Why can't we just add “nothing else is a set” as an axiom?
$begingroup$
The axioms of ZF define what a set is by:
$omega$ is a set- If $x$ and $y$ are sets, then ${x, y}$ is a set
- If $x$ is a set, then $bigcup x$ is a set
- If $x$ is a set, then $mathcal{P}(x)$ is a set
- 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
asked Jan 18 at 19:52
StefanStefan
1906
$endgroup$
|
show 14 more comments
$begingroup$
The axioms of ZF define what a set is by:
$omega$ is a set- If $x$ and $y$ are sets, then ${x, y}$ is a set
- If $x$ is a set, then $bigcup x$ is a set
- If $x$ is a set, then $mathcal{P}(x)$ is a set
- 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
asked Jan 18 at 19:52
StefanStefan
1906
$endgroup$
3
$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
Jan 18 at 20:36
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
Jan 18 at 20:40
1
$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
Jan 18 at 20:55
5
$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
Jan 18 at 22:30
1
$begingroup$
@Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)?
$endgroup$
– Asaf Karagila♦
Jan 20 at 0:45
|
show 14 more comments
$begingroup$
The axioms of ZF define what a set is by:
$omega$ is a set- If $x$ and $y$ are sets, then ${x, y}$ is a set
- If $x$ is a set, then $bigcup x$ is a set
- If $x$ is a set, then $mathcal{P}(x)$ is a set
- 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
asked Jan 18 at 19:52
StefanStefan
1906
$endgroup$
The axioms of ZF define what a set is by:
$omega$ is a set- If $x$ and $y$ are sets, then ${x, y}$ is a set
- If $x$ is a set, then $bigcup x$ is a set
- If $x$ is a set, then $mathcal{P}(x)$ is a set
- 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
set-theory
asked Jan 18 at 19:52
StefanStefan
1906
asked Jan 18 at 19:52
StefanStefan
1906
asked Jan 18 at 19:52
StefanStefan
1906
asked Jan 18 at 19:52
StefanStefan
1906
asked Jan 18 at 19:52
StefanStefan
1906
1906
3
$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
Jan 18 at 20:36
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
Jan 18 at 20:40
1
$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
Jan 18 at 20:55
5
$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
Jan 18 at 22:30
1
$begingroup$
@Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)?
$endgroup$
– Asaf Karagila♦
Jan 20 at 0:45
|
show 14 more comments
3
$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
Jan 18 at 20:36
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
Jan 18 at 20:40
1
$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
Jan 18 at 20:55
5
$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
Jan 18 at 22:30
1
$begingroup$
@Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)?
$endgroup$
– Asaf Karagila♦
Jan 20 at 0:45
3
3
$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
Jan 18 at 20:36
$begingroup$
You forgot collection/replacement
$endgroup$
– Not Mike
Jan 18 at 20:36
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
Jan 18 at 20:40
$begingroup$
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
Jan 18 at 20:40
1
1
$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
Jan 18 at 20:55
$begingroup$
See also en.wikipedia.org/wiki/Constructible_universe
$endgroup$
– Not Mike
Jan 18 at 20:55
5
5
$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
Jan 18 at 22:30
$begingroup$
"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
$endgroup$
– Jonathan
Jan 18 at 22:30
1
1
$begingroup$
@Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)?
$endgroup$
– Asaf Karagila♦
Jan 20 at 0:45
$begingroup$
@Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)?
$endgroup$
– Asaf Karagila♦
Jan 20 at 0:45
|
show 14 more comments
2 Answers
2
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oldest
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$begingroup$
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.)
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
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
Jan 18 at 20:57
1
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
add a comment |
$begingroup$
Your proposed rule would lead to contradictions. Take the continuum hypothesis, for example. We can't prove (in ZFC, assuming ZFC is consistent) that there is a set of cardinality between that of $mathbb{N}$ and $mathbb{R}$, so the rule says there isn't one. So, given the rule, the continuum hypothesis is true.
But wait! We also can't prove that there is a bijection between $omega_1$ and $mathbb{R}$, so the rule says there isn't one, and the continuum hypothesis is false.
