(Non?)-linearity of the consistency strength ordering in ZF - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:42:27Z http://mathoverflow.net/feeds/question/59717 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf (Non?)-linearity of the consistency strength ordering in ZF Marios Koulakis 2011-03-27T12:25:26Z 2011-11-17T15:43:20Z <p>Much of the research taking place in set theory, is related to the classification of sentences according to their consistency strength relative to ZF. In order to be more specific, we say that for all sentences $\sigma,\tau\in Sent$, $\sigma$ has less consistency strength than $\tau$ (or $\sigma\leq_{cons}\tau$) iff $cons(ZF+\tau)\rightarrow cons(ZF+\sigma)$. My main question (which I believe should have a negative answer) is the following: $$\mbox{ Is } \leq_{cons}\mbox{ linear? If not is there a natural counterexample?}$$ In fact I have several similar questions concerning $\leq_{cons}$ and I would be glad if someone could provide me with some references. Thanks! </p> http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/59749#59749 Answer by Stefan Geschke for (Non?)-linearity of the consistency strength ordering in ZF Stefan Geschke 2011-03-27T18:48:29Z 2011-03-28T16:51:17Z <p>There are some results of Woodin that indicate why large cardinal axioms of a certain kind might indeed be linearly ordered by consistency strength. In Kanamori's chart there are some possibilities for nonlinearity, but I think this is just lack of knowledge.</p> <p>I am not aware of any result that actually says that two natural statements are incomparable in consistency strength over ZF. Note however that this might be a problem of our current proof technology. </p> <p>We have good technology to show that the consistency of a statement $\sigma$ implies the consistency of $\tau$ (all over ZF). Namely, start with a model of ZF+$\sigma$ and use forcing and/or inner models to construct a model of ZF+$\tau$. The second incompleteness theorem tells us how we can get a stronger theory from a consistent theory $T$, namely by adding the axiom "$T$ is consistent".</p> <p>But how could we prove that two given statements are incomparable in consistency strength?</p> http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/59763#59763 Answer by Pakhomov Fedor for (Non?)-linearity of the consistency strength ordering in ZF Pakhomov Fedor 2011-03-27T20:22:59Z 2011-03-28T04:21:07Z <p>The order $\le_{cons}$ isn't linear. There are Godel Lob provability logic $GL$. It's a modal logic with one modality. $GL$ is complete with respect to arithmetic semantic. It was shown by Solovay. Formula $\square(\Diamond p \to \Diamond q) \lor \square(\Diamond q \to \Diamond p)$ is not theorem of $GL$. So by completeness theorem (it can be proved for $ZF$) there are propositions $A$ and $B$ such that $ZF \not \vdash (Con({\ulcorner} A{\urcorner}) \to Con({\ulcorner} B{\urcorner}))$ and $ZF\not \vdash (Con({\ulcorner} B{\urcorner}) \to Con({\ulcorner} A{\urcorner}))$ But such $A$ and $B$ provided by Solovay proof are complex. I don't know any "natural" example.</p> http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/59781#59781 Answer by Alex Lupsasca for (Non?)-linearity of the consistency strength ordering in ZF Alex Lupsasca 2011-03-28T00:32:35Z 2011-03-30T19:39:03Z <p>My understanding is that the large cardinals are indeed linearly ordered, which is a remarkable fact. Of course, it is not a theorem (nor could it ever be?), but merely an empirical fact; still, there is no known explanation for this completely surprising coincidence.</p> <p>On the other hand, there are sentences $\sigma$ such that $T+\sigma>T$ and $T+\neg\sigma>T$. These are called "Double jump sentences" and though they are rare, a number of them exist. Thus, $\le_{cons}$ is not linear, though the counterexamples are quite contrived (none of the known ones would be called "natural" by a mathematician).</p> <p>You can read more about Double jump sentences (and find references) here:</p> <p><a href="http://plato.stanford.edu/entries/independence-large-cardinals/#IntHie" rel="nofollow">http://plato.stanford.edu/entries/independence-large-cardinals/#IntHie</a></p> <p>I hope this answers your question to your satisfaction!</p> http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/59782#59782 Answer by Richard Borcherds for (Non?)-linearity of the consistency strength ordering in ZF Richard Borcherds 2011-03-28T00:35:53Z 2011-03-28T00:35:53Z <p>For theories one can define their proof theoretic ordinal, which is a countable ordinal that is (roughly) the smallest ordinal that the theory cannot prove is well ordered. If one orders theories by the size of the proof theoretic ordinal, this gives a total order on theories because ordinals are totally ordered. In practice the size of the proof-theoretic ordinal is a good measure of the consistency strength of a theory (roughly speaking, proving that the proof theoretic ordinal is well ordered tends to be about as strong as proving that the theory is consistent) so consistency strength usually gives a total order on theories. </p> http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/59800#59800 Answer by Andres Caicedo for (Non?)