Translating Grothendieck axiom UB into ZFC
$begingroup$
In SGA 4, Grothendieck introduced a set-theoretic device called a Grothendieck universe. He and his collaborators worked in Bourbaki set theory, which is practically similar to $mathsf{ZFC}$ but fundamentally different. To ease the work with universes, he introduced two addition axioms to the theory: they are mostly referred to as UA and UB.
While UA is widely known and doesn't specifically depend on Bourbaki set theory (it basically states that for any set $X$ there is a universe $mathscr{U}$ containing it), UB apparently heavily relies on that version of set theory. In particular, it uses such concepts as relations and and the operator $tau_x$, which $mathsf{ZFC}$ lacks.
My question is the following one: is it possible to get a version of UB with respect to $mathsf{ZFC}$ which would serve the same purposes for universes?
Here's the axiom UB in the language of Bourbaki set theory:
Let $R{x}$ be a relation and $mathscr{U}$ a universe. If there is $y in mathscr{U}$ so that we have $R{y}$, then $tau_xR{x} in mathscr{U}$.
P.S. I apologize if the question is not the best quality, I suspect it could even be a trivial one, but I personally don't understand Bourbaki set theory very well. Still, MO favors questions which a mathematical researcher potentially can ask and I can imagine a research who doesn't understand Bourbaki set theory but is interested in using universes à la Grothendieck in SGA.
ct.category-theory set-theory foundations
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
$endgroup$
|
show 2 more comments
$begingroup$
In SGA 4, Grothendieck introduced a set-theoretic device called a Grothendieck universe. He and his collaborators worked in Bourbaki set theory, which is practically similar to $mathsf{ZFC}$ but fundamentally different. To ease the work with universes, he introduced two addition axioms to the theory: they are mostly referred to as UA and UB.
While UA is widely known and doesn't specifically depend on Bourbaki set theory (it basically states that for any set $X$ there is a universe $mathscr{U}$ containing it), UB apparently heavily relies on that version of set theory. In particular, it uses such concepts as relations and and the operator $tau_x$, which $mathsf{ZFC}$ lacks.
My question is the following one: is it possible to get a version of UB with respect to $mathsf{ZFC}$ which would serve the same purposes for universes?
Here's the axiom UB in the language of Bourbaki set theory:
Let $R{x}$ be a relation and $mathscr{U}$ a universe. If there is $y in mathscr{U}$ so that we have $R{y}$, then $tau_xR{x} in mathscr{U}$.
P.S. I apologize if the question is not the best quality, I suspect it could even be a trivial one, but I personally don't understand Bourbaki set theory very well. Still, MO favors questions which a mathematical researcher potentially can ask and I can imagine a research who doesn't understand Bourbaki set theory but is interested in using universes à la Grothendieck in SGA.
ct.category-theory set-theory foundations
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
$endgroup$
$begingroup$
What is $tau_x$? It looks like a global choice function.
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:25
$begingroup$
Also what does it mean for $R{x}$ to be a relation here?
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:27
1
$begingroup$
@Asaf From what I know Bourbaki uses $tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $mathscr{U}$ is something like elementary in $V$.
$endgroup$
– Miha Habič
Nov 18 '18 at 14:28
5
$begingroup$
@MihaHabič UB looks considerably weaker than saying $mathcal U$ is elementary in $V$. It just says th chosen witness $tau_x,R{x}$ is in $mathcal U$ if some witness is in $mathcal U$ --- not if there's just some witness in $V$. So if $tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:17
3
$begingroup$
@AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $lambda<kappa$ with $V_lambdaprec V_kappa$.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:21
|
show 2 more comments
$begingroup$
In SGA 4, Grothendieck introduced a set-theoretic device called a Grothendieck universe. He and his collaborators worked in Bourbaki set theory, which is practically similar to $mathsf{ZFC}$ but fundamentally different. To ease the work with universes, he introduced two addition axioms to the theory: they are mostly referred to as UA and UB.
While UA is widely known and doesn't specifically depend on Bourbaki set theory (it basically states that for any set $X$ there is a universe $mathscr{U}$ containing it), UB apparently heavily relies on that version of set theory. In particular, it uses such concepts as relations and and the operator $tau_x$, which $mathsf{ZFC}$ lacks.
My question is the following one: is it possible to get a version of UB with respect to $mathsf{ZFC}$ which would serve the same purposes for universes?
