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It is a known theorem that for a model of $ZF$, $M$, if $M\models AC$ and $G$ is a $P$-generic filter over $M$, for some $P\in M$, then $M[G]\models AC$.

On the other hand, it is long known that other axioms, for example $GCH, CH, V=L, ...$ are not preserved by such extensions.

In a paper of Monro [1] the first paragraph speaks on a question by Dana Scott about why do some axioms get preserved by generic extensions and other not; it also says that there is an [almost obvious] equivalence between preservation in generic extensions, and preservation in Boolean-valued models.

The paper goes on to prove that some restricted versions of the axiom of choice are not preserved by generic extensions (and in a recent phone call it was explained to me how to break weak choice principles such $DC_\kappa$ quite easily in relatively simple generic extensions of models of $ZF$).

Lastly the paper says nothing about an answer to the question above, nor it cites any resource for a possible answer. However, the paper is quite old now, and in the past three decades some progress might have occurred.

Is there an answer for Scott's question: which axioms are preserved by all generic extensions?

Edit: If there is no "simple" and uniform answer to the above question, is there a possible answer to why is the Axiom of Choice preserved in generic extensions, while restricted versions are not?


Bibliography:

  1. G. P. Monro, On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic Vol. 48, No. 1 (Mar., 1983), pp. 39-52
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Just axioms or any sentences? –  François G. Dorais Oct 2 '11 at 18:57
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It's worth noting that the answer may depend on large cardinal assumptions. For example, by a result of Woodin, the existence of class many measurable Woodin cardinals implies that Sigma-2-1 formulas that are true in one forcing extension are true in all forcing extensions satisfying CH (see the paper of Illias Farah for further discussion). –  Haim Oct 2 '11 at 18:58
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I don't think there will be a uniform answer to this question. Several "axioms" are preserved by generic extensions for absoluteness reasons. But the reason that AC is preserved depends more on the details of the method of forcing. –  Stefan Geschke Oct 2 '11 at 18:58
    
@Francois: Any sentence, I'd assume that some triviality about such sentences is in order; and at the moment I'd be interested mostly in weak choice principles. However if you have any answer that includes non-trivial information on other sentences I'd be happy to hear about that too! –  Asaf Karagila Oct 2 '11 at 19:02
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I was wondering for a moment whether the ordering principle is preserved by generic extensions. I don't know and asked this here: mathoverflow.net/questions/77000 –  Stefan Geschke Oct 2 '11 at 21:41
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2 Answers

In the comments you consider the more general question of which sentences are necessarily preserved by forcing. One could cast the question as: what are the most general relations between set-theoretic truth and forceability?

This more general setting is the theme of my work with Benedikt Loewe on The modal logic of forcing, Trans. AMS, vol. 360, 2008. The idea here is to investigate the two operators:

  • $\varphi$ is forceable or possible, written $\lozenge\varphi$, if $\varphi$ holds in some forcing extension.
  • $\varphi$ is necessary, written $\square\varphi$, if $\varphi$ holds in all forcing extensions.

These are modal operators, but they are expressible in the language of set theory. One can view a set-theoretic universe in the context of all its forcing extensions as an enormous Kripke model of possible worlds, with each accessing its forcing extensions. Your question was, when can we expect $\varphi\to\square\varphi$?

Our inquiry was what are the most general valid principles of forcing. For example, one can easily verify the following validities, under the forcing interpretation:

  • (K) $\ \ \square(\varphi\to \psi)\to (\square\varphi\to \square\psi)$
  • (Dual) $\ \ \neg\lozenge\varphi\leftrightarrow\square\neg\varphi$
  • (S) $\ \ \square\varphi\to\varphi$
  • (4) $\ \ \square\varphi\to\square\square\varphi$
  • (.2) $\ \ \lozenge\square\varphi\to \square\lozenge\varphi$

The last axiom (.2) uses product forcing. The modal theory obtained from these axioms is known as S4.2. We define that a modal assertion $\varphi(p_0,\ldots,p_n)$ in the language of modal logic expresses a valid principle of forcing if $\varphi(\psi_0,\ldots,\psi_n)$ holds with the forcing interpretation of the modal operators for all set-theoretic assertions $\psi_0,\ldots,\psi_n$. The main theorem of our paper is:

Theorem. (Hamkins+Loewe) If ZFC is consistent, then the ZFC-provably valid principles of forcing are exactly those in the modal theory S4.2.

The proof uses the concept of buttons and switches, where a set theoretic sentence $\psi$ is a switch if both $\psi$ and $\neg\psi$ are necessarily possible, that is, forceable over any forcing extension, and $\varphi$ is a button if it is possibly necessary, that is, if it can be forced in such a way that it remains true in any forcing extension. Thus, switches can be turned on and off, but once you have pushed a button, you cannot unpush it. The arguments are a nice blend of easy forcing and modal logic.

