# Can we use proper classes in this way, to define a new infinity larger than |Ord|?

I believe there is a way to do this that makes sense, and I explain it below. I would like to know if I did some obvious mistake, or if the idea doesn't make sense for some reason I didn't figure it out.

We can generalize the idea of an ordinal number in the same way we did it to expand $\mathbb{N}$ to $\omega$ and beyond. Basically, you get the ordinals when you define a limit to the infinite sequence $1,2,...n,...$ which in its most naive definition has no limit, and call it $\omega$. Now we put one $\omega$ after another, in the same way that we had put one natural number after another, and we get not only countable infinite sets of various cofinalities, but we also reach uncountable sets. And not only of cardinality $2^{\omega}$, but also of any large cardinality (if we use the standard definition of cardinal). But we still reach a "limit". The limit is |V| (or |Ord|), the proper class of all ordinals. So, somehow, the infinities within ZFC (or most other popular variants of set theory) reach a limit, in the same sense that the naturals reach $\omega$.

So what happens if we put two proper classes one after another? (actually, we should start with a proper class and a set, one after another, and get $|V|+1, |V|+2, ... , |V|+|V|=|V|. 2, ...$ etc). We can try to name these infinities in the same way as we did with the transfinite ordinals (either countable or uncountable ones). Doing this gets infinities well beyond |V| (and presumably without any paradox). They are not sets, or course, they are other kind of object that generalizes what a set is. In the same way that an infinite ordinal generalizes what a natural number is (they are perhaps not proper classes either, but just put them a new name). Could we say that |V| is a fixed point of the power set operator? (you cannot reach |V|+1 from |V| within ZFC, I guess, or you can, but you would need to use some non-standard model, by analogy with the non-standard models of PA). Even if we do so, we will end up reaching some new limit, or fixed point. And when we reach it, we can start all over again. So we can have a hierarchy of theories that can reach higher and higher infinities. Each theory has a fixed point, in the sense that the point is the "infinity class" that the standard model of the theory can reach. Let us call them $S^n$, with $n \in \mathbb{N}$. Then, roughly: $S^0=PA$ (can reach $\omega$), $S^1=ZFC$ (can reach |Ord|), $S^2=?$ (reaches beyond the standard ordinals), ...., . But we can keep doing this forever, and define a new theory that is the limit of all previous theories. And so on, ad nauseam...

Do this makes sense at all???

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The question is whether or not you are trying to define extensions of ordinals or cardinals.

If you want larger cardinals than $|\sf Ord|$, note that in some models of $\sf ZFC$ we don't have a bijection between $V$ and $\sf Ord$ so $|V|$ is strictly larger in that sense. It's quite obvious that we cannot go beyond that limit from within the universe.

If you only care about order type then there are three points to make:

1. Despite that in $\sf PA$ we can only talk about the natural numbers themselves we can still define, and prove basic things about some infinite ordinals. For example it's quite easy to define a subset of $\Bbb N^2$ which is a well-order of order type $\omega^2$ on $\Bbb N$, and we can go even higher than that. Much higher.

Similarly, we can do the same with ordinals and go beyond $\sf Ord$. In both cases, however, the result is a definable class rather than an element of the universe. It matters to some things, but not to others. So it really depends on what you want to do with these order types.

2. You can talk about end-extensions, that is $M\subseteq N$ are two models of $\sf ZFC$, then $N$ is an end-extension of $M$ if there is some $\alpha\in{\sf Ord}^N$ such that $M=(V_\alpha)^N$. That is to say, all the new sets of $N$ come at stages that $M$ didn't know about. One trivial example is taking an inaccessible cardinal $\kappa$ and having $V$ as an end-extension of $V_\kappa$.

The ordinals of $N$ is a class which is strictly longer than that of $M$. So very much longer. And so you can talk about end-extensions in this sense. One can find those in the Tarski-Grothendieck set theory which has universes upon universes, and each universe is an end-extension of the previous one.

3. Terminology wise, using "non-standard" has a particular sound to it which implies that the new ordinals are non-standard ones, in the sense that they are not well-founded (externally, of course). Much like non-standard integers. We run into these kind of ordinals often when we talk about arbitrary models of set theory, because they are often there.

I don't believe that you meant something like that. I think that you wanted to suggest that the $\sf ZFC$ operations are insufficient to reach those kind of ordinals.

But we already know something like that in $\sf ZFC$. We know about inaccessible cardinals, and those are exactly the cardinals which are inaccessible from $\varnothing$ with the usual operations of set theory. So in some sense, it seems to me, that you are looking exactly for those. Which takes us back to the point about end-extensions.

You may also find it interesting to take a look at this MathOverflow question by me. While there is no actual answer given, it might give you some idea what to make of my first point above.

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I think you are right, and I think my mistake was an unconscious intuition (a wrong one), that somehow identified V with $\aleph_{\omega}$ . I think now that ZFC + large cardinals have been taking care of all the infinities I was talking about. So I have one question left: Is $\aleph_{\omega}$ the power set of a smaller cardinal? Or is it a limit of the power set operation? If that is the case, it is the first (I mean the smallest) cardinal with that property? –  julian fernandez May 12 '13 at 21:21
I just realized that I am talking about strong limit cardinals. Me only question is then is $\aleph_{\omega}$ the first (smallest) strong limit cardinal? –  julian fernandez May 12 '13 at 22:19
Julian, $\aleph_\omega$ is very consistently not a strong limit cardinal. The first strong limit cardinal is $\beth_\omega$. It is consistent that $\aleph_\omega=\beta_\omega$, but it is consistent that $\beth_\omega$ is much much much larger than $\aleph_\omega$. –  Asaf Karagila May 12 '13 at 22:21
I think I've got it, thanks! –  julian fernandez May 12 '13 at 22:25
To your first question, by the way, $\aleph_\omega$ is never the power set of a smaller cardinal. You see, the cofinality must increase when applying a power set, so there is no cardinal $\kappa$ such that $2^\kappa$ has cofinality $\omega$. –  Asaf Karagila May 12 '13 at 22:25