Recent claim that inaccessibles are inconsistent with ZF - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:13:02Z http://mathoverflow.net/feeds/question/73121 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73121/recent-claim-that-inaccessibles-are-inconsistent-with-zf Recent claim that inaccessibles are inconsistent with ZF Quinn Culver 2011-08-18T03:43:58Z 2011-08-20T02:47:25Z <p><a href="http://logiciansdoitwithmodels.com/2011/08/17/fom-posting-on-a-kiselevs-claim-that-there-are-no-weakly-inaccessible-cardinals-in-zf/" rel="nofollow">Here</a> it is mentioned that someone claims to have proven that there are no weakly inaccessibles in ZF.</p> <p><strong>Question 1:</strong> What reasons are there to believe that weakly inaccessibles exist?</p> <p><strong>Question(s) 2:</strong> Since all large cardinals are weakly inaccessible, this would have a profound effect on set theory. What are some of the most significant results <em>whose only known proof</em> assumes the existence of weakly inaccessibles? Might any of the arguments go through without their existence? For example, I've heard that the original proof of Fermat's Last Theorem (FLT) assumed (something equivalen to) a large cardinal, but it was then shown that the argument went through without such an assumption.</p> <p><strong>Edit.</strong> I just added the phrase "whose only known proof" to Question 2 above, which is what I intended originally. The point of course, is that I want to know which results, if any, would be "lost" if weakly inaccessibles were lost. FLT is not an example of that, but would have been before it was known that weakly inaccessibles are not necessary in its proof.</p> http://mathoverflow.net/questions/73121/recent-claim-that-inaccessibles-are-inconsistent-with-zf/73216#73216 Answer by François G. Dorais for Recent claim that inaccessibles are inconsistent with ZF François G. Dorais 2011-08-19T12:22:04Z 2011-08-19T21:54:40Z <p>As I pointed out in the meta thread, this question overlaps with a bunch of older MO questions.</p> <ul> <li><a href="http://mathoverflow.net/questions/44095/arguments-against-large-cardinals" rel="nofollow">http://mathoverflow.net/questions/44095/arguments-against-large-cardinals</a></li> <li><a href="http://mathoverflow.net/questions/48851/nonessential-use-of-large-cardinals" rel="nofollow">http://mathoverflow.net/questions/48851/nonessential-use-of-large-cardinals</a></li> <li><a href="http://mathoverflow.net/questions/35746/inaccessible-cardinals-and-andrew-wiless-proof" rel="nofollow">http://mathoverflow.net/questions/35746/inaccessible-cardinals-and-andrew-wiless-proof</a></li> <li><a href="http://mathoverflow.net/questions/29302/reasons-to-believe-vopenkas-principle-huge-cardinals-are-consistent" rel="nofollow">http://mathoverflow.net/questions/29302/reasons-to-believe-vopenkas-principle-huge-cardinals-are-consistent</a></li> </ul> <p>However, none of these questions directly address the particular case of the existence of inaccessible cardinals, which is of special interest as this is the weakest of all large cardinal hypotheses. This answer focuses on that case.</p> <p>Penelope Maddy gives several answers to Question 1 in &sect;III of <em>Believing the Axioms, I</em> [JSL 53 (1988), 481-511, <a href="http://www.ams.org/mathscinet-getitem?mr=947855" rel="nofollow">MR0947855</a>]. In this wonderful paper, Maddy justifies many set theoretic axioms and hypotheses using five widely believed "rules of thumb": <em>maximize</em>, <em>inexhaustibility</em>, <em>uniformity</em>, <em>whimsical identity</em>, and <em>reflection</em>. Here is a brief summary of these five arguments as it pertains to the existence of inaccessible cardinals.</p> <ul> <li><p><em>The maximization argument.</em> The <em>maximize</em> rule of thumb is perhaps best understood as the opposite of <a href="http://en.wikipedia.org/wiki/Occam%39s_razor" rel="nofollow">Occam's Razor</a>. However, blind application of this easily leads to contradictions. Thus, the rule is generally understood as a pair of statements: <em>thikness</em> &mdash; powersets are very large; and <em>tallness</em> &mdash; there are lots and lots of ordinals. The second easily leads to the existence of inaccessibles.