23
$\begingroup$

Is there any known axiomatization of set theory in which the real numbers are not a set, but the natural numbers and other infinite sets do exist?

Such a set theory would have an Axiom of Infinity, but not an Axiom of Power Set. I know that Kripke-Platek set theory has no Axiom of Power Set, but it is not clear to me whether the real numbers exist as a set in this theory or not.

In the type of set theory I am envisioning, the real numbers would exist as a class, as would the class of all subsets of the natural numbers, and they could be equivalent to the class of all ordinals, for example.

One advantage of such a set theory would be that it could admit the Axiom of Choice, and the Well-Ordering Theorem for all sets, without having to admit the well-ordering of the reals or the Banach-Tarski paradox.

$\endgroup$
  • 12
    $\begingroup$ Why the vote to close? This seems reasonable. $\endgroup$ – Noah Schweber May 6 '16 at 18:21
18
$\begingroup$

There is also Pocket set theory. In this set theory, infinite sets exists, and they are all equinumerous. So every infinite set is countable. But the reals exist as a proper class. This is a second-order theory in the sense that the objects are classes, and sets are classes which are elements of other classes.

In some sense this theory is somehow related to third-order arithmetic. There sets of natural numbers are still objects, and we can quantify over sets of sets of natural numbers (which in Pocket set theory make proper classes).

$\endgroup$
  • 1
    $\begingroup$ +1. Note that one crucial distinction between pocket set theory and KP is that models of KP in which $\mathbb{R}$ doesn't exist may still contain sets of very high rank - e.g. $L_{\omega_1}$ has elements of rank arbitrarily high below $\omega_1$, but $\mathbb{R}\cap L_{\omega_1}\not\in L_{\omega_1}$. And we can do better with forcing: for $\kappa$ arbitrary, consider $L_\kappa$ adjoin some Cohen reals. So we can avoid having a set of all reals, while still having no small bound on the rank of sets in our universe. $\endgroup$ – Noah Schweber May 6 '16 at 18:45
  • $\begingroup$ I think that the biggest difference between the two is that KP is a subtheory of ZFC, but Pocket set theory outright disproves the power set axiom. $\endgroup$ – Asaf Karagila May 7 '16 at 15:19
  • $\begingroup$ How does that really differ from taking the exact same theory and just replacing the word 'class' with 'set'? Given that 'class' seems to refer to 'sets that we don't want to call sets'... $\endgroup$ – Miles Rout May 15 '16 at 2:07
  • $\begingroup$ @Miles: How would you define classes in that case? Generally speaking, it's not different from calling the objects bananas and declaring that oranges are bananas which are elements of other bananas. It's just a naming convention; but it's hinting to us, the readers and users, what the objects are, and how to think about them. $\endgroup$ – Asaf Karagila May 15 '16 at 3:31
9
$\begingroup$

In KP, there need not be a set of all real numbers. The canonical example of this is $L_{\omega_1^{CK}}$, the first level of Godel's constructible universe which satisfies $KP$. There are reals with $L$-rank arbitrarily high below $\omega_1^{CK}$, so any set of reals in $L_{\omega_1^{CK}}$ is not the set of all reals in $L_{\omega_1^{CK}}$.


EDIT: Of course, $L_{\omega_1}$ also has this property, as does $L_\alpha$ for many countable $\alpha$s. However, if $L_\alpha$ is the Mostwoski collapse of an elementary substructure of $L_{\omega_2}$ (say), then $L_\alpha$ will think that there is a set of all reals, even if $\alpha$ is countable - basically, there are long "gaps" in the countable ordinals where no new reals enter $L$.

$\endgroup$
6
$\begingroup$

I think you can just take ZFC without the power set axiom.

The main thing you'd still want to use it for is constructing a Cartesian product, but that can still be done in smaller steps using Replacement.

And if your metatheory is sufficiently ZFC-like, the set of hereditarily countable sets ought to be a model, ensuring that you can have all of your usual reals (say, as Dedekind cuts) without getting set of all of them itself forced on you through some back door.

What this won't give you is a guarantee that all proper classes are equivalent -- but if you want, you can add that as an axiom to NBG$-$P, and then to make a model work within the constructible universe and take the set of all hereditarily countable sets, plus all of its subsets. This still gives you all the reals you could want (though perhaps not all the ones you had to begin with).

$\endgroup$
  • $\begingroup$ If you look at the hereditarily countable sets, and consider classes to be definable subsets, you get a model of Pocket set theory. $\endgroup$ – Asaf Karagila May 6 '16 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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