# Cardinal exponentiation without generalized continuum hypothesis

First I have to confess that I don't know about set theory language.

Let $$A$$ and $$B$$ be infinite cardinals with $$A>B$$.

My question is: $$A^B=A$$? (without assuming generalized continuum hypothesis)

Remark: assuming generalized continuum hypothesis (GCH briefly), this can be proved by the following (at least for unlimit cardinal).

Sps $$A$$ is a unlimit cardinal. Then $$A=2^C$$ for some $$C\ge B$$ by GCH. Therefore $$A^B = (2^C)^B=2^{CB}=2^C=A$$.

Unfortunately, I don't know how to prove for limit cardinal case. Please somebody help me!

This is a great question! It's totally reasonable to expect - assuming GCH - that $$A^B=A$$ when the base $$A$$ is larger than the exponent $$B$$ since that's true in all the "simply-imaginable" situations. However, that's not the whole picture. As you've noticed, limit cardinals pose an odd difficulty, and it turns out that a particular kind of limit cardinals break the pattern entirely - even if GCH holds.

## Some weirdness

Let me begin with a counterexample to your reasonable intuition, which works regardless of whether GCH holds, to motivate what follows:

$$(\aleph_{\omega})^{\aleph_0}>\aleph_\omega.$$

(Recall that $$\aleph_\omega$$ is the limit of the $$\aleph_n$$s ($$n\in\mathbb{N}$$). Even with GCH it's a bit of a weird object, in contrast with say $$\aleph_2$$ which is just the cardinality of the set of real functions under GCH.)

The fact above may look mysterious, but its proof is actually just a direct diagonalization argument.

First, let's replace $$(\aleph_{\omega})^{\aleph_0}$$ with something more meaningful. Specifically, it's not hard to show that $$(\aleph_\omega)^{\aleph_0}$$ is the cardinality of the set $$Seq$$ of increasing $$\omega$$-sequences of ordinals less than $$\aleph_\omega$$.

Now let's set up our diagonalization. No need to use proof by contradiction - let's be constructive! Suppose $$F:\aleph_\omega\rightarrow Seq$$; I want to produce an $$\omega$$-sequence $$S$$ of ordinals $$<\aleph_\omega$$ which is not in the range of $$F$$.

To do this, the trick is to "chop $$\aleph_\omega$$ into $$\omega$$-many blocks" (namely, "up to $$\aleph_0$$," "from $$\aleph_0$$ to $$\aleph_1$$," ..., "from $$\aleph_n$$ to $$\aleph_{n+1}$$," ...) - even though the blocks together cover all of $$\aleph_\omega$$, each individual block is "small" (= of size $$<\aleph_\omega$$).

Now just let the $$i$$th entry of our "antidiagonal sequence" $$S$$ be the smallest ordinal which isn't any of the first $$i$$ entries of any of the sequences $$F(\kappa)$$ for $$\kappa<\aleph_i$$. So, for example, to find $$S(2)$$ we look at the first $$\aleph_2$$-many (according to $$F$$) elements of $$Seq$$, and check all of the ordinals that occur as either the first or second terms of any of those; there are only $$\aleph_2$$-many of these, so there is some ordinal which doesn't appear in the first two terms of $$F(\kappa)$$ for any $$\kappa<\aleph_2$$, and the smallest of these is the ordinal we pick to be $$S(2)$$.

It's easy to check that the sequence $$S$$ so built is an element of $$Seq$$ not in the range of $$F$$, so we're done!

This is really weird. What makes $$\aleph_\omega$$ so different from, say, $$\aleph_{17}$$?

## Cofinality

The distinction between limit and successor (= non-limit) cardinals isn't all there is. The limit cardinals themselves split further into two types - the regular and singular limit cardinals - and it is the singular limit cardinals that often cause all the trouble.

Incidentally, it is consistent with ZFC+GCH that there are no regular limit cardinals at all - however, there are guaranteed to be lots of singular limit cardinals.

Intuitively, a limit cardinal $$\kappa$$ is singular if we can "count up to it" in fewer than $$\kappa$$-many steps. For example, the sequence $$\aleph_1,\aleph_2,\aleph_3,...$$ lets us count up to the cardinal $$\aleph_\omega$$ in $$\omega$$-many steps; since $$\aleph_\omega$$ is much bigger than $$\omega$$, this means that $$\aleph_\omega$$ is singular.

This is exactly the "block-chopping-into" thing we did above, but phrased a bit more abstractly.

By contrast, it's not hard to show that every successor (= non-limit) cardinal is regular (= non-singular): if $$(\alpha_\eta)_{\eta<\delta}$$ is an increasing sequence of ordinals with limit $$\beta=\gamma^+$$, then $$\beta$$ is the union of $$\delta$$-many sets of size $$\le\gamma$$, so $$\beta=\delta\times\gamma$$ and since $$\gamma<\beta$$ this means $$\delta=\beta$$.

The number of steps you need to count up to a given cardinal is called its cofinality, and the cofinality of $$\kappa$$ is denoted $$cf(\kappa)$$.

## Exponentiation

So what does this have to do with exponentiation?

Well, looking back at the proof that $$(\aleph_\omega)^{\aleph_0}>\aleph_\omega$$, the key point was that we were able chop the "base" (= $$\aleph_\omega$$) into "exponent-many" (= $$\aleph_0$$) small blocks; that is, the cofinality of the base was no larger than the exponent. Indeed, this turns out to be a fundamental issue - if the exponent $$\lambda$$ is large relative to the cofinality of the base $$\kappa$$ (not just the base itself!), we get $$\kappa^\lambda>\kappa$$ (a bit more snappily, we have $$\kappa^{cf(\kappa)}>\kappa$$ for all $$\kappa$$).

## Coda

Let me end by mentioning three points around this topic:

• The fact that $$\kappa^{cf(\kappa)}>\kappa$$ is a consequence of the more general Konig's theorem. If you want to get a handle on basic cardinal arithmetic, you should play around with this theorem until you're comfortable with it.

• Interestingly, in a precise sense Konig's theorem is basically the only nontrivial fact about cardinal exponentiation which ZFC can outright prove - this is a consequence of Easton's theorem. This is a very technical result, but I mention it only because knowing that something like it exists gives some additional "punch" to Konig's theorem.

• Easton's theorem (and its method of proof in general) suggests a rather bleak picture for ZFC: that basically any nontrivial question about cardinal arithmetic can't be decided from the ZFC axioms alone. This turns out to be false, and the ZFC-only investigation of cardinal arithmetic was pioneered by Shelah - I think this paper of his is a good, if quite hard, survey of the situation. I won't try to describe it here, but I'll mention one of his flashier results: that if $$\aleph_\omega$$ is a "strong limit cardinal" (that is, $$2^{\aleph_n}<\aleph_\omega$$ for all finite $$n$$ - this is implied by, but much weaker than, GCH up to $$\aleph_\omega$$), then $$2^{\aleph_\omega}<\aleph_{\omega_4}.$$ Incidentally, Shelah is on record as asking "Why the hell is it $$4$$?" (page $$4$$ of the above-linked article).

• Thank you for your detailed answer!!! I couldn't fully understand what you wrote because I have no basis about set theory, but I 'll read it carefully again. Have a nice day~ – MiRi_NaE Jan 18 at 0:20