</p></li> <li><p><em>The inexhaustibility argument.</em> Maddy describes this one very well: "The universe of sets is too complex to be exhausted by any handful of operations, in particular by power set and replacement, the two given by the axioms of Zermelo and Fraenkel. Thus there must be an ordinal number after all the ordinals generated by replacement and power set. This is an inaccessible." (p. 502)</p></li> <li><p><em>The uniformity argument.</em> Uniformity basically states that the richness of the universe should not concentrate in a small region, that if a certain property is found at a certain level of the cumulative hierarchy then analogue properties should also be found higher up. Thus, there should be many cardinals that share the same properties as $\aleph_0$, such as the fact that $2^k &lt; \aleph_0$ for every $k &lt; \aleph_0$. Combined with regularity, this leads to the existence of inaccessibles.</p></li> <li><p><em>The whimsical identity argument.</em> This rule of thumb states that there should be no accidental identities, "like the identity between 'human' and 'featherless biped'." (p. 499) It seems unlikely that $\aleph_0$ should be characterized as <em>the unique regular cardinal $\kappa$ such that $2^\mu &lt; \kappa$ for every $\mu &lt; \kappa$</em>. Therefore, there must be inaccessible cardinals.</p></li> <li><p><em>The reflection argument.</em> This powerful rule of thumb is a generalization of Montague's Reflection Theorem, which states that for every first-order formula $\phi(\bar{x})$ of $V \vDash \phi(\bar{x})$ then there are arbitrarily large ordinals $\alpha$ such that $V_\alpha \vDash \phi(\bar{x})$. The <a href="http://en.wikipedia.org/wiki/Reflection_principle" rel="nofollow">Reflection Principle</a> generalizes this from first-order properties to arbitrary properties. Thus, since $V$ is closed under replacement and powerset, there must be arbitrarily large ordinals $\alpha$ such that $V_\alpha$ is also closed under replacement and powerset. These ordinals are inaccessibles.</p></li> </ul> <p>These five arguments have a lot in common, but the basic principles behind them are quite different. I would contend that these are five distinct justifications for the existence of inaccessibles.</p> <p>Note that Maddy's paper has a sequel <em>Believing the Axioms, II</em> [JSL 53 (1988), 736-764, <a href="http://www.ams.org/mathscinet-getitem?mr=960996" rel="nofollow">MR0960996</a>]). Let me also point out nother highly relevant paper: Kanamori and Magidor, <em>The evolution of large cardinal axioms in set theory</em> [LNM 669, 99-275, <a href="http://www.ams.org/mathscinet-getitem?mr=520190" rel="nofollow">MR0520190</a>]. Of course, detailed information can be found in Kanamori's <em>The Higher Infinite</em> [Perspectives in Mathematical Logic. Springer-Verlag, Berlin, 1994].</p> http://mathoverflow.net/questions/73121/recent-claim-that-inaccessibles-are-inconsistent-with-zf/73256#73256 Answer by Joel David Hamkins for Recent claim that inaccessibles are inconsistent with ZF Joel David Hamkins 2011-08-20T01:28:46Z 2011-08-20T02:47:25Z <p>François has excellently addressed your question 1; allow me to address question 2. I understand the question to be: what will be the mathematical effects if someone were to show that there are no (weakly) inaccessible cardinals? A similar question would apply to any of several large cardinals. So let me list some consequences.</p> <p>First, let me note that the existence of a weakly inaccessible cardinal is provably equiconsistent with the existence of a (strongly) inaccessible cardinal, since any weakly inaccessible cardinal is strongly inaccessible in $L$, and so the issue about weakly or strongly inaccessible is entirely irrelevant when it comes to consistency.