Existence of saturated models of a theory.
$begingroup$
In Tent, Ziegler: A Course in Model Theory it is stated on page 89, that
Saturated structures need not exist (think about why not), but
by considering special models instead, we can preserve many of the important
properties – and prove their existence.
I do not see why this statement holds in this generality. Finite structures are always saturated, so consider a (complete) $L$-theory $T$ with infinite models. By Löwenheim-Skolem we find a model $MvDash T$ with $|M|=kappageq |T|^+$. Then for $Asubseteq M$ with $|A|leq|T|$ we have $|S_n^M(A)|leq 2^{|T|}$. If we assume the continuum hypothesis $2^{|T|}=|T|^+$, then we may add at most $|T|^+=2^{|T|}$ additional elements to realize all omitted types of $S_n^M(A)$ to the structure $M$ and by compactness get an elementary extension $Nsucc M$ of the same cardinality. So it should always be possible to find a saturated model at some sufficiently large cardinality in this setting.
Is my guess that it is about the continuum hypothesis correct or am I missing the point?
set-theory model-theory
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
$endgroup$
add a comment |
$begingroup$
In Tent, Ziegler: A Course in Model Theory it is stated on page 89, that
Saturated structures need not exist (think about why not), but
by considering special models instead, we can preserve many of the important
properties – and prove their existence.
I do not see why this statement holds in this generality. Finite structures are always saturated, so consider a (complete) $L$-theory $T$ with infinite models. By Löwenheim-Skolem we find a model $MvDash T$ with $|M|=kappageq |T|^+$. Then for $Asubseteq M$ with $|A|leq|T|$ we have $|S_n^M(A)|leq 2^{|T|}$. If we assume the continuum hypothesis $2^{|T|}=|T|^+$, then we may add at most $|T|^+=2^{|T|}$ additional elements to realize all omitted types of $S_n^M(A)$ to the structure $M$ and by compactness get an elementary extension $Nsucc M$ of the same cardinality. So it should always be possible to find a saturated model at some sufficiently large cardinality in this setting.
Is my guess that it is about the continuum hypothesis correct or am I missing the point?
set-theory model-theory
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
$endgroup$
add a comment |
$begingroup$
In Tent, Ziegler: A Course in Model Theory it is stated on page 89, that
Saturated structures need not exist (think about why not), but
by considering special models instead, we can preserve many of the important
properties – and prove their existence.
I do not see why this statement holds in this generality. Finite structures are always saturated, so consider a (complete) $L$-theory $T$ with infinite models. By Löwenheim-Skolem we find a model $MvDash T$ with $|M|=kappageq |T|^+$. Then for $Asubseteq M$ with $|A|leq|T|$ we have $|S_n^M(A)|leq 2^{|T|}$. If we assume the continuum hypothesis $2^{|T|}=|T|^+$, then we may add at most $|T|^+=2^{|T|}$ additional elements to realize all omitted types of $S_n^M(A)$ to the structure $M$ and by compactness get an elementary extension $Nsucc M$ of the same cardinality. So it should always be possible to find a saturated model at some sufficiently large cardinality in this setting.
Is my guess that it is about the continuum hypothesis correct or am I missing the point?
set-theory model-theory
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
$endgroup$
In Tent, Ziegler: A Course in Model Theory it is stated on page 89, that
Saturated structures need not exist (think about why not), but
by considering special models instead, we can preserve many of the important
properties – and prove their existence.
I do not see why this statement holds in this generality. Finite structures are always saturated, so consider a (complete) $L$-theory $T$ with infinite models. By Löwenheim-Skolem we find a model $MvDash T$ with $|M|=kappageq |T|^+$. Then for $Asubseteq M$ with $|A|leq|T|$ we have $|S_n^M(A)|leq 2^{|T|}$. If we assume the continuum hypothesis $2^{|T|}=|T|^+$, then we may add at most $|T|^+=2^{|T|}$ additional elements to realize all omitted types of $S_n^M(A)$ to the structure $M$ and by compactness get an elementary extension $Nsucc M$ of the same cardinality. So it should always be possible to find a saturated model at some sufficiently large cardinality in this setting.
Is my guess that it is about the continuum hypothesis correct or am I missing the point?
set-theory model-theory
set-theory model-theory
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
edited Dec 12 '18 at 16:29
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
asked Dec 11 '18 at 20:23
ZikrunumeaZikrunumea
709
709
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your argument is correct, assuming the instance $2^{|T|} = |T|^+$ of the generalized continuum hypothesis.
