Large cardinals without the ambient set theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:49:56Z http://mathoverflow.net/feeds/question/101700 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101700/large-cardinals-without-the-ambient-set-theory Large cardinals without the ambient set theory? Mirco Mannucci 2012-07-08T19:59:09Z 2012-08-24T15:32:20Z <p>In an attempt to understand a bit better large cardinals, I have been thinking along the following lines, which could be summarized under the slogan</p> <blockquote> <p><strong>Talk about cardinals without the (ambient) set theory</strong></p> </blockquote> <p>the class ON is first-order axiomatizable, and thus it looks like I can carve out of ON the subclass CARD (for instance, one could add to the theory an equivalence relation, $\alpha \equiv \beta$ formalizing equinomerosity, and then define a cardinal in the usual way as the min ordinal in the equivalence class). </p> <p>Once I have my definable predicate $CARD(\alpha)$, I can proceed to introduce cardinal arithmetics. For instance, I can define successor as the minimal cardinal greater than the given cardinal. </p> <p>Obviously, I need to make some assumptions as to the basic cardinal arithmetics, so that it looks like the standard one in $ZFC$ + (possibly) generalized continuum hypothesis. </p> <p>Now, assuming one has done all of the above, it appears that the "small" large cardinals, such as weak inaccessible, Mahlo, etc are definable in this theory (even in standard presentations, such as Drake, their definition is arithmetic ). </p> <p>But what about the others, the heavy-weight ones? Do I necessarily have to resort to the ambient set theory ( stationary points, elementary embeddings, etc ) to talk about very large cardinals, or there is always a direct (algebraic/arithmetical/topological) way to provide their definition? </p> <p>Prima facie, it looks like the answer is no, but maybe there is a clever path to answer in the affirmative. Or perhaps, there is some kind of intrinsic boundary, beyond which you need to think of cardinals within the context of set theory</p> <p>Any thought, refs, or known fact? </p> http://mathoverflow.net/questions/101700/large-cardinals-without-the-ambient-set-theory/101709#101709 Answer by Joel David Hamkins for Large cardinals without the ambient set theory? Joel David Hamkins 2012-07-08T22:20:27Z 2012-07-08T22:20:27Z <p>Most of the large cardinals have a variety of equivalent formulations in ZFC, and some of them are characterized by long lists of diverse equivalent properties. For a few examples, take a look at the <a href="http://cantorsattic.info" rel="nofollow">Cantor's Attic</a> entries for <a href="http://cantorsattic.info/Weakly_compact" rel="nofollow">weakly compact</a>, <a href="http://cantorsattic.info/Strongly_compact" rel="nofollow">strongly compact</a> and <a href="http://cantorsattic.info/Weakly_measurable" rel="nofollow">weakly measurable</a> cardinals.</p> <p>There is a phenomenon, however, that in weaker set theories such as ZF some of these characterizations are no longer equivalent. For example, one cannot expect to prove the embedding characterization of measurable cardinals from the ultrafilter characterization in ZF, since one needs the axiom of choice in order to establish the Los theorem on ultrapowers. A similar situation arises with most all of the larger large cardinals. There has been some work understanding the various large cardinals in ZF worlds, for example, under the axiom of determinacy, which implies that many successor cardinals including $\omega_1$ are measurable according to the ultrafilter definition. </p> <p>Secondly, most of the largest large cardinals are defined by second order properties, such as the existence of proper class objects, often embeddings of a particular kind. These definitions cannot be made in ZFC, which has no capacity for second-order quantifiers, but rather are usually made in G&ouml;del-Bernays set theory or Kelly-Morse set theory. It turns out, however, that these second-order definitions in every case (except <a href="http://cantorsattic.info/Reinhardt#Reinhardt_cardinal" rel="nofollow">Reinhardt cardinals</a>) have a first-order equivalent. For example, a cardinal $\kappa$ is measurable if and only if (second-order) it is the critical point of an elementary embedding $j:V\to M$, if and only if (first-order) there is a $\kappa$-complete ultrafilter on $\kappa$. These equivalences are not provable in ZFC, and not even statable in ZFC, but are proved in GBC.</p> <p>Meanwhile, the theory $\text{ZFC}^-$, meaning ZFC without the power set axiom (see my paper, <a href="http://jdh.hamkins.org/what-is-the-theory-zfc-without-power-set/" rel="nofollow">What is the theory of ZFC without power set axiom?</a> for some subtleties about how to axiomatize this theory properly) is sufficient to formalize the basic properties of those large cardinals with a $\Sigma_2$ definition, such as inaccessible, Mahlo, weakly compact, measurable, superstrong, Woodin and huge cardinals, whose existence is absolute between $H_\delta$ and $V$. Meanwhile, notions such as tall, strong and supercompact cardinals involve another quantifier through all the ordinals, which makes them not usually absolute between $H_\delta$ and $V$.</p> <p>Although you seem to object to the embedding characterization of large cardinals, I would say that it is the embedding characterizations that have proved the most fruitful in their use and analysis. It is surely the embedding outlook that unifies an enormous part of the theory.</p> http://mathoverflow.net/questions/101700/large-cardinals-without-the-ambient-set-theory/105347#105347 Answer by o a for Large cardinals without the ambient set theory? o a 2012-08-23T19:37:49Z 2012-08-24T15:32:20Z <p><em>But anyway, my point is this: just look at ON and totally forget that is the spine of V, and see it as a system of ordered numbers. Now axiomatize what you see, much in the same way as you axiomatize its initial segment N. – Mirco Mannucci Jul 8 at 22:21</em></p> <p><em>for instance, one could add to the theory an equivalence relation, α≡β formalizing equinomerosity, and</em></p> <p>talk about cardinals without the (ambient) set theory</p> <p>Well, <a href="http://mishap.sdf.org" rel="nofollow">this construction</a> [Gavrilovich and Hasson’s <em>Exercices de Style</em>, a homotopy theory of set theory] attempts to talk about cardinal invariants and use as little set theory as possible: instead it uses axioms of a model category to do that. and indeed, as you suggest, there equinomerosity (up to some fixed $\kappa$) is introduced, under the name of cofibrant. The construction does not work with the skeleton, though; but perhaps it would be closer to using the skeleton if you modify the defitinions and replace everywhere 'inclusion' aka 'subset' by 'injective map'; then you'd lose limits in your model category. </p> <p>How powerful the language is, is unclear. But you can indeed say $\kappa$ is measurable: that's when the corresponding homotopy caetgory is not dense as a partial order. </p>