Question Is there any forcing free proof for hard independence results? talks about the use of forcing for independence results such as: $Con(ZFC)\longrightarrow Con (ZFC+\neg CH) $. For that case (and its companions $V=L$ and $AC$) constructibility is used for the other half of the independence: $Con(ZFC)\longrightarrow Con (ZFC+ CH) $.

Is this typical? Do later independence results still tend to use forcing one way, and $V=L$ the other? Are there independence results where forcing is used in both directions?

Are there statements $S$ where forcing is used for both $Con(ZF+S)$ and $Con(ZF+\neg S)$?

Question Is every class that does not add sets necessarily added by forcing? mentions use of forcing over a model of $NBG$ to add a universal choice function. But I do not know for whether you would also use forcing to eliminate that version of global choice in that context.

  • $\begingroup$ So your question is, essentially, what statement are independent of $V=L$ that we can force to be true? $\endgroup$ – Asaf Karagila Aug 10 '13 at 17:38
  • $\begingroup$ Here's a contrived example (which is why I'm not posting it as an answer): $S$ is the statement "$\sf CH$ holds and there exists a real number $r$ which is Cohen over $L$". We can force this statement to be true by adding one Cohen real over $L$; and we can force its negation by adding many Cohen reals. $\endgroup$ – Asaf Karagila Aug 10 '13 at 18:01
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    $\begingroup$ Asaf mentioned "there is a real Cohen generic over $L$". This is an example of what Joel calls a "button". (Another is "$V\ne L$", of course, but this one is more interesting, in my opinion.) $\endgroup$ – Andrés E. Caicedo Aug 10 '13 at 19:30
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    $\begingroup$ @Asaf The statment $S$ that you proposed is false in $L$, so we need forcing only for the consistency of $S$. $\endgroup$ – Andreas Blass Aug 11 '13 at 4:42
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    $\begingroup$ @Asaf Still refutable in $L$. Propositional combinations of sentences decided by $V=L$ are themselves decided. I don't think there's much real set theory undecided in $L$, though of course there are Gödel sentences and such. $\endgroup$ – Andreas Blass Aug 12 '13 at 1:26

There is a wide variety of statements $S$ for which forcing can be used to prove the (relative) consistency of both $S$ and $\lnot S$ with $\mathsf{ZFC}$.

For example, Cohen developed forcing to prove the consistency of $\lnot\mathsf{CH}$, but we can also prove $\mathsf{CH}$ consistent this way: We can add to the universe a well-ordering of the reals of type $\omega_1$ without adding any reals. This, by the way, shows that any projective statement provable from $\mathsf{CH}$ is provable without it.

There are some restrictions: $V=L$ cannot be forced. This is because $L$ is forcing invariant (we can also say generically absoute): If $W$ is a forcing extension of $V$, then $L$ in the sense of $W$ is the same as $L$ in the sense of $V$. This is true of larger core models, and is one of their key features. It allows us to prove results about the universe by arguing about forcing extensions, and is particularly useful when proving consistency strength lower bounds (and in the development of the general theory of core models).

Similarly, if $\alpha$ is a countable ordinal, then we cannot use forcing to make $\alpha$ uncountable (the idea being that if we already see a surjection from $\omega$ onto $\alpha$, adding more sets is not going to make this go away).

Other examples of statements $S$ that can be forced either way are: $2^{\aleph_1}=\aleph_{17}$, $\mathfrak b=\omega_1<\mathfrak c$, any algebraic homomorphism between Banach algebras is continuous, etc. The list can go on forever. If we assume the presence of large cardinals in the universe, then even more examples can be given.

Joel Hamkins and his collaborators (notably, Benedikt Loewe) have worked on this, making explicit interesting connections with modal logic (think of $\diamondsuit\phi$ as "it is forceable that $\phi$" and of $\square\phi$ as "$\phi$ holds in all extensions"). See here for a starting point.

Now, an interesting fact about forcing is how ubiquitous it is. This distinguishes it from other techniques. Woodin's $\Omega$-conjecture can be informally described as stating that any $\Pi_2$ statement we can verify consistent (with large cardinals) in fact can be forced (over arbitrary models with large cardinals). (This is an informal description. A more technical presentation would perhaps obscure the point here. See this survey by Bagaria, Castells, and Larson, for details.)

