$ are sets and $P$ is a statement free variables $x, y, z_1, ..., z_n$, then ${z in x : P}$ is a set


and also defines equality between sets by the axiom of extensionality.



But I don't understand why we can't just add "nothing else is a set" as an axiom to the list, as we usually do when defining a certain mathematical object.



Wouldn't adding that axiom solve the problem of some statements being independent? If we can't prove a set exists, then it means it can't be produced using the rules listed above, so it's not a set.










share|cite|improve this question









$endgroup$












  • $begingroup$
    This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing.
    $endgroup$
    – Robert Wolfe
    4 hours ago








  • 1




    $begingroup$
    You forgot collection/replacement
    $endgroup$
    – Not Mike
    3 hours ago






  • 1




    $begingroup$
    For example, there is a set called $mathcal P(omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set").
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    See also en.wikipedia.org/wiki/Constructible_universe
    $endgroup$
    – Not Mike
    3 hours ago








  • 2




    $begingroup$
    "The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
    $endgroup$
    – Jonathan
    1 hour ago
















5












$begingroup$


The axioms of ZF define what a set is by:





  1. $omega$ is a set

  2. If $x$ and $y$ are sets, then ${x, y}$ is a set

  3. If $x$ is a set, then $bigcup x$ is a set

  4. If $x$ is a set, then $mathcal{P}(x)$ is a set

  5. If $x, y_1, ..., y_n$ are sets and $P$ is a statement free variables $x, y, z_1, ..., z_n$, then ${z in x : P}$ is a set


and also defines equality between sets by the axiom of extensionality.



But I don't understand why we can't just add "nothing else is a set" as an axiom to the list, as we usually do when defining a certain mathematical object.



Wouldn't adding that axiom solve the problem of some statements being independent? If we can't prove a set exists, then it means it can't be produced using the rules listed above, so it's not a set.










share|cite|improve this question









$endgroup$












  • $begingroup$
    This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing.
    $endgroup$
    – Robert Wolfe
    4 hours ago








  • 1




    $begingroup$
    You forgot collection/replacement
    $endgroup$
    – Not Mike
    3 hours ago






  • 1




    $begingroup$
    For example, there is a set called $mathcal P(omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set").
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    See also en.wikipedia.org/wiki/Constructible_universe
    $endgroup$
    – Not Mike
    3 hours ago








  • 2




    $begingroup$
    "The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
    $endgroup$
    – Jonathan
    1 hour ago














5












5








5





$begingroup$


The axioms of ZF define what a set is by:





  1. $omega$ is a set

  2. If $x$ and $y$ are sets, then ${x, y}$ is a set

  3. If $x$ is a set, then $bigcup x$ is a set

  4. If $x$ is a set, then $mathcal{P}(x)$ is a set

  5. If $x, y_1, ..., y_n$ are sets and $P$ is a statement free variables $x, y, z_1, ..., z_n$, then ${z in x : P}$ is a set


and also defines equality between sets by the axiom of extensionality.



But I don't understand why we can't just add "nothing else is a set" as an axiom to the list, as we usually do when defining a certain mathematical object.



Wouldn't adding that axiom solve the problem of some statements being independent? If we can't prove a set exists, then it means it can't be produced using the rules listed above, so it's not a set.










share|cite|improve this question









$endgroup$




The axioms of ZF define what a set is by:





  1. $omega$ is a set

  2. If $x$ and $y$ are sets, then ${x, y}$ is a set

  3. If $x$ is a set, then $bigcup x$ is a set

  4. If $x$ is a set, then $mathcal{P}(x)$ is a set

  5. If $x, y_1, ..., y_n$ are sets and $P$ is a statement free variables $x, y, z_1, ..., z_n$, then ${z in x : P}$ is a set


and also defines equality between sets by the axiom of extensionality.



But I don't understand why we can't just add "nothing else is a set" as an axiom to the list, as we usually do when defining a certain mathematical object.