Meanwhile, it is consistent that a model can exhibit more than merely the S4.2 validities. For example, the maximality principle is the scheme asserting:

$$\lozenge\square\varphi\to\square\varphi$$

for all sentences $\varphi$. (See J. D. Hamkins, A simple maximality principle, JSL, vol. 68, 2003, also independently investigated by Stavi and Vaananen.) Thus, the maximality principle asserts that any sentence that could be forced in such a way that it remains true in all subsequent forcing extensions, is already necessarily true in this way. Thus, it asserts that all buttons have been pushed. The maximality principle is equiconsistent with ZFC, but it is not true that every model of set theory has a forcing extension with MP. If one allows parameters into the scheme, the strength rises, and the necessary maximality principle with real parameters implies $\text{AD}^{L(\mathbb{R})}$.

The fact that different models of set theory can exhibit different valid principles of forcing implies that even when one looks at the general modal form of a set-theoretic sentence, the question of which sentences are invariant by forcing will depend on the model of set theory.

Incidently, there are numerous open questions in the modal logic of forcing. For example, if one restricts to ccc forcing or to proper forcing or any of several other natural forcing classes, we don't yet know the exact modal theory of validites. What are the ZFC-provably valid principles of ccc forcing?

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Joel, many thanks for the answer. It seems that the axiom of choice is used here aplenty. How much of this holds without it? –  Asaf Karagila Nov 9 '11 at 7:29
    
I guess it is still correct to say that the ZF-provably valid principles of forcing are still S4.2, since these are valid also for forcing over ZF models, and for the upper bound, one can still use the ZFC model, since it suffices to have a single ZF model whose validities are contained in S4.2. But perhaps one might attack the question about intermediate logics between S4.2 and S5 by means of a ZF model in which AC fails badly. I'm not sure. –  Joel David Hamkins Nov 10 '11 at 17:29
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Forward: This answer does not address the main question "which statements are preserved for every generic extension." Which I suspect has no simple answer.

Lets start by defining what this answer is meant to address: Preservation results fall into two types (modulo fine details):

  1. Large structural preservation. An example of this kind is Shoenfield's classic result concerning the absoluteness of $\Sigma^1_2$ statements. These generally take the form: For some statement(s) $\varphi$ and "canonical" inner model $M$, we can show that $ZFC \vdash $"$\varphi $ is absolute for $M,V$ ". This is rapidly transformed into a forcing absoluteness result by cleverly setting up the notion of "canonical" so as to guarantee: if $V \subset N$ are models of $ZFC$ and $V\vDash M$ is "canonical", then $N \vDash M$ is "canonical." (In the case of Shoenfield's result this last bit is trivial since $V\subset N \implies L^V = L^N$.)

  2. Small combinatorial preservation. These results are numerous and deal with individual statements assumed to hold in the ground model and then using this assumption produce that the statement is true in the extension (using either the definability of forcing or constructing explicit names of objects in the extension, or a combination of both.) Moreover, results of this form can be considered as the reason a certain statement holds in an extension. These types of proofs can be understood as being essentially internal to $V$.

The contents of this answer will attempt to illustrate two examples of the second method and explain why $AC$ is preserved when passing to any generic extension.

Now for a horrible and vague answer to a strange question: What does using a partial order $\mathbb{P}$ to force over a model $V$ actually do? It lets people with imaginations strictly confined to $V$ imagine/build an exterior universe $V[G]$ (whose properties are bound to $\mathbb{P}$, the forcing relation $\Vdash_{\mathbb{P}}$, and $V$.)

From this perspective the original question can be recast as: How does one prove certain statements are preserved by a particular $\mathbb{P}$? The answer: it depends, but in general there are two methods which appear most. From here we will consider a couple examples.

Consider the case of $AC$: Now, $AC$ is equivalent to the statement $ \forall X\ \exists \alpha \in ON\,\ f\subset X \times \alpha\ (\ f$ is injective$)$ and we want to show that this statement is preserved between forcing extensions.

First Method: (Brute force combinatorics) (paraphrasing what is found in Kunens' text)

We must show $\forall p\in \mathbb{P}\ \exists q \le p\ [q \Vdash \forall \dot{X}\ \exists\check{\alpha}\in ON,\ \dot{f} \subset \dot{X} \times \check{\alpha}\ (\dot{f} $ is an injection with domain $\dot{X})]$. Using the standard facts about the forcing relation $\Vdash$, this becomes: for every $\dot{X} \in V^{\mathbb{P}}$ and $p \in \mathbb{P}$, there exists some $q \le p$, $\alpha \in ON$, and $\dot{f}\in V^{\mathbb{P}}$ such that $q \Vdash [\ \dot{f} $ is an injection with domain $\dot{X}\ ].$

In order to show this assertion, we are going to explicitly build the name $\dot{f}$, and produce the condition $q$ which forces "$\dot{f} $ is an injection with domain $\dot{X}$" only using statements $ZF+AC$ can prove.

