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I used to think that to avoid philosophical problems in forcing one can assume consistency of ZFC. Then one obtains a set model $M$ contained in the von Neumann universe $V$. And then the generic filter as well as the extension $M[G]$ are also in $V$.

But apparently one cannot do this. Apparently $M$ will only ever be a model of a finite fragment of ZFC. Why? Why, if ZFC is consistent, does there not exist a set model $M$ of ZFC?

Thanks.

Asaf Karagila
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user73486
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  • I tried to give a slightly more meaningful title to the question, rather than "Forcing question". I'm not 100% sure the title is accurate enough, and if someone feels it can be improved please do so! – Asaf Karagila Apr 22 '13 at 12:20

3 Answers3

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$\mathsf{ZFC}$ cannot prove that it has models (thanks to Gödel), and even if you assume that $\mathsf{ZFC}$ has models, you still cannot assume that it has nice models (transitive "standard" models). (Augment the language of set theory by adding constant symbols $c_0 , c_1 , \ldots$. If $\mathsf{ZFC}$ is consistent, then by Compactness so is $T = \mathsf{ZFC} + ( c_{i+1} \in c_i : i \in \omega)$, and $T$ has no models in which the interpretation of $\in$ is well-founded.)

However, Levy Reflection says that given any finite list of axioms of $\mathsf{ZFC}$, there is a $\theta$ such that $V_\theta$ satisfies each of these axioms. Then using Löwenheim-Skolem you can take a countable elementary submodel of a suitable $V_\theta$. Collapsing this to a transitive set, you get a countable transitive model of the original list of axioms.

Of course, after this technically is recognised, you begin to think of your forcing as being done over $\mathbf{V}$, or perhaps $\mathbf{L}$ if you want $\mathsf{(G)CH}$ to hold in the ground model.

user642796
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    Ok, I get that ZFC cannot prove that it has models. But Goedel's completeness theorem says "Every consistent first-order theory with a well-orderable language has a model.". ZFC is a first order theory and the language of sets is well-orderable. Could you elaborate a bit on why Goedel's completeness theorem cannot be applied? As stated it does not seem to exclude ZFC. – user73486 Apr 20 '13 at 09:31
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    @SomeDude: The basic idea is that the Levy Reflection is not a formal theorem of ZFC, but actually a meta-theorem which yields a formal theorem of ZFC for each sentence $\psi$ (or, equivalently, each finite list of sentences) of the language of set theory. Thus you cannot apply Compactness within ZFC to conclude that ZFC is consistent. – user642796 Apr 20 '13 at 09:38
  • Regarding your first paragraph: why is it that if the augmented theory $T$ cannot have a standard model that then $ZFC$ cannot have a standard model either? I believe this is what your first paragraph is saying, correct? – user73486 Apr 29 '13 at 14:07
  • And what is $\theta$ in your second paragraph? – user73486 Apr 29 '13 at 14:14
  • And finally, I'm very sorry to be so persistent: regarding my first comment, I am asking "Assume consistency of ZFC. Why can we not apply Goedel's completeness theorem to deduce that there exists a set model". It seems to me that in your comment/response to it you speak of why we cannot apply compactness theorem to obtain consistency if we have a model for each finite fragment. – user73486 Apr 29 '13 at 14:24
  • @SomeDude: (re the first message of the trifecta). Basically what I was saying was the following. Since there is an extension $T$ of $\mathsf{ZFC}$ which is consistent if $\mathsf{ZFC}$ is, and which, if consistent, has no standard set models (i.e., set models in which $\in$ is interpreted as real membership), the even under the assumption of its consistency, $\mathsf{ZFC}$ may not have standard models. So the existence of these nice models is a strictly stronger assumption that simply consistency. – user642796 Apr 29 '13 at 15:08
  • @SomeDude (re the second message) $\theta$ is just some uncountable regular cardinal. (re the third message) That was talking about the connection between Levy Reflection and the Compactness Theorem of logic. The Compactness Theorem says that if each finite subtheory is satisfiable, so is the entire theorem. Levy Reflection seems to say that every finite subtheory of $\mathsf{ZFC}$ is satisfiable. So the natural question is: why can't you put these together and say that $\mathsf{ZFC}$ is consistent. The comment (attempted to) expound on this fine point. – user642796 Apr 29 '13 at 15:11
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    @SomeDude: Never apologise for being persistent!!! ;) – user642796 Apr 29 '13 at 15:11
  • Thank you! .-) So I will persist: why is it that if the augmented theory $T$ cannot have a standard model then $ZFC$ cannot have a standard model either? – user73486 Apr 29 '13 at 18:30
  • @SomeDude: I think you are mis-understanding me. What I am trying to say is only that the assumption of "nice" (standard) models for $\sf{ZFC}$ is strictly stronger than simple consistency of $\sf{ZFC}$ (having any ol' model). For perhaps a better explanation of the details, see this thread. We darn-well hope that $\sf{ZFC}$ has standard models, as these are the ones we have some intuitive feeling for! – user642796 Apr 29 '13 at 19:24
  • Yes, I understand. But in your first paragraph, as I understand it, you are giving an argument why we cannot make the stronger assumption of existence of standard (nice) models even if we assume consistency of $\mathsf ZFC$. But I don't understand your argument yet. It seems to say: assume consistency of $\mathsf ZFC$. Then consider the augmented theory $\mathsf T$. Note that $\mathsf T$ cannot have a standard model. Therefore $\mathsf ZFC$ cannot have a standard model either. – user73486 Apr 30 '13 at 06:05
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    @SomeDude: It's not that the consistency of $T$ implies that $\sf{ZFC}$ cannot have standard models, but only that it might not have standard models. – user642796 May 01 '13 at 09:10
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You are correct, and in fact one assumes more than just the consistency of $\sf ZFC$. In fact one assumes the consistency of a standard model of $\sf ZFC$ (which is a stronger assumption than just consistency).