More generally, you can prove that an arbitrary theory $T$ has a saturated model of cardinality $kappa$ if $kappa geq |T|^+$, $kappa$ is regular, and $kappa^{<kappa} = kappa$. Note that:
- If $kappa = lambda^+ = 2^{lambda}$ for some cardinal $lambdageq |T|$, then $kappa$ satisfies the hypothesis.
- If $kappa$ is strongly inaccessible (and greater than $|T|$), then $kappa$ satisfies the hypotheses.
But both of these hypotheses on $kappa$ go beyond ZFC.
It's also possible to prove that if $T$ is stable in $kappageq |T|$ (meaning that for any $Mmodels T$ and $Asubseteq M$ with $|A|leq kappa$, $|S_n^M(A)| = kappa$), then $T$ has a saturated model of cardinality $kappa$. We say that $T$ is stable if it is stable in some cardinal $kappageq |T|$, and (of course) not every theory is stable.
When Tent and Ziegler say "saturated structures need not exist", they mean "we can't prove in ZFC that every theory has a saturated model". Indeed, it's consistent with ZFC that no unstable theory has a saturated model. For example, it's consistent that the theory $text{Th}(mathbb{R};+,-,cdot,0,1)$ of real closed fields has no saturated models.
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
$endgroup$
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
add a comment |
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$begingroup$
Your argument is correct, assuming the instance $2^{|T|} = |T|^+$ of the generalized continuum hypothesis.
More generally, you can prove that an arbitrary theory $T$ has a saturated model of cardinality $kappa$ if $kappa geq |T|^+$, $kappa$ is regular, and $kappa^{<kappa} = kappa$. Note that:
- If $kappa = lambda^+ = 2^{lambda}$ for some cardinal $lambdageq |T|$, then $kappa$ satisfies the hypothesis.
- If $kappa$ is strongly inaccessible (and greater than $|T|$), then $kappa$ satisfies the hypotheses.
But both of these hypotheses on $kappa$ go beyond ZFC.
It's also possible to prove that if $T$ is stable in $kappageq |T|$ (meaning that for any $Mmodels T$ and $Asubseteq M$ with $|A|leq kappa$, $|S_n^M(A)| = kappa$), then $T$ has a saturated model of cardinality $kappa$. We say that $T$ is stable if it is stable in some cardinal $kappageq |T|$, and (of course) not every theory is stable.
When Tent and Ziegler say "saturated structures need not exist", they mean "we can't prove in ZFC that every theory has a saturated model". Indeed, it's consistent with ZFC that no unstable theory has a saturated model. For example, it's consistent that the theory $text{Th}(mathbb{R};+,-,cdot,0,1)$ of real closed fields has no saturated models.
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
$endgroup$
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
add a comment |
$begingroup$
Your argument is correct, assuming the instance $2^{|T|} = |T|^+$ of the generalized continuum hypothesis.
More generally, you can prove that an arbitrary theory $T$ has a saturated model of cardinality $kappa$ if $kappa geq |T|^+$, $kappa$ is regular, and $kappa^{<kappa} = kappa$. Note that:
- If $kappa = lambda^+ = 2^{lambda}$ for some cardinal $lambdageq |T|$, then $kappa$ satisfies the hypothesis.
- If $kappa$ is strongly inaccessible (and greater than $|T|$), then $kappa$ satisfies the hypotheses.
But both of these hypotheses on $kappa$ go beyond ZFC.
It's also possible to prove that if $T$ is stable in $kappageq |T|$ (meaning that for any $Mmodels T$ and $Asubseteq M$ with $|A|leq kappa$, $|S_n^M(A)| = kappa$), then $T$ has a saturated model of cardinality $kappa$. We say that $T$ is stable if it is stable in some cardinal $kappageq |T|$, and (of course) not every theory is stable.
When Tent and Ziegler say "saturated structures need not exist", they mean "we can't prove in ZFC that every theory has a saturated model". Indeed, it's consistent with ZFC that no unstable theory has a saturated model. For example, it's consistent that the theory $text{Th}(mathbb{R};+,-,cdot,0,1)$ of real closed fields has no saturated models.
