# Cardinality of infinite product?

Is the cardinality of an infinite product easily expressed in simpler terms?

The specific problem is this. Suppose $\alpha$ is a countable limit ordinal. The set I want to count has size equal to $\Pi_{\beta<\alpha} \beth_\beta$.

I would like the output to be $\beth_\alpha$ but honestly I'm not sure how this works out. Finite cardinal multiplication is very easy, but I don't know anything about infinite products like this. Certainly I can bound it above; $\beth_\beta<\beth_\alpha$, so the product has size at most $\beth_\alpha^\omega$, but by Konig's theorem this is strictly larger than $\beth_\alpha$ and doesn't sit well with me.

• Nothing much can be said about infinite product of cardinals. Sure we can say this and that. But something concrete? Not very often. – Asaf Karagila Sep 10 '15 at 19:20
• So in this case, is it likely to say "it's at most $\beth_\alpha^\omega$" and leave it there? – Richard Rast Sep 10 '15 at 20:23
• I have to disagree with Asaf, though I figure it is really that we interpret the quantifier "nothing much" in different ways. For a general result in particular containing the computation you are asking, see theorem 13 in this blog post (which refers to this one). – Andrés E. Caicedo Sep 23 '15 at 15:25

It turns out this does have a nice answer. I originally had a much more complicated proof, and almost immediately afterward, a simple one was given to be by Douglas Ulrich. The answer is that for any limit ordinal $\alpha$, $\prod_{\beta<\alpha} \beth_\beta=\beth_{\alpha+1}$. What follows is a proof in more detail than is maybe necessary.
The essential reason is the $\prod_i 2^{\kappa_i}=2^{\sum_i \kappa_i}$, which is easily verified. This gives us the following:
$\prod_{\beta<\alpha}\beth_\beta=\prod_{\beta<\alpha}2^{\beth_\beta}$; this is seen by seeing both sides inject naturally into the other. Left to right is that $\beth_\beta\leq 2^{\beth_\beta}$, while right to left is because this is a product of a smaller set (since $\alpha$ is a limit ordinal).
Then $\prod_{\beta<\alpha}2^{\beth_\beta}=2^{\sum_\beta\beth_\beta}$. Clearly $\sum_\beta\beth_\beta=\beth_\alpha$, so this last is $2^{\beth_\alpha}=\beth_{\alpha+1}$, as desired.