One can do so, and many indeed do that.

But if we want that our proofs will go through in $\sf ZFC$ and not in $\sf ZFC+$"there is a standard model", then we cannot use models of the full theory at all. Luckily $\sf ZFC$ prove the existence of transitive models of any finite fragment of $\sf ZFC$, and we can take an arbitrary "large enough fragment" and work with that.

One can also use Boolean-valued [class] models and rid of all need reference to set models.

Asaf Karagila
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  • Hello Asaf. When you say, “the consistency of a standard model of $ \mathsf{ZFC} $ is stronger than just the consistency of $ \mathsf{ZFC} $,” do you mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM})? $$ Arthur says (and I totally agree) that $$ \mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \neg \mathsf{SM}), $$ but I do not see how it should imply $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM}). $$ – Leonard Huang Jul 21 '13 at 06:47
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    Leonard, $\operatorname{Con}(\sf ZFC+SM)$ implies $\operatorname{Con}(\sf ZFC+\operatorname{Con}(ZFC))$. If we can prove the implication then $\sf ZFC$ can prove its own consistency, but we know it can't. – Asaf Karagila Jul 21 '13 at 06:59
  • Thanks for the clarification! I have a further question. If I apply Gödel’s Second Incompleteness Theorem, then what I have is: If $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent, then $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}) $. However, are we not supposed to just show: $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}) $? Many set-theory texts do not really make precise the statement ‘one can prove that the consistency of $ \mathsf{ZFC} $ does not imply the existence of a standard model’. – Leonard Huang Jul 21 '13 at 07:18
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    Leonard, note that $\sf ZFC\vdash\operatorname{Con}(ZFC+SM)\rightarrow \operatorname{Con}(ZFC+\operatorname{Con}(ZFC))$. This means, amongst other things, that if $\sf ZFC+\operatorname{Con}(\sf ZFC)$ were to prove $\operatorname{Con}(\sf ZFC+SM)$ then it would prove its own consistency, and therefore would be inconsistent to begin with. – Asaf Karagila Jul 21 '13 at 07:21
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    Yes, this is also true because $\sf ZFC\not\vdash\operatorname{Con}(ZFC)$ to begin with. – Asaf Karagila Jul 21 '13 at 07:48
  • I apologize for deleting the question because I thought I had asked something stupid. Anyway, thanks! – Leonard Huang Jul 21 '13 at 07:49
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    No problem! :-) – Asaf Karagila Jul 21 '13 at 07:49
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A standard way to get around this ‘difficulty’ is to work in the following conservative extension of $ \mathsf{ZFC} $: $$ \mathsf{ZFC} \cup \{ \text{$ \mathbf{M} $ is countable and transitive} \} \cup \left\{ \phi^{\mathbf{M}} ~ \Big| ~ \phi \in \mathsf{ZFC} \right\}, $$ where $ \mathbf{M} $ is a constant symbol. This ‘innovation’ is due to Joseph Shoenfield and is discussed in detail in Kenneth Kunen’s book Set Theory: An Introduction to Independence Proofs. Of course, one can work with a syntactic forcing relation over the universe $ \mathbb{V} $ or with Boolean-valued models or ‘what not’, but I do not think that is where real posets come from. (Or do they?)

Berrick Caleb Fillmore
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hot_queen
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  • While it is much more illuminating to think about posets when forcing, it can be much easier to work with Boolean valued models for some things. For example the theorem stating every countable forcing is Cohen, is absolutely trivial with BVM, but completely impossible to see when talking about posets. – Asaf Karagila Apr 24 '13 at 06:13
  • By the way, I think that the idea to add a symbol for a ctm is due to Feferman, not Shoenfield. – Asaf Karagila Jun 22 '14 at 08:49