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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 $M\vDash T$ with $|M|=\kappa\geq |T|^+$. Then for $A\subseteq M$ 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 $N\succ M$ 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?

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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 $\lambda\geq |T|$, 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 $\kappa\geq |T|$ (meaning that for any $M\models T$ and $A\subseteq M$ 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 $\kappa\geq |T|$, 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. (Edit: I was a bit glib here. As Andrés points out in the comments, given an unstable theory $T$, in order to prove that it is consistent that $T$ has no saturated models, we need to prove that it's consistent for GCH to fail at every cardinal, and we need to assume consistency of some large cardinals to do that. See this recent MathOverflow question for more information.)

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  • $\begingroup$ 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

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