To this end, note that since we are assuming $AC$, we may assume $\dot{X} = \bigcup_{\gamma\in\mu}\{ \langle \sigma_\gamma, q \rangle: q \in A_\gamma \}$, where each $\sigma_\gamma$ is a $\mathbb{P}$-name, $\mu$ is some cardinal, and $A_\gamma$ is an anti-chain such that $\forall q \le p$: we have $q \Vdash [\sigma_\gamma \in \dot{X}]$, if and only if $\{ s \in A_\gamma: s \not\perp q \}$ is maximal below $q$.

Let $\dot{g}= \bigcup_{\alpha\in\mu} \{ \langle \rho_\alpha(q), q \rangle: q \in A_\alpha \}$ where $\rho_\alpha(q) = \{\langle \sigma_\alpha, q \rangle, \langle \{\langle \sigma_\alpha, q \rangle, \langle \check{\alpha}, q \rangle, \}, q \rangle \}$. Then, taking $q=p$, $\alpha = \mu$ and $\dot{f} = \dot{g}$ establishes the result.

Second Method: (Sage like and Model Theoretic)

The statement "$f$ is an injection" is $\Delta_0$. Moreover, if $f$ is an injection, then $ZF(?C)$ proves that the same holds for all of its subsets. Noting that for every $\dot{X} \in V^{\mathbb{P}}$ there exists some $Y \in V$ such that $ 1 \Vdash \exists \dot{g} \subset \dot{X} \times \check{Y}\ (\dot{g} $ is an injection $)$ (namely $Y = \{\langle \sigma, q \rangle^\check{\ }: \langle \sigma, q \rangle \in \dot{X} \}$) and by $AC$ there exists some injection $f : Y \to \mu$ (some cardinal $\mu$), the result follows.

Conclusion for $AC$: The reason $AC$ is preserved when passing to a generic extension is essentially because it is equivalent to a "well-positioned" statement. Where "well-positioned" in this case means: it asserts the existence of an object with a $\Delta_0$ property and this $\Delta_0$ property is maintained by each of its subsets, all of which can be turned into a witness for some particular instance of $AC$. In this way, the truth or falsity of $AC$ in the extension does not depend on $\mathbb{P}$ and depends only on whether or not it held in $V$.

To contrast this with other weak forms of $AC$: consider $DC_\omega$. In order for $DC_\omega$ to be preserved we must avoid something. In particular we must avoid adding some $R \subset X \times X$ which is an entire binary relation and witnesses the failure of $DC_\omega$ (i.e. there is no $\{x_n:n\in\omega\}$ so that $\langle x_n, x_{n+1} \rangle \in R$) and we have no reason to expect this is the case without knowing something about $\mathbb{P}$.

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The ordering principle can be written as $\forall X\exists R\subseteq X\times X: R\text{ is a total order on } X$, where as being a relation is $\Delta_0$, and being a total order is also $\Delta_0$ (all quantifiers are bounded). How is the ordering principle different than the axiom of choice in that matter? As I answered Stefan Geschke's question - the ordering principle is not preserved under forcing. –  Asaf Karagila Oct 3 '11 at 9:46
    
@Asaf every subset of $R$ need not be a total order. The trick that makes the AC result hold is that it is provably the case that every subset of $f$ must also be an injection. –  Michael Blackmon Oct 3 '11 at 10:02
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Michael, your mystical proof is a bit too mystical for me. You wrote "for every $\dot{X} \in V^{\mathbb{P}}$ there exists some $Y \in V$ such that $\forall p\in\mathbb{P}\ ( p \Vdash \dot{X} \subset \check{Y})$." This would mean that every set in the forcing extension is a subset of a set from the ground model, which isn't true --- consider for example the singleton of the generic filter. I think you need to work with a ground model set $Y$ that contains enough names for elements of $X$ so that the forcing extension sees a surjection from $\check{Y}$ onto $X$. –  Andreas Blass Oct 3 '11 at 12:55
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And ev'ryone will say / As you walk your mystic way, / "If this young man expresses himself / In terms too deep for me, / Why, what a very singularly deep young man / This deep young man must be!" –  Todd Trimble Oct 3 '11 at 18:36
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Very ironic.... –  Michael Blackmon Oct 5 '11 at 4:08
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