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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.

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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 $$ 2^\lambda=2^{\sum_{\alpha<\kappa}\lambda_\alpha}=\prod_{\alpha<\kappa}2^{\lambda_\alpha}=\lambda^\kappa. $$

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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 E. 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 E. 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 $\lambda^{\aleph_0}$.

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

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