Wouldn't adding that axiom solve the problem of some statements being independent? If we can't prove a set exists, then it means it can't be produced using the rules listed above, so it's not a set.







set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









StefanStefan

1825




1825












  • $begingroup$
    This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing.
    $endgroup$
    – Robert Wolfe
    4 hours ago








  • 1




    $begingroup$
    You forgot collection/replacement
    $endgroup$
    – Not Mike
    3 hours ago






  • 1




    $begingroup$
    For example, there is a set called $mathcal P(omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set").
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    See also en.wikipedia.org/wiki/Constructible_universe
    $endgroup$
    – Not Mike
    3 hours ago








  • 2




    $begingroup$
    "The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
    $endgroup$
    – Jonathan
    1 hour ago


















  • $begingroup$
    This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing.
    $endgroup$
    – Robert Wolfe
    4 hours ago








  • 1




    $begingroup$
    You forgot collection/replacement
    $endgroup$
    – Not Mike
    3 hours ago






  • 1




    $begingroup$
    For example, there is a set called $mathcal P(omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set").
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    See also en.wikipedia.org/wiki/Constructible_universe
    $endgroup$
    – Not Mike
    3 hours ago








  • 2




    $begingroup$
    "The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
    $endgroup$
    – Jonathan
    1 hour ago
















$begingroup$
This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing.
$endgroup$
– Robert Wolfe
4 hours ago






$begingroup$
This would be akin to a minimal, countable model of ZF. It wouldn't solve incompleteness as Godel's theorem would still apply. Being able to prove that a set exists is a different matter than a set existing.
$endgroup$
– Robert Wolfe
4 hours ago






1




1




$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
3 hours ago




$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
3 hours ago




1




1




$begingroup$
For example, there is a set called $mathcal P(omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set").
$endgroup$
– GEdgar
3 hours ago




$begingroup$
For example, there is a set called $mathcal P(omega)$, but it has only countably many elements (only countably many can be proved from these axioms, and "nothing else is a set").
$endgroup$
– GEdgar
3 hours ago




1




1




$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
3 hours ago






$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
3 hours ago






2




2




$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
1 hour ago




$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
1 hour ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

It's not clear why we need to. One reason we "can't" has to do with a technicality. ZF is a theory in "first-order logic". Saying that nothing is a set except that which the axioms imply is a set is not a "first-order" statement, so if we tried to add it we'd no longer have a first-order theory. That would be too bad, because first-order logic works out a lot nicer than higher-order logics.



There's no way to say "nothing is a set except things that the axioms imply are sets" using just $forall$, $exists$ and $in$.



(In fact, the axioms don't quite say what you say they say! There's no mention of "is a set" in the actual axioms. What you say is maybe how one thinks off what the axioms mean, but in fact, for example the actual axiom (2) is $$forall xforall yexists z(forall t(tin ziff (t= xlor t= y))).$$The axioms talk about sets, not about what is or is not a set; adding "is a set" to the formalism makes it a totally different sort of thing.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
    $endgroup$
    – Arturo Magidin
    3 hours ago






  • 1




    $begingroup$
    @ArturoMagidin Yes of course, wasn't thinking.
    $endgroup$
    – David C. Ullrich
    3 hours ago



















-1












$begingroup$

The approach of defining a recursive data structure $S$ by giving some rules for constructing members of $S$ and saying "anything constructed by these rules is an $S$ and nothing else is an $S$" is a shorthand for saying that $S$ is the intersection of all sets that are closed under the rules. This approach is usually used in contexts where the candidates for membership of $S$ are already known, so the intersection is over a well-defined set. E.g., if the rules are given by a grammar over some set of symbols, $S$ will be a subset of the set of all strings of symbols.



To apply this approach to a first-order axiomatization of set theory, you'd need some mechanism allowing the language of set theory to talk about itself, so that you could say in the language of set theory that the only sets are those that can described following your rules. This isn't impossible, but it would be technically tricky: the axiomatization would depend on something equivalent to a Gödel numbering of the language and you would have to presuppose some notion of the satisfaction relation in order to give meaning to the set comprehensions in your rule 5.



