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!


1 Answer 1


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:


(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, $\aleph_{17}$?

The answer is:


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 $cf(\kappa)$.


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 $\kappa$).


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).

  • 1
    $\begingroup$ 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~ $\endgroup$
    – MiRi_NaE
    Jan 18, 2019 at 0:20

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