Here's the axiom UB in the language of Bourbaki set theory:
Let $R{x}$ be a relation and $mathscr{U}$ a universe. If there is $y in mathscr{U}$ so that we have $R{y}$, then $tau_xR{x} in mathscr{U}$.
P.S. I apologize if the question is not the best quality, I suspect it could even be a trivial one, but I personally don't understand Bourbaki set theory very well. Still, MO favors questions which a mathematical researcher potentially can ask and I can imagine a research who doesn't understand Bourbaki set theory but is interested in using universes à la Grothendieck in SGA.
ct.category-theory set-theory foundations
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
$endgroup$
In SGA 4, Grothendieck introduced a set-theoretic device called a Grothendieck universe. He and his collaborators worked in Bourbaki set theory, which is practically similar to $mathsf{ZFC}$ but fundamentally different. To ease the work with universes, he introduced two addition axioms to the theory: they are mostly referred to as UA and UB.
While UA is widely known and doesn't specifically depend on Bourbaki set theory (it basically states that for any set $X$ there is a universe $mathscr{U}$ containing it), UB apparently heavily relies on that version of set theory. In particular, it uses such concepts as relations and and the operator $tau_x$, which $mathsf{ZFC}$ lacks.
My question is the following one: is it possible to get a version of UB with respect to $mathsf{ZFC}$ which would serve the same purposes for universes?
Here's the axiom UB in the language of Bourbaki set theory:
Let $R{x}$ be a relation and $mathscr{U}$ a universe. If there is $y in mathscr{U}$ so that we have $R{y}$, then $tau_xR{x} in mathscr{U}$.
P.S. I apologize if the question is not the best quality, I suspect it could even be a trivial one, but I personally don't understand Bourbaki set theory very well. Still, MO favors questions which a mathematical researcher potentially can ask and I can imagine a research who doesn't understand Bourbaki set theory but is interested in using universes à la Grothendieck in SGA.
ct.category-theory set-theory foundations
ct.category-theory set-theory foundations
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
edited Nov 18 '18 at 13:53
Jxt921
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
asked Nov 18 '18 at 11:30
Jxt921Jxt921
564612
564612
$begingroup$
What is $tau_x$? It looks like a global choice function.
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:25
$begingroup$
Also what does it mean for $R{x}$ to be a relation here?
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:27
1
$begingroup$
@Asaf From what I know Bourbaki uses $tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $mathscr{U}$ is something like elementary in $V$.
$endgroup$
– Miha Habič
Nov 18 '18 at 14:28
5
$begingroup$
@MihaHabič UB looks considerably weaker than saying $mathcal U$ is elementary in $V$. It just says th chosen witness $tau_x,R{x}$ is in $mathcal U$ if some witness is in $mathcal U$ --- not if there's just some witness in $V$. So if $tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:17
3
$begingroup$
@AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $lambda<kappa$ with $V_lambdaprec V_kappa$.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:21
|
show 2 more comments
$begingroup$
What is $tau_x$? It looks like a global choice function.
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:25
$begingroup$
Also what does it mean for $R{x}$ to be a relation here?
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:27
1
$begingroup$
@Asaf From what I know Bourbaki uses $tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $mathscr{U}$ is something like elementary in $V$.
$endgroup$
– Miha Habič
Nov 18 '18 at 14:28
5
$begingroup$
@MihaHabič UB looks considerably weaker than saying $mathcal U$ is elementary in $V$. It just says th chosen witness $tau_x,R{x}$ is in $mathcal U$ if some witness is in $mathcal U$ --- not if there's just some witness in $V$. So if $tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:17
3
$begingroup$
@AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $lambda<kappa$ with $V_lambdaprec V_kappa$.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:21
$begingroup$
What is $tau_x$? It looks like a global choice function.
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:25
$begingroup$
What is $tau_x$? It looks like a global choice function.
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:25
$begingroup$
Also what does it mean for $R{x}$ to be a relation here?
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:27
$begingroup$
Also what does it mean for $R{x}$ to be a relation here?
$endgroup$
– Asaf Karagila
Nov 18 '18 at 14:27
1
1
$begingroup$
@Asaf From what I know Bourbaki uses $tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $mathscr{U}$ is something like elementary in $V$.
$endgroup$
– Miha Habič
Nov 18 '18 at 14:28
$begingroup$
@Asaf From what I know Bourbaki uses $tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $mathscr{U}$ is something like elementary in $V$.