The end result is going to be unsatisfactory. In ZF, we can prove (1) that uncountable sets exist and (2) that the language of ZF is countable. Hence your modification of ZF in which all the sets are expressible by formulas in the language would be inconsistent. To regain consistency, you would have to weaken ZF. There is no obvious weakening that would achieve consistency and would respect the underlying intuitions behind ZF.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
    $endgroup$
    – Noah Schweber
    41 mins ago












  • $begingroup$
    @NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
    $endgroup$
    – Rob Arthan
    31 mins ago












  • $begingroup$
    Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
    $endgroup$
    – Rob Arthan
    7 mins ago











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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

It's not clear why we need to. One reason we "can't" has to do with a technicality. ZF is a theory in "first-order logic". Saying that nothing is a set except that which the axioms imply is a set is not a "first-order" statement, so if we tried to add it we'd no longer have a first-order theory. That would be too bad, because first-order logic works out a lot nicer than higher-order logics.



There's no way to say "nothing is a set except things that the axioms imply are sets" using just $forall$, $exists$ and $in$.



(In fact, the axioms don't quite say what you say they say! There's no mention of "is a set" in the actual axioms. What you say is maybe how one thinks off what the axioms mean, but in fact, for example the actual axiom (2) is $$forall xforall yexists z(forall t(tin ziff (t= xlor t= y))).$$The axioms talk about sets, not about what is or is not a set; adding "is a set" to the formalism makes it a totally different sort of thing.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
    $endgroup$
    – Arturo Magidin
    3 hours ago






  • 1




    $begingroup$
    @ArturoMagidin Yes of course, wasn't thinking.
    $endgroup$
    – David C. Ullrich
    3 hours ago
















7












$begingroup$

It's not clear why we need to. One reason we "can't" has to do with a technicality. ZF is a theory in "first-order logic". Saying that nothing is a set except that which the axioms imply is a set is not a "first-order" statement, so if we tried to add it we'd no longer have a first-order theory. That would be too bad, because first-order logic works out a lot nicer than higher-order logics.



There's no way to say "nothing is a set except things that the axioms imply are sets" using just $forall$, $exists$ and $in$.



(In fact, the axioms don't quite say what you say they say! There's no mention of "is a set" in the actual axioms. What you say is maybe how one thinks off what the axioms mean, but in fact, for example the actual axiom (2) is $$forall xforall yexists z(forall t(tin ziff (t= xlor t= y))).$$The axioms talk about sets, not about what is or is not a set; adding "is a set" to the formalism makes it a totally different sort of thing.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
    $endgroup$
    – Arturo Magidin
    3 hours ago






  • 1




    $begingroup$
    @ArturoMagidin Yes of course, wasn't thinking.
    $endgroup$
    – David C. Ullrich
    3 hours ago














7












7








7





$begingroup$

It's not clear why we need to. One reason we "can't" has to do with a technicality. ZF is a theory in "first-order logic". Saying that nothing is a set except that which the axioms imply is a set is not a "first-order" statement, so if we tried to add it we'd no longer have a first-order theory. That would be too bad, because first-order logic works out a lot nicer than higher-order logics.



There's no way to say "nothing is a set except things that the axioms imply are sets" using just $forall$, $exists$ and $in$.



(In fact, the axioms don't quite say what you say they say! There's no mention of "is a set" in the actual axioms. What you say is maybe how one thinks off what the axioms mean, but in fact, for example the actual axiom (2) is $$forall xforall yexists z(forall t(tin ziff (t= xlor t= y))).$$The axioms talk about sets, not about what is or is not a set; adding "is a set" to the formalism makes it a totally different sort of thing.)






share|cite|improve this answer











$endgroup$



It's not clear why we need to. One reason we "can't" has to do with a technicality. ZF is a theory in "first-order logic". Saying that nothing is a set except that which the axioms imply is a set is not a "first-order" statement, so if we tried to add it we'd no longer have a first-order theory. That would be too bad, because first-order logic works out a lot nicer than higher-order logics.



There's no way to say "nothing is a set except things that the axioms imply are sets" using just $forall$, $exists$ and $in$.