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
$endgroup$
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.)
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
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
Jan 18 at 20:57
1
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
add a comment |
$begingroup$
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.)
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
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
Jan 18 at 20:57
1
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
add a comment |
$begingroup$
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.)
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
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.)
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
edited Jan 18 at 20:59
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 18 at 20:37
David C. UllrichDavid C. Ullrich
60.9k43994
60.9k43994
$begingroup$
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
Jan 18 at 20:57
1
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
add a comment |
$begingroup$
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
Jan 18 at 20:57
1
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
$begingroup$
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
Jan 18 at 20:57
$begingroup$
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
Jan 18 at 20:57
1
1
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
$begingroup$
@ArturoMagidin Yes of course, wasn't thinking.
$endgroup$
– David C. Ullrich
Jan 18 at 21:00
add a comment |
$begingroup$
Your proposed rule would lead to contradictions. Take the continuum hypothesis, for example. We can't prove (in ZFC, assuming ZFC is consistent) that there is a set of cardinality between that of $mathbb{N}$ and $mathbb{R}$, so the rule says there isn't one. So, given the rule, the continuum hypothesis is true.
But wait! We also can't prove that there is a bijection between $omega_1$ and $mathbb{R}$, so the rule says there isn't one, and the continuum hypothesis is false.
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
$endgroup$
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
add a comment |
$begingroup$
Your proposed rule would lead to contradictions. Take the continuum hypothesis, for example. We can't prove (in ZFC, assuming ZFC is consistent) that there is a set of cardinality between that of $mathbb{N}$ and $mathbb{R}$, so the rule says there isn't one. So, given the rule, the continuum hypothesis is true.
But wait! We also can't prove that there is a bijection between $omega_1$ and $mathbb{R}$, so the rule says there isn't one, and the continuum hypothesis is false.
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
$endgroup$
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
add a comment |
$begingroup$
Your proposed rule would lead to contradictions. Take the continuum hypothesis, for example. We can't prove (in ZFC, assuming ZFC is consistent) that there is a set of cardinality between that of $mathbb{N}$ and $mathbb{R}$, so the rule says there isn't one. So, given the rule, the continuum hypothesis is true.
But wait! We also can't prove that there is a bijection between $omega_1$ and $mathbb{R}$, so the rule says there isn't one, and the continuum hypothesis is false.
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
$endgroup$
Your proposed rule would lead to contradictions. Take the continuum hypothesis, for example. We can't prove (in ZFC, assuming ZFC is consistent) that there is a set of cardinality between that of $mathbb{N}$ and $mathbb{R}$, so the rule says there isn't one. So, given the rule, the continuum hypothesis is true.
But wait! We also can't prove that there is a bijection between $omega_1$ and $mathbb{R}$, so the rule says there isn't one, and the continuum hypothesis is false.
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
answered Jan 19 at 4:00
Eric M. SchmidtEric M. Schmidt
2,56111228
2,56111228
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
add a comment |
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
$begingroup$
No, it is not that simple. The system with the proposed axiom added certainly isn't ZFC anymore, and it is not even first-order.
$endgroup$
– Andrés E. Caicedo
Jan 19 at 12:59
add a comment |
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You forgot collection/replacement
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– Not Mike
Jan 18 at 20:36
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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").
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– GEdgar
Jan 18 at 20:40
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See also en.wikipedia.org/wiki/Constructible_universe
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– Not Mike
Jan 18 at 20:55
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"The axioms of ZF define what a set is..." Where does this misconception come from? I've seen it so many times...
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– Jonathan
Jan 18 at 22:30
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@Jonathan: Yes, but if you describe the structure of a vector space, then you could say that you're talking about properties of vectors (e.g., closed under addition and scalar multiplication). So in that sense, why wouldn't it be correct to say that the axioms of ZF describe the properties of sets (e.g., if $x$ is a set, then it has a power set, and subclasses of $x$ are sets, etc.)?
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– Asaf Karagila♦
Jan 20 at 0:45