</p> <p>Second, let me note that set theorists are not generally satisfied by claims of the sort "the only known proof uses such-and-such," but rather they use the concepts of consistency strength and equiconsistency, which allow for precise claims to be proved about exactly which large cardinals are required to prove which statements. The situation is that for many mathematical assertions, we can prove that any proof <em>must</em> use a certain type of large cardinal or something just as strong, in the sense that the consistency of the statement itself implies the consistency of the large cardinal in question. In this way, we avoid any problematic issue about knowledge concerning whether a better proof is simply not yet discovered.</p> <p>As a result, if inaccessible cardinals should be refuted, then using the known results we immediately gain an enormous number of positive theorems. So it isn't really a case of losing theorems, but rather gaining.</p> <p><b>Theorem.</b> If inaccessible cardinals are inconsistent, then (we can prove that) we can construct a non-Lebesgue measurable set of reals without using the axiom of choice.</p> <p>This follows from the fact that Solovay and Shelah have proved that the possibility of constructing a non-Lebesgue measurable set of reals (in the context of ZF+DC) without using AC is exactly equivalent to the inconsistency of inaccessible cardinals.</p> <p>Most people believe that one must use AC in any Vitali-type construction of a non-Lebesgue measurable set, and the theorem above shows that this belief is provably equivalent to the consistency of inaccessible cardinals. Perhaps many mathematicians would find their confidence in the consistency of inaccessible cardinals to increase upon learning of this, and in this sense, this is also an answer to question 1. In any case, many well-known set theorists have emphasized enormous confidence in the consistency of large cardinals, and have stated quite explicitly that if inaccessible cardinals should become known to be inconsistent, then we should expect further inconsistency much lower in ZFC itself or in the low levels of PA.</p> <p><b>Theorem.</b> If inaccessible cardinals are inconsistent (and even merely if we can refute infinitely many Woodin cardinals), then (we can prove that) there is a projective set of reals $A\subset\mathbb{R}$ whose corresponding two-person game of perfect information has no winning strategy for either player. In other words, the infinitary de Morgan law $$\neg\forall n_0\exists n_1\forall n_2\exists n_3\cdots A(\vec n)\iff\exists n_0\forall n_1\exists n_2\forall n_3\cdots\neg A(\vec n)$$ will fail for some projective set $A$.</p> <p>The projective sets of reals are those reals that are definable by a property involving quantification only over real numbers and integers. The reason for the theorem is that projective determinacy is equiconsistent over ZFC with infinitely many Woodin cardinals, and so if we refute the large cardinals in ZFC, then we similarly refute projective determinacy.</p> <p><b>Theorem.</b> If inaccessible cardinals are inconsistent (and even if merely measurable cardinals are inconsistent), then (we can prove that) there is an analytic set (a continuous image of a Borel set) that is not determined. </p> <p><b>Theorem.</b> If inaccessible cardinals are inconsistent, then we can prove that the full set-theoretic universe is very close to the constructible universe in the sense of covering. In particular, $L$ computes the successors of singular cardinals correctly.</p> <p>This shocking conclusion follows in this case from Jensen's covering lemma, since refuting inaccessible cardinals implies a refutation of $0^\sharp$.</p> <p><b>Theorem.</b> If inaccessible cardinals are inconsistent, then on no set is there a countably complete real-valued measure measuring all subsets of the set and giving points no mass.</p> <p>This is simply because any real-valued measurable cardinal is measurable and hence inaccessible in an inner model.</p> <p><b>Theorem.</b> If inaccessible cardinals are inconsistent, then (we can prove that) there are no uncountable Grothendieck universes and the axiom of universes in category theory is false.</p> <p>An uncountable Grothendieck universe is exactly $H_\kappa$ for an inaccessible cardinal $\kappa$, and the axiom of universes asserts that every set is in such a universe.</p> <p>There are many more examples. (I invite any knowledgeable person to edit the answer with additional examples.)</p>