Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $\lambda$ is a strong limit cardinal, i.e. $\forall \alpha<\lambda \ 2^\alpha<\lambda$, and the cofinality of $\lambda$: $cf(\lambda)=\omega$. How do we show that $2^\lambda \leq \lambda^{\aleph_0}$?

This occurs during my reading of Kunen's Theorem about the nonexistence of nontrivial elementary embedding of $V$ into itself. I tried to reduce subsets of $\lambda$ 1-1 to functions from $\omega$ to $\lambda$ and of course by the property of strong limit cardinals, only unbounded subsets of $\lambda$ need to be considered. Then I am stuck. Thanks in advance.

share|cite|improve this question

2 Answers 2

up vote 3 down vote accepted

There is a nice argument for this:

Let $\lambda$ be a limit cardinal of cofinality $\kappa$. Write $\kappa$ as the disjoint union $\bigcup_{\alpha<\kappa}A_\alpha$ of $\kappa$ disjoint sets, each of size $\kappa$, using that $\kappa\times\kappa$ has the same size as $\kappa$. Note that each $A_\alpha$ is cofinal in $\kappa$.

Fix a sequence $\lambda_\alpha$, $\alpha<\kappa$, strictly increasing, cofinal in $\lambda$, and consisting of nonzero cardinals.

Now, given any set $A$ cofinal in $\kappa$, consider $\prod_{\alpha\in A}\lambda_\alpha$. First, this product is clearly at most $\prod_\alpha\lambda=\lambda^\kappa$. On the other hand, the product is at least $\lambda_\alpha$ for any $\alpha\in A$, and therefore it is at least $\lambda$.

It follows that $\lambda^\kappa\ge\prod_{\alpha<\kappa}\lambda_\alpha=\prod_{\alpha<\kappa}\prod_{\beta\in A_\alpha}\lambda_\beta\ge\prod_{\alpha<\kappa}\lambda=\lambda^\kappa$, and we have equality.

Note we have actually shown that if we have any strictly increasing $\kappa$-sequence of nonzero cardinals cofinal in $\lambda$, then their product is $\lambda^\kappa$.

OK. Now, suppose that $\lambda$ is in addition strong limit, and note that this gives us that $$ 2^\lambda=2^{\sum_{\alpha<\kappa}\lambda_\alpha}=\prod_{\alpha<\kappa}2^{\lambda_\alpha}=\lambda^\kappa. $$

share|cite|improve this answer
Thanks! The trick really is the bijective correspondence between $\kappa$ and $\kappa\times \kappa$. – Jing Zhang Mar 24 '14 at 15:02
Yes. It feels like a trick the first time one sees it, but I tried not to call it that way, as really all ideas in this argument reappear frequently through cardinal arithmetic. – Andrés Caicedo Mar 24 '14 at 15:04
If we only consider strong limit cardinal, would not the last line of the proof above suffices: $2^\lambda\leq 2^{\Sigma_{\alpha<\kappa}\lambda_\alpha}=\Pi_{\alpha<\kappa}2^{\lambda_\alpha} \leq \Pi_{\alpha<\kappa} \lambda = \lambda^\kappa$? – Jing Zhang Mar 24 '14 at 15:14
Yes. My point in splitting the argument as I did was to show that it is really a different result than the pursued one that makes everything work, and that the strong limit assumption is almost incidental. A minimal proof is as you suggest. – Andrés Caicedo Mar 24 '14 at 15:24

This is similar to Andres Caicedo's argument but it might be a bit easier. Since $\lambda$ has cofinality $\omega$, it is the supremum of an $\omega$-sequence of smaller caridnals, say $(\kappa_n)$. Notice that any subset $X$ of $\lambda$ is completely determined by the $\omega$-sequence of intersections $(X\cap\kappa_n)$. Indeed, if $Y$ is a different subset of $\lambda$, then there is some $\alpha<\lambda$ that belongs to one of $X$ and $Y$ but not to the other. Since the $\kappa_n$ are cofinal in $\lambda$, $\alpha$ is in some $\kappa_n$, and then $\alpha$ witnesses that $X\cap\kappa_n\neq Y\cap\kappa_n$. So we have a one-to-one function $X\mapsto(X\cap\kappa_n)_{n\in\omega}$ from the power set of $\lambda$, which has cardinality $2^\lambda$, into the product of the power sets of the $\kappa_n$'s. Since $\lambda$ is a strong limit, the power set of each $\kappa_n$ has size at most $\lambda$ (in fact, strictly smaller than $\lambda$, but I don't need that), so the product of these countably many power sets has cardinality at most $\lambda^{\aleph_0}$.

share|cite|improve this answer
(And this is how the argument is presented in Kunen's article on combinatorics, in the Handbook of mathematical logic.) – Andrés Caicedo Mar 24 '14 at 15:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.