$endgroup$
– Miha Habič
Nov 18 '18 at 14:28
5
5
$begingroup$
@MihaHabič UB looks considerably weaker than saying $mathcal U$ is elementary in $V$. It just says th chosen witness $tau_x,R{x}$ is in $mathcal U$ if some witness is in $mathcal U$ --- not if there's just some witness in $V$. So if $tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:17
$begingroup$
@MihaHabič UB looks considerably weaker than saying $mathcal U$ is elementary in $V$. It just says th chosen witness $tau_x,R{x}$ is in $mathcal U$ if some witness is in $mathcal U$ --- not if there's just some witness in $V$. So if $tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:17
3
3
$begingroup$
@AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $lambda<kappa$ with $V_lambdaprec V_kappa$.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:21
$begingroup$
@AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $lambda<kappa$ with $V_lambdaprec V_kappa$.
$endgroup$
– Andreas Blass
Nov 18 '18 at 16:21
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $tau$.
Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows:
$tau$ always chooses in the smallest possible universe.
Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
$endgroup$
$begingroup$
Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
$endgroup$
– user21820
Nov 18 '18 at 17:47
$begingroup$
@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 17:57
$begingroup$
Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
$begingroup$
With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
$begingroup$
I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
add a comment |
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$begingroup$
If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $tau$.
Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows:
$tau$ always chooses in the smallest possible universe.
Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
$endgroup$
$begingroup$
Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
$endgroup$
– user21820
Nov 18 '18 at 17:47
$begingroup$
@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 17:57
$begingroup$
Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
$begingroup$
With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
$begingroup$
I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
add a comment |
$begingroup$
If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $tau$.
Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows:
$tau$ always chooses in the smallest possible universe.
Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
$endgroup$
$begingroup$
Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
$endgroup$
– user21820
Nov 18 '18 at 17:47
$begingroup$
@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 17:57
$begingroup$
Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
$begingroup$
With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
$begingroup$
I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
add a comment |
$begingroup$
If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $tau$.
Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows:
$tau$ always chooses in the smallest possible universe.
Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
$endgroup$
If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $tau$.
Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows:
$tau$ always chooses in the smallest possible universe.
Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
answered Nov 18 '18 at 15:52
Fred RohrerFred Rohrer
4,38111734
4,38111734
$begingroup$
Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
$endgroup$
– user21820
Nov 18 '18 at 17:47
$begingroup$
@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 17:57
$begingroup$
Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
$begingroup$
With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
$begingroup$
I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
add a comment |
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Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
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– user21820
Nov 18 '18 at 17:47
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@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
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– Fred Rohrer
Nov 18 '18 at 17:57
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Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
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With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
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I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
$begingroup$
Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
$endgroup$
– user21820
Nov 18 '18 at 17:47
$begingroup$
Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right?
$endgroup$
– user21820
Nov 18 '18 at 17:47
$begingroup$
@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 17:57
$begingroup$
@user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $tau$.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 17:57
$begingroup$
Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
$begingroup$
Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator?
$endgroup$
– user21820
Nov 18 '18 at 18:12
$begingroup$
With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
$begingroup$
With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true.
$endgroup$
– Fred Rohrer
Nov 18 '18 at 20:35
$begingroup$
I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
$begingroup$
I see. I can't read that language so never mind I'll take your word for it. =)
$endgroup$
– user21820
Nov 19 '18 at 6:12
add a comment |
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$begingroup$
What is $tau_x$? It looks like a global choice function.
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– Asaf Karagila
Nov 18 '18 at 14:25
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Also what does it mean for $R{x}$ to be a relation here?
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– Asaf Karagila
Nov 18 '18 at 14:27
1
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@Asaf From what I know Bourbaki uses $tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $mathscr{U}$ is something like elementary in $V$.
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– Miha Habič
Nov 18 '18 at 14:28
5
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@MihaHabič UB looks considerably weaker than saying $mathcal U$ is elementary in $V$. It just says th chosen witness $tau_x,R{x}$ is in $mathcal U$ if some witness is in $mathcal U$ --- not if there's just some witness in $V$. So if $tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied.
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– Andreas Blass
Nov 18 '18 at 16:17
3
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@AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $lambda<kappa$ with $V_lambdaprec V_kappa$.
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– Andreas Blass
Nov 18 '18 at 16:21