(In fact, the axioms don't quite say what you say they say! There's no mention of "is a set" in the actual axioms. What you say is maybe how one thinks off what the axioms mean, but in fact, for example the actual axiom (2) is $$forall xforall yexists z(forall t(tin ziff (t= xlor t= y))).$$The axioms talk about sets, not about what is or is not a set; adding "is a set" to the formalism makes it a totally different sort of thing.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 3 hours ago









David C. UllrichDavid C. Ullrich

59.7k43893




59.7k43893












  • $begingroup$
    Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
    $endgroup$
    – Arturo Magidin
    3 hours ago






  • 1




    $begingroup$
    @ArturoMagidin Yes of course, wasn't thinking.
    $endgroup$
    – David C. Ullrich
    3 hours ago


















  • $begingroup$
    Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
    $endgroup$
    – Arturo Magidin
    3 hours ago






  • 1




    $begingroup$
    @ArturoMagidin Yes of course, wasn't thinking.
    $endgroup$
    – David C. Ullrich
    3 hours ago
















$begingroup$
Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
$endgroup$
– Arturo Magidin
3 hours ago




$begingroup$
Are you sure the Axiom of pairs is what you write? Shouldn’t that be $t=x lor t=y$? Otherwise, what you get is $xcup y$.
$endgroup$
– Arturo Magidin
3 hours ago




1




1




$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
3 hours ago




$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
3 hours ago











-1












$begingroup$

The approach of defining a recursive data structure $S$ by giving some rules for constructing members of $S$ and saying "anything constructed by these rules is an $S$ and nothing else is an $S$" is a shorthand for saying that $S$ is the intersection of all sets that are closed under the rules. This approach is usually used in contexts where the candidates for membership of $S$ are already known, so the intersection is over a well-defined set. E.g., if the rules are given by a grammar over some set of symbols, $S$ will be a subset of the set of all strings of symbols.



To apply this approach to a first-order axiomatization of set theory, you'd need some mechanism allowing the language of set theory to talk about itself, so that you could say in the language of set theory that the only sets are those that can described following your rules. This isn't impossible, but it would be technically tricky: the axiomatization would depend on something equivalent to a Gödel numbering of the language and you would have to presuppose some notion of the satisfaction relation in order to give meaning to the set comprehensions in your rule 5.



The end result is going to be unsatisfactory. In ZF, we can prove (1) that uncountable sets exist and (2) that the language of ZF is countable. Hence your modification of ZF in which all the sets are expressible by formulas in the language would be inconsistent. To regain consistency, you would have to weaken ZF. There is no obvious weakening that would achieve consistency and would respect the underlying intuitions behind ZF.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
    $endgroup$
    – Noah Schweber
    41 mins ago












  • $begingroup$
    @NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
    $endgroup$
    – Rob Arthan
    31 mins ago












  • $begingroup$
    Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
    $endgroup$
    – Rob Arthan
    7 mins ago
















-1












$begingroup$

The approach of defining a recursive data structure $S$ by giving some rules for constructing members of $S$ and saying "anything constructed by these rules is an $S$ and nothing else is an $S$" is a shorthand for saying that $S$ is the intersection of all sets that are closed under the rules. This approach is usually used in contexts where the candidates for membership of $S$ are already known, so the intersection is over a well-defined set. E.g., if the rules are given by a grammar over some set of symbols, $S$ will be a subset of the set of all strings of symbols.



To apply this approach to a first-order axiomatization of set theory, you'd need some mechanism allowing the language of set theory to talk about itself, so that you could say in the language of set theory that the only sets are those that can described following your rules. This isn't impossible, but it would be technically tricky: the axiomatization would depend on something equivalent to a Gödel numbering of the language and you would have to presuppose some notion of the satisfaction relation in order to give meaning to the set comprehensions in your rule 5.



The end result is going to be unsatisfactory. In ZF, we can prove (1) that uncountable sets exist and (2) that the language of ZF is countable. Hence your modification of ZF in which all the sets are expressible by formulas in the language would be inconsistent. To regain consistency, you would have to weaken ZF. There is no obvious weakening that would achieve consistency and would respect the underlying intuitions behind ZF.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
    $endgroup$
    – Noah Schweber
    41 mins ago












  • $begingroup$
    @NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
    $endgroup$
    – Rob Arthan
    31 mins ago












  • $begingroup$
    Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
    $endgroup$
    – Rob Arthan
    7 mins ago














-1












-1








-1





$begingroup$

The approach of defining a recursive data structure $S$ by giving some rules for constructing members of $S$ and saying "anything constructed by these rules is an $S$ and nothing else is an $S$" is a shorthand for saying that $S$ is the intersection of all sets that are closed under the rules. This approach is usually used in contexts where the candidates for membership of $S$ are already known, so the intersection is over a well-defined set. E.g., if the rules are given by a grammar over some set of symbols, $S$ will be a subset of the set of all strings of symbols.



