On statements independent of ZFC + V=L - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:45:49Z http://mathoverflow.net/feeds/question/11480 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11480/on-statements-independent-of-zfc-vl On statements independent of ZFC + V=L Ashutosh 2010-01-12T00:20:20Z 2010-05-23T16:44:49Z <p>Let $V=L$ denote the axiom of constructibility. Are there any <strong><em>interesting</em></strong> examples of set theoretic statements which are independent of $ZFC + V=L$? And how do we construct such independence proofs? The (apparent) difficulty is as follows: Let $\phi$ be independent of $ZFC + V=L$. We want models of $ZFC + V=L + \phi$ and $ZFC + V=L + \neg\phi$. An inner model doesn't work for either one of these since the only inner model of $ZFC + V = L$ is $L$ and whatever $ZFC$ can prove to hold in $L$ is a consequence of $ZFC + V=L$. Forcing models are of no use either, since all of them satisfy $V \neq L$.</p> http://mathoverflow.net/questions/11480/on-statements-independent-of-zfc-vl/11482#11482 Answer by François G. Dorais for On statements independent of ZFC + V=L François G. Dorais 2010-01-12T00:50:11Z 2010-01-12T00:50:11Z <p>This is a very interesting question. Here is what Shelah says about this in <a href="http://shelah.logic.at/E16/E16.html" rel="nofollow">The Future of Set Theory</a>:</p> <blockquote> <p>ISSUE: Where does the truth lie between the following two extremes</p> <ol> <li><p>Every combinatorial statement is decidable in L.</p></li> <li><p>We should have a forcing-like technique to get independence from ZFC+V=L (or for PA, and similar cases, e.g. the twin prime conjecture).</p></li> </ol> <p>I have strong intuition in favor of both positions, but little knowledge. “Combinatorial” means not syntactical but semantical; consistency strength is discounted as well, as disguised versions of it.</p> </blockquote> <p>As far as I know no significant progress has been made. There are however little gems like this <a href="http://mathoverflow.net/questions/3462/characterizations-of-non-wellfounded-models/10075#10075" rel="nofollow">neat trick of Adrian Mathias</a> to show that every real can be constructible which was posted here by Joel Hamkins not long ago.</p> http://mathoverflow.net/questions/11480/on-statements-independent-of-zfc-vl/11491#11491 Answer by Joel David Hamkins for On statements independent of ZFC + V=L Joel David Hamkins 2010-01-12T02:14:21Z 2010-01-12T02:14:21Z <p>There are numerous examples of such statements. Let me organize some of them into several categories.</p> <p>First, there is the hierarchy of <a href="http://en.wikipedia.org/wiki/Large%5Fcardinal" rel="nofollow">large cardinal</a> axioms that are relatively consistent with V=L. See the <a href="http://en.wikipedia.org/wiki/List%5Fof%5Flarge%5Fcardinal%5Fproperties" rel="nofollow">list of large cardinals</a>. All of the following statements are provably indpendent of ZFC+V=L, assuming the consistency of the relevent large cardinal axiom.</p> <ul> <li><p>There is an inaccessible cardinal. </p></li> <li><p>There is a Mahlo cardinal.</p></li> <li><p>There is a weakly compact cardinal.</p></li> <li><p>There is an indescribable cardinal. </p></li> <li><p>and so on, for all the large cardinals that happen to be relatively consistent with V=L. </p></li> </ul> <p>These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, because if we have such a large cardinal in V, then in each of these cases (and many more), the large cardinal retains its large cardinal property in L, so we get consistency with V=L. Conversely, it is consistent with V=L that there are no large cardinals, since we might chop the universe off at the least inaccessible cardinal. </p> <p>Second, even for those large cardinal properties that are not consistent with V=L, we can still make the consistency statement, which is an arithmetic statement having the same truth value in V as in L. </p> <ul> <li><p>Con(ZFC)</p></li> <li><p>Con(ZFC+there is an inaccessible cardinal')</p></li> <li><p>Con(ZFC+there is a Mahlo cardinal')</p></li> <li><p>Con(ZFC+there is a measurable cardinal') </p></li> <li><p>Con(ZFC+there is a supercompact cardinal'). </p></li> <li><p>and so on, for any large cardinal property. Con(ZFC+large cardinal property). </p></li> </ul> <p>These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, since on the one hand, if W is a model of