# How many weak/strong limit cardinals exist under different assumptions?

I am trying to hastily answer the question in the title because I need it for another problem. I have been consulting multiple sources and am confused about the standard meanings of the following terms and the relations among them: successor cardinal, weak/strong limit cardinal, weakly/strongly inaccessible cardinal, large cardinal.

On the terminology front, is there a term analogous to successor cardinal but for the powerset operation instead, e.g., something like powerset cardinal? When repeated applications of an operation are being considered, does this always mean countably many applications or something more general?

From what I understand so far, assuming GCH should not affect the number of weak limit cardinals because it says only that every weak limit cardinal is also a strong limit cardinal and vice versa. Is this correct?

Finally, how many weak limit cardinals are there assuming and not assuming GCH, if this makes a difference? Does GCH change the number of strong limit cardinals?

If anyone wants to point out some cool questions and results surrounding this question, that would also be welcome.

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1. First the terminology: Let $\kappa$ be a cardinal,

• $\kappa$ is a successor cardinal if there exists a cardinal $\lambda$ such that, if $\alpha<\kappa$ then $|\alpha|\leq\lambda$.
• $\kappa$ is a weak limit cardinal if it is $\aleph_\alpha$ for a limit ordinal $\alpha$. Namely, it is the limit of a sequence of cardinals.
• $\kappa$ is a strong limit cardinal if it is a limit cardinals which have the property: if $\lambda<\kappa$, then $\beth_1(\lambda)<\kappa$.
• $\kappa$ is a weakly inaccessible cardinal if it is a limit cardinal, and it is regular. Namely $\kappa=\aleph_\kappa$, and $\kappa$ is not the limit of a sequence of $<\kappa$ cardinals.
• $\kappa$ is a strongly inaccessible cardinal if it is weakly inaccessible and a strong limit cardinal.
2. There is, albeit not too common, notation for $\beth_\alpha(\kappa)$. Where $\beth_\alpha$ denotes the $\alpha$-th iteration of power set of $\aleph_0$ (and limits at limit ordinals), using $\beth_\alpha(\kappa)$ we do the same and simply begin from $\kappa$.

If you use this notation, I suggest you define it as well ($\beth_0(\kappa)=\kappa$; $\beth_{\alpha+1}(\kappa)=2^{\beth_\alpha(\kappa)}$; and limits at limit ordinals).

This also tells you that when suggestion "repeated" operation we may refer to transfinitely many operations.

3. Yes, GCH will not affect the weak limit cardinals. Those are simply cardinals whose index is a limit ordinal. There will always be a proper class of these.

GCH, however, says a lot more than just equivalence between weak and strong limit cardinals. In fact GCH can fail for every successor cardinal, but every limit cardinal is a strong limit cardinal. This is an immediate consequence of Easton's theorem.

4. There are always a proper class of weak limit cardinals. Simply because every limit ordinal defines one, and there are class many of those.

How many strong limit cardinals is also a proper class, but it could be a proper subclass of the proper class of weak limits. Confused? Good. Let's clear this up.

Consider $\beth$ numbers, whenever $\kappa=\beth_\alpha$ for some limit ordinal $\alpha$, it means that if $\lambda<\kappa$ then $2^\lambda<\kappa$. This means that there is a proper class of strong limit cardinals. Always.

However this class can be rather "thin" compared to the weak limit cardinals. We may have that every limit cardinal is a strong limit cardinal (e.g. assuming GCH) or we may have that between every two strong limit cardinals there are infinitely many weak limit cardinals. Again, this is a consequence of Easton's theorem.

5. For sake of completeness, I will point out that $\aleph_0$ is a strong limit cardinal, however the existence of an uncountable weakly inaccessible cardinals cannot be proved from ZFC, let alone the existence of an uncountable strongly inaccessible cardinals.

However if there is a model of ZFC in which there exists an uncountable weakly inaccessible cardinal, then there is a model of ZFC in which there exists an uncountable strongly inaccessible one.

From this notion large cardinal axioms are often called "strong infinity axioms", the existence of $\aleph_0$ cannot be deduced without the axiom of infinity, much like the existence of an [uncountable] inaccessible cardinal cannot be deduced without an additional axiom.

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Does it make sense to think about the "order type" of the class of weak limit cardinals? I really am interested in the idea of somehow iterating through this class, i.e., starting at the least one and reaching them all by repeated application of some operation. – Rachel Aug 31 '12 at 21:17
@Rachel: The order type of any proper class of ordinals is $\mathbf{On}$. Always. You can always talk about $F(\alpha)$ being the $\alpha$-th limit ordinal, and then $\kappa_\alpha=\aleph_{F(\alpha)}$ being the $\alpha$-th limit cardinal. – Asaf Karagila Aug 31 '12 at 21:18
So among weak limit cardinals, there are limit weak limit cardinals that are analogous to limit ordinals in the order of On? – Rachel Aug 31 '12 at 21:51
@Rachel: Consider $\aleph_{\omega_1}$, the first cardinal which has uncountably many cardinals below it. This cardinal is a weak limit; it is also the limit of weak limit cardinals; it is also the limit of limits of weak limit cardinals; it is also the limit of limits of limits ... of weak limit cardinals; ... and with all that said, it is possible that $2^{\aleph_0}$ is even larger than $\aleph_{\omega_1}$ making it, and all those below it, weak limit cardinals which are not strong limits. – Asaf Karagila Aug 31 '12 at 21:55
Pardon the vagueness, but I am trying to get a general feel for these structures. So it's the limits that are responsible for the size of all of these classes (ordinals, cardinals, weak-limit cardinals, etc.)? That is, there are few objects between limits relative to the number of limits? You can keep taking limits (of limits of limits...) and always have a proper class? – Rachel Aug 31 '12 at 22:11