Beth cardinals and inacceesible cardinals

Question

Since my last question here about the Alephs was too imprecise and thus went over like a lead balloon, I am trying a new and simpler question which asks what I probably should have asked before.

Under "Beth Numbers" in Wikipedia I read:

"In ZF, for any cardinals $\kappa$ and $\mu$, there is an ordinal $\alpha$ such that:

$\kappa \leq \beth_\alpha(\mu)$."

But under "Inaccessible Cardinals" I read:

"a cardinal $\kappa$ is strongly inaccessible if it is uncountable, it is not a sum of fewer than $\kappa$ cardinals that are less than $\kappa$, and $\alpha < \kappa$ implies $2^\alpha < \kappa$."

These two passages are troubling to me since they seem to be contradictory. The first one seems to imply that for ANY cardinal one can always find a Beth number which exceeds it. While the second one clearly seems to imply that the first inaccesible and any cardinal larger than it, of which there are uncountably many, of course, are all vastly larger than any Beth cardinal generated by even $\omega$ applications of the Power Set operation could ever be.

I assume that I am simply missing something important here, and that both statements from the Wikipedia are actually true. But what exactly am I missing??

Answer 1

It seems like the issue you're having is with understanding what $\beth_\alpha$ means when $\alpha \geq \omega$. The definition is by recursion: $\beth_0 = \aleph_0$, for any ordinal $\alpha$ we define $\beth_{\alpha+1} = 2^{\beth_\alpha}$, and when $\alpha$ is a limit ordinal we have $\beth_\alpha = \sup\{\beth_\beta : \beta < \alpha\}$. This allows us to continue the powerset operation transfinitely.

Notice that, for every $\alpha$, $\aleph_\alpha \leq \beth_\alpha$. Thus, if $\kappa = \aleph_\alpha$, then $\beth_\alpha \geq \aleph_\alpha = \kappa$, so it is indeed true that there is a $\beth$ number larger than $\kappa$ (if you want strictly larger, go for $\beth_{\alpha+1}$).

The reason the previous paragraph doesn't contradict the definition of $\kappa$ being inaccessible is that if $\kappa$ is inaccessible then the $\alpha$ for which $\kappa = \aleph_\alpha$ is $\kappa$ itself, i.e., $\kappa = \aleph_\kappa$. Thus, unlike in the definition of inaccessibility, we're not in the position of calculating $2^{\alpha}$ with $\alpha < \kappa$.

Answer 2

Your comments indicate that you are dubious about iterating operations on ordinals more than finitely many times, and very dubious about iterating them uncountably many times. This is a fundamental point in set theory. The danger is in thinking of recursive definitions as processes which need to be carried out, in which case our "finitary biases" get in the way. Instead, you should think of a recursive definition as "happening all at once." Essentially, we can show that every recursive description corresponds to a unique function, and what lets us do this is transfinite induction. (It shouldn't be surprising that "we can do recursion for as long as we can do induction.")

Specifically, suppose that $F:Ord\rightarrow Ord$ is a function on ordinals (or rather, class function; for simplicity I'm assuming we're working in a theory like NBG that makes all this much simpler to say). For $\theta>0$ an ordinal, say that a function $G$ iterates $F$ along $\theta$ starting at $\alpha$ iff

• The domain of $G$ is $\theta$,

• $G(0)=\alpha$,

• for $\beta+1<\theta$ we have $G(\beta+1)=F(G(\beta))$, and

• for $\lambda<\theta$ a limit we have $G(\lambda)=\sup\{G(\beta): \beta<\lambda\}$.

Incidentally, this last condition is really only a natural thing to do if $F$ is nondecreasing, but strictly speaking this works for any $F$.

In principle, there could be many $G$ with this property, or none at all. However, it turns out that there is only ever exactly one:

For every $F:Ord\rightarrow Ord$ (= the function to be iterated), $\theta>0$ (= the iteration length), and $\alpha$ (= the starting value), there is exactly one $G$ which iterates $F$ along $\theta$ starting at $\alpha$.

Moreover, the $G$s "cohere" in the sense that if $G$ iterates $F$ along $\theta$ starting at $\alpha$ and $G'$ iterates $F$ along $\theta'$ starting at $\alpha$, with $\theta<\theta'$, then for each $\eta<\theta$ we have $G(\eta)=G'(\eta)$. So in some sense there is a unique way to iterate $F$ along $Ord$.

The proof is by transfinite induction: fixing an arbitrary $F$ and $\alpha$, consider some $\theta$ such that the claim holds for all iteration lengths $<\theta$. Intuitively, if $\theta=\gamma+1$ we just take the $G$ for $\gamma$ and "stick one more value onto it," and if $\theta$ is a limit we "glue the earlier $G$s together." It's a good exercise to turn this vague hint into an actual proof.

The sequence of $\beth$ numbers can be constructed in this way:

• $F$ is the map sending an ordinal $\alpha$ to the cardinality of the powerset of $\alpha$ (which, remember, is itself an ordinal - cardinals are just initial ordinals).

• The starting value $\alpha$ is $\omega$: this amounts to setting $\beth_0=\omega$.

• To determine what $\beth_\eta$ should be, we set $\theta=\eta+1$ - or really we pick any $\theta>\eta$, by the "coherence" point above it doesn't affect the answer.

Answer 3

Say that a natural number $n$ is "large" if for any number $k$ you'd think of in the next 24 hours, $n>10^k$. Let's go bigger, say $n>k\uparrow^k k$, using the Knuth notation.

Pretty much by definition, $n$ is unimaginably large. So large that for the next 24 hours there is no way you can even imagine it. But tomorrow evening, you'd be sitting with a beer and realize that you can imagine an even larger number. Why is that even possible? Because even though this $n$ is large, almost all the other natural numbers are larger.

Inaccessible cardinals are ordinals. They are incredibly large, yes. But at the end of the day, most ordinals are in fact larger. More cardinals are larger.

If $\kappa$ is inaccessible and $\mu<\kappa$, then we can prove that the smallest $\alpha$ for which $\kappa\leq\beth_\alpha(\mu)$ is in fact $\kappa$ itself. Namely, $\kappa\leq\beth_\kappa(\mu)$, and they are in fact equal, and if $\alpha<\kappa$, then $\beth_\alpha(\mu)<\kappa$ as well. But since $\kappa$ is an ordinal, taking $\alpha=\kappa$ is perfectly valid.

So there is no contradiction there. You can move to $\aleph$ fixed points, whose existence is provable in $\sf ZFC$, and replace the $\beth_\alpha(\mu)$ by $\mu^{+\alpha}$, the $\alpha$th successor of $\mu$. If $\kappa=\aleph_\kappa$, and $\mu,\alpha<\kappa$, then $\mu^{+\alpha}<\kappa$. Nevertheless, there is some $\beta$ such that $\mu^{+\beta}\geq\kappa$, in fact a proper class of those $\beta$s. And specifically, $\kappa$ itself works for that.