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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?


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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
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3 Answers

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.

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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 6:47
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 6:59
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 7:18
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 7:21
Yes, this is also true because $\sf ZFC\not\vdash\operatorname{Con}(ZFC)$ to begin with. –  Asaf Karagila Jul 21 '13 at 7:48
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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.

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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. –  SomeDude Apr 20 '13 at 9:31
@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. –  Arthur Fischer Apr 20 '13 at 9:38
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? –  SomeDude Apr 29 '13 at 14:07
@SomeDude: Never apologise for being persistent!!! ;) –  Arthur Fischer Apr 29 '13 at 15:11
@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. –  Arthur Fischer May 1 '13 at 9:10
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A standard way to get around this "difficulty" is to work in the following conservative extension of ZFC: $ZFC \cup \{M \text{ is countable and transitive}\} \cup \{\phi^{M}: \phi \in ZFC\}$ where $M$ is a constant symbol. This "innovtaion" is due to Shoenfield and is discussed in detail in Kunen's book. Of course one can work with syntactic forcing relation over V or boolean valued models or "what not" but I do not think that is where real posets come from (or do they?).

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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 6:13
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