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

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

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.

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    $\begingroup$ This is a teaching moment. What would you think an analogue for products should be? $\endgroup$ – Asaf Karagila May 5 at 8:50
  • $\begingroup$ from a purely aesthetic parallel to the sum one, sup^lambda. I know this is looked down upon here, however I would like to know the statement without the enlightenment part of knowing the formal proof $\endgroup$ – kyary May 5 at 9:00
  • $\begingroup$ Well, do you know how to prove the formula for summations? $\endgroup$ – Asaf Karagila May 5 at 9:08
  • $\begingroup$ yes, though it doesn't shed light on the product proof. I can easily show sup^lambda is an upper bound of the infinite product but there are two problems $\endgroup$ – kyary May 5 at 9:14
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    $\begingroup$ "I'm not trying to become the world's greatest set theorist." What an absurd thing to say! $\endgroup$ – Andrés E. Caicedo May 5 at 12:10
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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 $$ \prod_{i<\lambda}\kappa_i \ge \prod_{j<\lambda}\sup_{i<\lambda} \kappa_i = (\sup_{i<\lambda}\kappa_i)^\lambda$$

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    $\begingroup$ @kyary, just to be clear: The "non-decreasing sequence" condition may sound innocuous, but it's a significant limitation on the applicability of this formula. Given a family of cardinals $\{\kappa_i\mid i\in I\}$, it's a non-trivial thing to ask that they can be reindexed by the cardinal $\lambda = |I|$ in a non-decreasing way. For example, this implies that the family either has no maximum element, or that almost all of the elements of the family are equal to the maximum element $\kappa$, in the sense that $|\{i\in I\mid \kappa_i\neq \kappa\}| < |I|$. $\endgroup$ – Alex Kruckman May 5 at 22:49
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    $\begingroup$ Another consequence is that if some cardinal $\kappa$ occurs $|I|$-many times in the family, it must be the maximum element of the family (since it must be cofinal when the family is indexed by $\lambda = |I|$). $\endgroup$ – Alex Kruckman May 5 at 22:54
  • $\begingroup$ @AlexKruckman Yes, despite the notation (which in my defense, I copied from Jech :)), $i$ is far from being an arbitrary index here (whereas it very much is for the sum case). It is also easy to find counterexamples where $\lambda$ is a limit ordinal but not a cardinal. $\endgroup$ – spaceisdarkgreen May 5 at 23:04
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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.


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.

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    $\begingroup$ On the other hand, if the $\kappa_\alpha$ are increasing (and nonzero, of course), then the formula works. $\endgroup$ – spaceisdarkgreen May 5 at 17:40
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    $\begingroup$ @spaceisdarkgreen Maybe you should give a proof of this fact as an additional answer to the question? $\endgroup$ – Alex Kruckman May 5 at 17:52
  • $\begingroup$ Indeed, my counterexample with the $1$'s was silly; but I was too tired to make explicit the connection with cofinality - you did a great job pointing to it anyways ! $\endgroup$ – Max May 5 at 20:40
  • $\begingroup$ @Max Silly counterexamples are my favorite kind! $\endgroup$ – Alex Kruckman May 5 at 20:42
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    $\begingroup$ @DanielWainfleet Interesting. I'm not so familiar with Kunen's book, but in my experience, "König's Lemma" almost always refers to the result about trees, while "König's Theorem" refers to a variety of results, among them the inequality I quoted. Wikipedia agrees. But more importantly, so does the book by Hrabacek and Jech, which is what the OP is reading - see König's Theorem on p. 158 and König's Lemma on p. 227. $\endgroup$ – Alex Kruckman May 6 at 17:17

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