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
$endgroup$
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
add a comment |
$begingroup$
Your argument is correct, assuming the instance $2^{|T|} = |T|^+$ of the generalized continuum hypothesis.
More generally, you can prove that an arbitrary theory $T$ has a saturated model of cardinality $kappa$ if $kappa geq |T|^+$, $kappa$ is regular, and $kappa^{<kappa} = kappa$. Note that:
- If $kappa = lambda^+ = 2^{lambda}$ for some cardinal $lambdageq |T|$, then $kappa$ satisfies the hypothesis.
- If $kappa$ is strongly inaccessible (and greater than $|T|$), then $kappa$ satisfies the hypotheses.
But both of these hypotheses on $kappa$ go beyond ZFC.
It's also possible to prove that if $T$ is stable in $kappageq |T|$ (meaning that for any $Mmodels T$ and $Asubseteq M$ with $|A|leq kappa$, $|S_n^M(A)| = kappa$), then $T$ has a saturated model of cardinality $kappa$. We say that $T$ is stable if it is stable in some cardinal $kappageq |T|$, and (of course) not every theory is stable.
When Tent and Ziegler say "saturated structures need not exist", they mean "we can't prove in ZFC that every theory has a saturated model". Indeed, it's consistent with ZFC that no unstable theory has a saturated model. For example, it's consistent that the theory $text{Th}(mathbb{R};+,-,cdot,0,1)$ of real closed fields has no saturated models.
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
$endgroup$
Your argument is correct, assuming the instance $2^{|T|} = |T|^+$ of the generalized continuum hypothesis.
More generally, you can prove that an arbitrary theory $T$ has a saturated model of cardinality $kappa$ if $kappa geq |T|^+$, $kappa$ is regular, and $kappa^{<kappa} = kappa$. Note that:
- If $kappa = lambda^+ = 2^{lambda}$ for some cardinal $lambdageq |T|$, then $kappa$ satisfies the hypothesis.
- If $kappa$ is strongly inaccessible (and greater than $|T|$), then $kappa$ satisfies the hypotheses.
But both of these hypotheses on $kappa$ go beyond ZFC.
It's also possible to prove that if $T$ is stable in $kappageq |T|$ (meaning that for any $Mmodels T$ and $Asubseteq M$ with $|A|leq kappa$, $|S_n^M(A)| = kappa$), then $T$ has a saturated model of cardinality $kappa$. We say that $T$ is stable if it is stable in some cardinal $kappageq |T|$, and (of course) not every theory is stable.
When Tent and Ziegler say "saturated structures need not exist", they mean "we can't prove in ZFC that every theory has a saturated model". Indeed, it's consistent with ZFC that no unstable theory has a saturated model. For example, it's consistent that the theory $text{Th}(mathbb{R};+,-,cdot,0,1)$ of real closed fields has no saturated models.
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
edited Dec 12 '18 at 0:17
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
answered Dec 11 '18 at 22:12
Alex KruckmanAlex Kruckman
28.3k32758
28.3k32758
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
add a comment |
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
When I took my first (and really only) course in model theory, we heard the same thing "these assumptions go beyond ZFC". And that sort of scared me at the time. Like "Oh no, it's stuff you cannot prove! Inaccessible blasphemy!!!", but later it turns out that it's not that. It's something different. It's something I can't put my finger on, but the sentence "goes beyond ZFC" is scary, whereas ZFC+GCH is really not.
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 0:03
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
@AsafKaragila Right, I think it's important to emphasize that you need some set theoretic assumptions to construct saturated models, but that these assumptions are relatively tame: GCH or strongly inaccessible cardinals. Of course, the level to which these assumptions seem scary vs tame depends on how comfortable you are with set theory, and (to some extent) your view on the philosophy of mathematics. Personally, I'd rather assume a proper class of inaccessibles than assume GCH.
$endgroup$
– Alex Kruckman
Dec 12 '18 at 0:15
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
On the other hand, to ensure that some theories have no saturated models you need a global failure of GCH, an assumption which has some large cardinal strength.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 2:12
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
$begingroup$
Alex, of course it's important. But the way it was presented as "going beyond ZFC" is a bit scary to someone who is new to model theory and/or set theory. It puts a mystic twist on this, as something we can't really prove. And yes, it's something we can't really prove, but the point is that this can be better presented by model theorists. :-)
$endgroup$
– Asaf Karagila♦
Dec 12 '18 at 7:34
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
