Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Talk about cardinals without the (ambient) set theory

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).

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.

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.

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 ).

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?

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

Any thought, refs, or known fact?

share|improve this question
    
Mirco, I think you need to expand on what it "looks" like to you. To me, there is no visible first-order axiomatization of ON and, even then, there is no visible definition of CARD since there is no way to encode injections and bijections using ordinals alone... –  François G. Dorais Jul 8 '12 at 21:23
    
Perhaps you're thinking of the skeleton of the category Set? All sets with the same cardinality are isomorphic, so a skeleton for Set consists of exactly one set for each cardinal together with all functions between these sets. –  François G. Dorais Jul 8 '12 at 21:29
2  
The book "Algebraic Set theory" of Joyal, Moerdik, is a attempt to generalize and treat set theory in a categorical (topos theory) framework –  Buschi Sergio Jul 9 '12 at 8:30
1  
I can't seem to find it right now, but I remember Harvey Friedman having done some work on theories with very high consistency strength in a language with two binary relations < and << that are required to be linear orderings by the theory, with one refining the other. I might have the details wrong though. –  Trevor Wilson Jul 17 '12 at 19:24
1  
Looks like it can be found in a manuscript on H. Friedman's web page called "concept calculus." The relations < and << are called "better than" and "much better than" respectively. I failed to confirm at a glance that they are linear orderings though. –  Trevor Wilson Jul 18 '12 at 21:57

2 Answers 2

up vote 2 down vote accepted

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

for instance, one could add to the theory an equivalence relation, α≡β formalizing equinomerosity, and

talk about cardinals without the (ambient) set theory

Well, this construction [Gavrilovich and Hasson’s Exercices de Style, 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.

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.

share|improve this answer
    
Thanks Peter! I will have to check it out on two counts: one is because it seems in line with the basic questions of this post, and the other because I am looking for applications of homotopical ideas to set theory and model theory –  Mirco Mannucci Aug 27 '12 at 10:14

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 Cantor's Attic entries for weakly compact, strongly compact and weakly measurable cardinals.

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.

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 Gödel-Bernays set theory or Kelly-Morse set theory. It turns out, however, that these second-order definitions in every case (except Reinhardt cardinals) 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.

Meanwhile, the theory $\text{ZFC}^-$, meaning ZFC without the power set axiom (see my paper, What is the theory of ZFC without power set axiom? 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 $V$.

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.

share|improve this answer
2  
To complement the point about measurability and ultrafilters: It is consistent with ZFC that there is a model of ZF+There is an infinite set with a $\sigma$-complete ultrafilter. Namely, every amorphous set has a $\sigma$-complete ultrafilter: the cofinite subsets. Note that the ultrafilter itself is a D-finite set, therefore every sequence of sets in the ultrafilter is finite and has an intersection there. One can require DC to hold, and then we can switch to an $\aleph_1$-amorphous sets, and the co-countable subsets will have a similar role. (cont.) –  Asaf Karagila Jul 8 '12 at 22:30
1  
(cont) One can ask, how would an ultrapower by such $\sigma$-complete ultrafilter would look like, and the answer is not good. It was proved (by Spector and by Howard) that such ultrapower gives either a non-elementary embedding (which results in a model in which Extensionality fails); or the identity if we take external ultrapower (the ultrafilter is not in the model). Either way, if we require Los' theorem to hold, we have to have enough choice to kill those sets, and otherwise we have non-elementary embedding and the ultraprod. are not models of ZF at all. –  Asaf Karagila Jul 8 '12 at 22:33
    
Joeal, as usual thank! Let me process your (dense) answer. Meanwhile, just one quick comment: I do not object to the characterization of large cardinals in terms of embedding, I realize that this is crucial to set theorists, I am simply curious as to how one could grasp cardinal arithmetics in a completely algebraic way (much in the same fashion as one can axiomatize N without talking of its ambient set theory, V_omega). Perhaps there is no such as way, at least for huge cardinals, but perhaps there is (I am thinking of some transfinite version of Ackermann Yoga.....) –  Mirco Mannucci Jul 8 '12 at 22:39
    
PS Super-thanks for Cantor's Attic! Fantastic resource..most definitely in my bookmarks honor list –  Mirco Mannucci Jul 8 '12 at 22:45
    
Thanks, Mirco, Cantor's Attic is a work-in-progress. –  Joel David Hamkins Jul 8 '12 at 23:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.