To apply this approach to a first-order axiomatization of set theory, you'd need some mechanism allowing the language of set theory to talk about itself, so that you could say in the language of set theory that the only sets are those that can described following your rules. This isn't impossible, but it would be technically tricky: the axiomatization would depend on something equivalent to a Gödel numbering of the language and you would have to presuppose some notion of the satisfaction relation in order to give meaning to the set comprehensions in your rule 5.



The end result is going to be unsatisfactory. In ZF, we can prove (1) that uncountable sets exist and (2) that the language of ZF is countable. Hence your modification of ZF in which all the sets are expressible by formulas in the language would be inconsistent. To regain consistency, you would have to weaken ZF. There is no obvious weakening that would achieve consistency and would respect the underlying intuitions behind ZF.






share|cite|improve this answer











$endgroup$



The approach of defining a recursive data structure $S$ by giving some rules for constructing members of $S$ and saying "anything constructed by these rules is an $S$ and nothing else is an $S$" is a shorthand for saying that $S$ is the intersection of all sets that are closed under the rules. This approach is usually used in contexts where the candidates for membership of $S$ are already known, so the intersection is over a well-defined set. E.g., if the rules are given by a grammar over some set of symbols, $S$ will be a subset of the set of all strings of symbols.



To apply this approach to a first-order axiomatization of set theory, you'd need some mechanism allowing the language of set theory to talk about itself, so that you could say in the language of set theory that the only sets are those that can described following your rules. This isn't impossible, but it would be technically tricky: the axiomatization would depend on something equivalent to a Gödel numbering of the language and you would have to presuppose some notion of the satisfaction relation in order to give meaning to the set comprehensions in your rule 5.



The end result is going to be unsatisfactory. In ZF, we can prove (1) that uncountable sets exist and (2) that the language of ZF is countable. Hence your modification of ZF in which all the sets are expressible by formulas in the language would be inconsistent. To regain consistency, you would have to weaken ZF. There is no obvious weakening that would achieve consistency and would respect the underlying intuitions behind ZF.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 4 mins ago

























answered 3 hours ago









Rob ArthanRob Arthan

29.1k42966




29.1k42966












  • $begingroup$
    "Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
    $endgroup$
    – Noah Schweber
    41 mins ago












  • $begingroup$
    @NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
    $endgroup$
    – Rob Arthan
    31 mins ago












  • $begingroup$
    Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
    $endgroup$
    – Rob Arthan
    7 mins ago


















  • $begingroup$
    "Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
    $endgroup$
    – Noah Schweber
    41 mins ago












  • $begingroup$
    @NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
    $endgroup$
    – Rob Arthan
    31 mins ago












  • $begingroup$
    Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
    $endgroup$
    – Rob Arthan
    7 mins ago
















$begingroup$
"Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
$endgroup$
– Noah Schweber
41 mins ago






$begingroup$
"Hence ZF can prove that there are sets that are not definable in the language of ZF." That's not actually true ... The point where the naive argument breaks down is that satisfaction (and hence definability) is not definable, so it can't even express "there is an undefinable set."
$endgroup$
– Noah Schweber
41 mins ago














$begingroup$
@NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
$endgroup$
– Rob Arthan
31 mins ago






$begingroup$
@NoahSchweber: I meant that in the modification of ZF that the OP is proposing one could prove that there are at most countably many sets. I have edited my answer. Please let me know if you still disagree.
$endgroup$
– Rob Arthan
31 mins ago














$begingroup$
Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
$endgroup$
– Rob Arthan
7 mins ago




$begingroup$
Note that ZF can quantify over possible definitions of the satisfaction relation even though it can't prove one exists. So ZF can prove the family of sets closed under rules 1 to 5 in the question is countable for any possible definition of the satisfaction relation.
$endgroup$
– Rob Arthan
7 mins ago


















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