What is the analogue to this cardinal arithmetic theorem for infinite products?

Question

$\begin{array}{l}{\text { 1.3 Theorem Let } \lambda \text { be an infinite cardinal, let } \kappa_{\alpha}(\alpha<\lambda) \text { be nonzero cardinal}} \\ {\text {numbers, and let } \kappa=\sup \left\{\kappa_{\alpha} | \alpha<\lambda\right\} . \text { Then }}\end{array}$

$\sum_{\alpha<\lambda} \kappa_{\alpha}=\lambda \cdot \kappa=\lambda \cdot \sup \left\{\kappa_{\alpha} | \alpha<\lambda\right\}$

This is a theorem from Hrbacek + Jech. I know that there is an analogue to this theorem for infinite cardinal products, but it's not in this book.

I've been looking elsewhere and surprisingly, most online sources on cardinal arithmetic don't even mention infinite cardinal products at all.

If anyone could provide a statement to the analogue for infinite products, that would be greatly appreciated.

Answer 1

At the top of p. 158 of your book, the authors write the following:

Infinite products are more difficult to evaluate than infinite sums. In some special cases ... some simple rules can be proved.

What they're telling you is that you shouldn't expect a simple formula that applies to all infinite products like the one they give for infinite sums. And this is not so surprising: repeated cardinal addition is related to cardinal multiplication, which is a very simple operation. But repeated cardinal multiplication is related to cardinal exponentiation, which is extremely complicated.

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In the comments, you made the reasonable guess that $$\prod_{\alpha<\lambda} \kappa_\alpha = (\sup\{\kappa_\alpha\mid \alpha<\lambda\})^\lambda.$$

The formula for infinite sums assumes that all of the $\kappa_\alpha$ are non-zero. To have a hope of the above formula holding, we should assume that all of the $\kappa_\alpha$ are not equal to $0$ or $1$. This handles silly counterexamples like the one in Max's comment. And it's not so bad, because we can remove all $1$ terms from the product without changing its value, while if a single $0$ term appears, the whole product is $0$.

Ok, your guess is reasonable because we have an obvious upper bound: $$\prod_{\alpha<\lambda} \kappa_\alpha \leq (\sup\{\kappa_\alpha\mid \alpha<\lambda\})^\lambda.$$ And equality is achieved sometimes, e.g. as computed on the same page of your book: $$\prod_{n<\aleph_0} n = 2^{\aleph_0} = \aleph_0^{\aleph_0} = (\sup\{n\mid n<\aleph_0\})^{\aleph_0}.$$

On the other hand, it can fail, e.g. assume the continuum hypothesis (we actually only need $2^{\aleph_0} < \aleph_\omega$), let $\kappa_0 = \aleph_\omega$, and let $\kappa_n = 2$ for all $1\leq n<\aleph_0$:

$$\prod_{n<\aleph_0} \kappa_n = \aleph_\omega \cdot \prod_{1\leq n<\aleph_0} 2 = \aleph_\omega \cdot 2^{\aleph_0} = \aleph_\omega < (\aleph_\omega)^{\aleph_0} = (\sup\{\kappa_n\mid n<\aleph_0\})^{\aleph_0}.$$

The fact that $\aleph_\omega < (\aleph_\omega)^{\aleph_0}$ follows from König's Theorem: $$\aleph_\omega = \sum_{n< \aleph_0} \aleph_n < \prod_{n<\aleph_0} \aleph_\omega = (\aleph_\omega)^{\aleph_0}.$$

If you don't want to assume the continuum hypothesis, you can replace $\aleph_\omega$ with any cardinal $\kappa>2^{\aleph_0}$ such that $\text{cf}(\kappa) = \aleph_0$ (if $2^{\aleph_0} = \aleph_\alpha$, then $\kappa = \aleph_{\alpha+\omega}$ works), and the same argument goes through.

Answer 2

An analogue for products is given in Jech's Set Theory 3rd edition, Lemma 5.9:

If $\lambda$ is an infinite cardinal and $\langle \kappa_i: i <\lambda\rangle $ is nondecreasing sequence of nonzero cardinals, then $$ \prod_{i<\lambda}\kappa_i =(\sup_{i<\lambda}\kappa_i)^\lambda.$$

Note that the nondecreasing requirement makes this consistent with the counterexample given by Alex Kruckman.

As you mention, it is clear that the RHS is an upper bound. For the other direction, we can partition $\lambda$ into $\lambda$-many subsets that are cofinal in $\lambda.$ Write $\lambda = \bigcup_{j<\lambda}A_j.$ Then we have $$ \prod_{i<\lambda}\kappa_i = \prod_{j<\lambda}\prod_{i\in A_j}\kappa_i.$$ Since the $A_j$ are cofinal, $$ \prod_{i\in A_j}\kappa_i \ge \sup_{i\in A_j}\kappa_i =\sup_{i<\lambda}\kappa_i,$$ so $$ \prod_{i<\lambda}\kappa_i \ge \prod_{j<\lambda}\sup_{i<\lambda} \kappa_i = (\sup_{i<\lambda}\kappa_i)^\lambda$$