When is it okay to intersect infinite families of proper classes?
$begingroup$
For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.
Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.
So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?
Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?
[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]
set-theory lo.logic ra.rings-and-algebras
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
$endgroup$
|
show 2 more comments
$begingroup$
For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.
Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.
So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?
Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?
[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]
set-theory lo.logic ra.rings-and-algebras
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
$endgroup$
2
$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago
$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago
1
$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago
$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago
1
$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago
|
show 2 more comments
$begingroup$
For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.
Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.
So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?
Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?
[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]
set-theory lo.logic ra.rings-and-algebras
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
$endgroup$
For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.
Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.
So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?
Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?
[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]
set-theory lo.logic ra.rings-and-algebras
set-theory lo.logic ra.rings-and-algebras
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
asked 1 hour ago
Pace NielsenPace Nielsen
7,15223175
7,15223175
2
$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago
$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago
1
$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago
$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago
1
$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago
|
show 2 more comments
2
$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago
$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago
1
$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago
$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago
1
$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago
2
2
$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago
$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago
$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago
$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago
1
1
$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago
$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago
$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago
$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago
1
1
$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago
$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago
|
show 2 more comments
1 Answer
1
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oldest
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$begingroup$
You will have to come up with trickier examples.
There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
$endgroup$
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
add a comment |
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$begingroup$
You will have to come up with trickier examples.
There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
$endgroup$
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
add a comment |
$begingroup$
You will have to come up with trickier examples.
There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
$endgroup$
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
add a comment |
$begingroup$
You will have to come up with trickier examples.
There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
$endgroup$
You will have to come up with trickier examples.
There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
answered 48 mins ago
Andrej BauerAndrej Bauer
30.3k477166
30.3k477166
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
add a comment |
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago
add a comment |
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$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago
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@Max See my response to Andrej below.
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– Pace Nielsen
37 mins ago
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I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
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– Max
27 mins ago
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@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
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– Pace Nielsen
17 mins ago
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The definition of an $n$-DF ring is internal to the universe of sets. Not external.
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– Asaf Karagila
8 mins ago