One of the reasons why the $\Omega$-conjecture is of interest is that it holds in all known "$L$-like" models. It is also forcing absolute, so its consistency cannot be verified by forcing (unless we already know it is true). Another reason is that there are a few examples of $\Pi_2$ statements we do not know how to force over arbitrary models, though we can prove consistent with appropriate large cardinals (by forcing over special ground models). One is: "There is a $\mathbf{\Sigma}^2_1$-well-ordering of the reals, and the continuum is a real-valued measurable cardinal", see here.

In his book The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal, Hugh suggests a few statements of large cardinal character (stronger than "there is a limit $\lambda$ and a nontrivial $j:V_\lambda\to V_\lambda$") that appear to be fragile, meaning that it is not clear that they are preserved by small forcing. Whether these statements can be proved consistent with $\mathsf{CH}$ or with $\lnot\mathsf{CH}$ does not seem to be a matter of forcing. Something different would be needed here.

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    $\begingroup$ And it is true that many statements can be decided using $V=L$, while their negation can be decided using forcing, but in quite a few situations, we do not want that. For example, we may only be interested in a statement in the presence of large cardinals that are incompatible with $V=L$ (for example, in the presence of a measurable cardinal). $\endgroup$ – Andrés E. Caicedo Aug 10 '13 at 17:58
  • $\begingroup$ A related question is whether we can find statements $S$ such that $V=L+S$ and $V=L+\lnot S$ are both (relatively) consistent. In this case, forcing cannot be used. Obvious examples are Gödel like sentences. Harvey Friedman has produced many natural combinatorial statements of this kind. See here. $\endgroup$ – Andrés E. Caicedo Aug 10 '13 at 18:04
  • $\begingroup$ For how complex a statement needs to be for there to even be a chance of forcing it, you may want to read on Shoenfield absoluteness. Several questions on this site address extensions of this important result in the presence of large cardinals. $\endgroup$ – Andrés E. Caicedo Aug 10 '13 at 18:06
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    $\begingroup$ @ColinMcLarty Yes, definitely. The consistency of $\mathsf{GCH}$ with the existence of a measurable cardinal was first shown by Silver using forcing. (Starting from a model with a measurable. Of course, from the same assumptions, we can also force the existence of a measurable, and the failure of $\mathsf{GCH}$.) Nowadays, we establish this result appealing to inner model theory: The model $L[\mu]$ of Kunen and Silver is a model of $\mathsf{GCH}$ where there is a measurable (this is an example of a core model). (Cont.) $\endgroup$ – Andrés E. Caicedo Aug 11 '13 at 3:28
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    $\begingroup$ @ColinMcLarty In general, the core model for a large cardinal statement $\phi$ is a model of $\mathsf{GCH}$ where $\phi$ holds. This is a nice, forcing absolute model, with many of the properties of $L$. The problem is that the existence of these models is not known for all large cardinal assumptions (proving this existence is the goal of inner model theory), so for cardinals at the level of superstrong or beyond, we currently need forcing arguments to establish their relative consistency with $\mathsf{GCH}$. Luckily, these arguments are typically "standard". $\endgroup$ – Andrés E. Caicedo Aug 11 '13 at 3:31

There are plenty of results where one uses forcing to make both parts of the independence. Even the case of CH can be done this way, since Solovay proved shortly after forcing was invented that one can force CH as well as $\neg$CH. Thus, CH is an example of what Benedikt Löwe and I call a switch, a statement $\varphi$ such that $\varphi$ and $\neg\varphi$ are each forceable over any forcing extension, so that you can turn it on and off again at will. A button, in contrast, is a statement that you can force in such a way that it remains true in all further extensions. These concepts were important in our work on The modal logic of forcing.

Meanwhile, there are several examples where the only known proofs use forcing on both sides, particularly in the context of very large cardinals. For example:

  • The existence of a supercompact cardinal $\kappa$ is relatively consistent with $2^\kappa=\kappa^+$ and with $2^\kappa\gt\kappa^+$. Both sides of these proofs use forcing.

  • Similar consistency results of the very largest large cardinals with CH and GCH and their negations are only known by forcing.

  • Many other consistency results concerning the very largest large cardinals are only known by forcing, because there is no inner model theory analogue of $L$ for those cardinals.

  • $\begingroup$ Ah, I see Andres has posted an answer mentioning some of this... $\endgroup$ – Joel David Hamkins Aug 10 '13 at 18:01

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