ZFC+Con(ZFC+phi), then L<sup>W</sup> is a model of ZFC+V=L+Con(ZFC+phi), as Con(ZFC+phi) is an arithmetic statement. And on the other hand, by the Incompleteness theorem, there must be models of ZFC+&not;Con(ZFC+phi), and the L of such a model will have ZFC+V=L+&not;Con(ZFC+phi). </p> <p>Third, there is an interesting trick related to the theorem of Mathias that Dorais mentioned in his answer. For any statement phi, the assertion that there is a countable well-founded model of ZFC+phi is a Sigma<sup>1</sup><sub>2</sub> statement, and hence absolute between V and L. And the existence of a countable well-founded model of a theory is equivalent by the Lowenheim-Skolem theorem to the existence of a well-founded model of the theory. Thus, the truth of each of the following statements is the same in V as in L.</p> <ul> <li><p>There is a well-founded set model of ZFC. This is equivalent to the assertion: there is an ordinal &alpha; such that L<sub>&alpha;</sub> is a model of ZFC.</p></li> <li><p>There is a well-founded set model of ZFC with &not;CH. (This is also equivalent to the previous statement.) </p></li> <li><p>There is a well-founded set model of ZFC with Martin's Axiom.</p></li> <li><p>and so on. For all the statements known to be forceable, you can ask for a well-founded set model of the theory. </p></li> <li><p>There is a well-founded set model of ZFC with an inaccessible cardinal. </p></li> <li><p>There is a well-founded set model of ZFC with a measurable cardinal. </p></li> <li><p>There is a well-founded set model of ZFC with a supercompact cardinal. </p></li> <li><p>and the same for any large cardinal notion. </p></li> </ul> <p>These are all independent of ZFC=V=L, since they are independent of ZFC, and their truth is the same in V as in L. I find it quite remarkable that there can be a model of V=L that has a transitive model of ZFC+'there is a supercompact cardinal'. The basic lesson is that the L of a model with enormous large cardinals has very different properties and kinds of objects in it than a model of V=L arising elsewhere. And I believe that this gets to the heart of your question.</p> <p>Since all these statements are studied very much in set theory, and are very interesting, and are independent of ZFC+V=L, I find them to be positive instances of what was requested. </p> <p><b>However</b>, how does this relate to Shelah's view in Dorais's excellent answer? He seems there to dismiss the entire class of consistency strength statements as combinatorics in disguise. What does he mean exactly? Since we set theorists are very interested in these statements, I don't think that he means to dismiss them as silly tricks with the Incompleteness theorem. Perhaps he means something like: to the extent that we believe that a large cardinal property LC is consistent, then we don't really want to consider the theory ZFC+V=L, but rather, the theory ZFC+V=L+Con(LC). That is, we aren't so interested in models having the wrong arithmetic theory, so we insist that Con(LC) if we are comitted to that. And none of the examples I have given exhibit independence from <em>that</em> corresponding theory.</p> http://mathoverflow.net/questions/11480/on-statements-independent-of-zfc-vl/25688#25688 Answer by Andres Caicedo for On statements independent of ZFC + V=L Andres Caicedo 2010-05-23T16:44:49Z 2010-05-23T16:44:49Z <p>What I would call the <em>standard</em> source of examples is the series of very nice "finitary" combinatorial statements that Harvey Friedman has been working on. You can see plenty of such statements in his numbered series of <a href="http://cs.nyu.edu/mailman/listinfo/fom" rel="nofollow">FOM</a> posts, and for details of some of the arguments, see for example his nice paper "Finite Functions and the Necessary Use of Large Cardinals", Annals of Math., Vol. 148, No. 3, 1998, pp. 803-893.</p> <p>The examples Harvey examines are arithmetic. In the "true" model (which is well-founded) they are decided one way, and there are ill-founded (in fact, not $\omega$-) models where they are decided the other way.</p> <p>It would be highly desirable to have examples of nice arithmetic statements that are not just independent but for which we can produce differing answers in different <em>well-founded</em> models. I don't think anybody has any clue at the moment on how to do such thing. To achieve this would be akin to the invention of forcing.</p>