-linearity of the consistency strength ordering in ZF Andres Caicedo 2011-03-28T05:34:37Z 2011-03-30T18:44:47Z <p>Marios, this is indeed a fascinating topic. </p> <p>The consistency strength hierarchy is <em>not</em> linearly ordered. One can produce counterexamples by variants of Gödel sentences or of <a href="http://en.wikipedia.org/wiki/Rosser%2527s_trick" rel="nofollow">Rosser sentences</a>. It is actually an interesting exercise to produce explicit examples of a pair of Rosser-like sentences $\phi,\psi$ both independent of ZFC, such that ZFC$+\phi$ and ZFC$+\psi$ are incomparable, by letting $\phi$ refer to the length of proofs of consistency of ZFC$+\psi$, and letting $\psi$ refer to the length of proofs of the consistency of ZFC$+\phi$. Explicit examples can be found, I believe, in Per Lindström's "Aspects of Incompleteness" (Lecture Notes in Logic 10).</p> <p>Harvey Friedman has worked on this question, and expanded on the idea above. I believe it is his result that one can even embed a universal countable partial order in the consistency strength degrees or in the degrees of <em>interpretability</em>. I'm pretty sure he has commented on this in the <a href="http://www.cs.nyu.edu/mailman/listinfo/fom/" rel="nofollow">FOM</a> list, but I don't seem to be able to find his relevant postings right now. (I once gave a talk on this years ago, but cannot find my slides either. Old age...)</p> <p>On the other hand, everybody agrees that these examples are not <em>natural</em> (see for example, the beginning of the section "The Gödel hierarchy" in <a href="http://www.math.psu.edu/simpson/papers/gh/" rel="nofollow">this paper</a> by Simpson, or page 54 of "<a href="http://books.google.com/books?id=qqdXsato74QC&amp;printsec=frontcover&amp;dq=fixing+frege&amp;hl=en&amp;src=bmrr&amp;ei=BR2QTbGGDZScsQO6q_CDCQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCgQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">Fixing Frege</a>" by John P. Burgess). It is a remarkable empirical phenomenon that we indeed have comparability for natural theories. We expect this to always be the case, and a significant amount of work in inner model theory is guided by this belief. But although we have many striking partial results in this direction, it is not quite a theorem, in part because <em>natural</em> is not a formal concept. </p> <p>The standard way by which we establish comparability of statements is by using forcing and the technique of inner models to measure a given statement against the large cardinal hierarchy, identifying this way the <em>large cardinal companion</em> of the statement. Different statement can then be compared by comparing their companions. </p> <p>Hugh Woodin, in his <a href="http://www.amazon.com/Determinacy-Forcing-Nonstationary-Gruyter-Applications/dp/3110197022" rel="nofollow">book</a> on Pmax suggests a way of formalizing this observation, by explaining how to associate <a href="http://en.wikipedia.org/wiki/Universally_Baire_set" rel="nofollow">Universally Baire sets</a> to theories, the point being that under appropriate determinacy assumptions, the Wadge degrees of any two such sets are comparable, and the resulting ordering coincides precisely with the interpretability or consistency strength orderings. (The universally Baire sets ultimately trace back to the complexity of certain iteration strategies.) This, of course, requires that we work under additional large cardinal assumptions beyond ZFC, so it only applies to theories of high consistency strength (past $\omega$ Woodin cardinals, for example). This remarkable approach has thus the drawback that it does not explain why we find the phenomenon in the low levels of the hierarchy, the ones that are precisely amenable to current inner model theoretic techniques.</p> <p>At the moment there are a few notable examples of theories that we do not quite know how to fit into the linear ordering of consistency strength. This is most likely due to our present technical limitations. For example, <em>strongly compact cardinals</em> are expected to either coincide or sit just below supercompact cardinals consistency-wise. However, their behavior is rather atypical and they are not really as well understood as other (strong) large cardinals.</p> <p>For a nice introduction to these ideas, examples, and a far reaching program that touchesof Woodin's recent work on "ultimate $L$", see these <a href="http://math.berkeley.edu/~steel/talks/phila2010.pdf" rel="nofollow">slides</a> of a talk by John Steel at the Workshop on Set Theory and the Philosophy of Mathematics, Oct. 2010.</p> http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/81182#81182 Answer by Joost Lambregts for (Non?)-linearity of the consistency strength ordering in ZF Joost Lambregts 2011-11-17T15:43:20Z 2011-11-17T15:43:20Z <p>Here is the relevant posting on the FOM list by Harvy Friedman can be found here: <a href="http://www.cs.nyu.edu/pipermail/fom/1998-August/001929.html" rel="nofollow">http://www.cs.nyu.edu/pipermail/fom/1998-August/001929.html</a> The result that the consistency strength ordering has an infinite set of incomparable elements is found in corollary 4. It seems to me by the way that his proof of corollary 4 only holds for models of PA+Con(PA), not models of PA+notCon(PA)</p>