Is multiplication of infinite cardinals defined in ZF without Choice?

2This may sound silly, but what exactly do you mean by multiplication? Multiplication of (real/natural) numbers? Multiplication of cardinals? – Srivatsan Jan 25 '12 at 17:07

I'm sorry. I meant multiplication of infinite cardinals. – mathNotebook Jan 25 '12 at 17:20
2 Answers
I assume that you mean multiplication of cardinals.
Finite multiplication is always defined, since it is just Cartesian product which is defined regardless to the axiom of choice.
So if $A$ and $B$ are sets the set $A\times B$ exists, and $A\cdot B=A\times B$. This is indeed the definition.
Furthermore, if $B=\varnothing$, then $A\times B=\varnothing$, why? Note that $\langle a,b\rangle\in A\times B$ if and only if $a\in A$ and $b\in B$. In the case where one of the sets is empty the product is empty. This means that regardles to the axiom of choice $A\cdot 0 = 0$.
What the axiom of choice does tell us that we can define infinite products, that is $I$ is some infinite index set, if for all $i\in I$ we have that $X_i\neq\varnothing$ then the axiom of choice assures that $\prod_{i\in I}X_i$ is not empty.
Without the axiom of choice there are such families whose product is empty, and other families of sets whose product is nonempty. However we always have to require that $X_i\neq\varnothing$ for all $i\in I$ since otherwise the result is simply $\varnothing$.
It is also worth noting that without the axiom of choice $\aleph_\alpha\cdot\aleph_\beta = \aleph_{\max\{\alpha,\beta\}} = \max\{\aleph_\alpha,\aleph_\beta\}$. When assuming the axiom of choice every infinite set has the cardinality of $\aleph_\alpha$, so $A\cdotB=\max\{A,B\}$. Without the axiom of choice there are cardinals which are not wellordered and are not equal to any $\aleph$. In particular there are cardinals which are incomparable, so $\max\{\frak p,q\}$ becomes undefined.
Added: (after the comment)
Assuming the axiom of choice, if $\lambda_i$ is an infinite cardinal, the $\prod\lambda_i =I\cdot\sup\{\lambda_i\mid i\in I\}$. This means that if $X_i$ is a set whose cardinality is $\lambda_i$ we have:
$$\left\prod X_i\right = \prodX_i = \prod\lambda_i = I\cdot\sup\{\lambda_i\mid i\in I\}$$
Without the axiom of choice this equality does not hold for every family. Therefore it is not well defined. We can have a model in which there is a countable family of pairs without a choice function. So we have the following situation:
$$0=\varnothing=\left\prod_{n\in\omega} P_n\right\neq\prod_{n\in\omega} P_n = \left\prod_{n\in\omega}\{2n,2n+1\}\right = 2^{\aleph_0}$$
Showing that an infinite product of cardinals is not welldefined, since cardinality is an equivalence relation and operation on cardinals should not be dependent on choice of representatives. In fact, in certain situation there is no possible choice of representatives. So all in all we cannot say that the operation of infinite products (or sums) is welldefined without the axiom of choice.
Further reading on this site:
 There's nonAleph transfinite cardinals without the axiom of choice?
 About a paper of Zermelo
 For every infinite $S$, $S=S\times S$ implies the Axiom of choice
 Defining cardinality in the absence of choice
 Somewhat similar question: Multiplying Infinite Cardinals (by Zero Specifically)
 Also the comments here: The Axiom of Choice and the Cartesian Product.
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1No. It is possible that there is an *infinite family of sets* and $\prod X_i=\varnothing$, so the cardinality $\prod X_i\neq\prodX_i$. – Asaf Karagila Jan 25 '12 at 17:21


@user23648: If $A$ is a set that cannot be well ordered, we can have $\alpha$ the minimal ordinal for which there is no injection into $A$ (remember that finite subsets of $A$ can be well ordered, and possible *some* infinite subsets). It appears in several of the links I've added to my answer that this $\alpha$ is a cardinal (i.e. an ordinal which has no bijection with previous ordinals) and $A$ is incomparable with $\alpha$. – Asaf Karagila Jan 25 '12 at 17:33

Thanks for these excellent answers. I've read them, but I've also bookmarked them so I can study this at a later date after I've learned more. This has been a valuable resource. I now realize that there is much more to this subject than I had anticipated. – mathNotebook Jan 25 '12 at 17:47

@user23648: Tell me about it... :) If you have further questions, I will be glad to answer. – Asaf Karagila Jan 25 '12 at 17:49

@Asaf: I have concerns about your statement that an infinite product is *undefined*. The argument in your answer simply shows that you can't compute the product as $I \sup \lambda_i$, not that the cardinal $\prod X_i$ depends on how you choose the $X_i$ to represent the $\lambda_i$'s. – Jan 25 '12 at 18:28

1@Hurkyl: First, there might not be a supremum for $\lambda_i$. In fact it is perfectly possible that we cannot even *choose representatives* to calculate such supremum (which obviously depends on the choice of representatives)! Secondly, the cardinality of the product of the sets is indeed defined, but not as a product of cardinalities. Cardinals are equivalence classes, and the product of them is defined on the equivalence classes. If the operation depends on the choice of representatives (and if we sometimes cannot even choose those) then it is simply undefined as a whole. – Asaf Karagila Jan 25 '12 at 19:15

@Asaf: Hi Asaf. Do you happen to know if the identity $ \omega_{1} \times \mathbb{R} = \mathbb{R} $ requires the Axiom of Choice or some weaker version thereof? – Berrick Caleb Fillmore Oct 10 '17 at 09:38

@BerrickCalebFillmore: Yes. It requires some choice. It is consistent (e.g. in Solovay's model, but also in other models without needing large cardinals) that $\omega_1\times\Bbb R>\omega_1\cup\Bbb R>\Bbb R$. – Asaf Karagila Oct 10 '17 at 09:39

@Asaf: Thanks! I thought that I could prove this without $ \mathsf{AC} $ by applying the SchröderBernstein Theorem (which doesn’t require $ \mathsf{AC} $) to the obvious injection from $ \mathbb{R} $ into $ \omega_{1} \times \mathbb{R} $ and some injection from $ \omega_{1} \times \mathbb{R} $ into $ \mathbb{R} \times \mathbb{R} $ (which we know to be in bijection with $ \mathbb{R} $ even in the absence of $ \mathsf{AC} $). However, I guess that the problem with this approach is that injecting $ \omega_{1} $ into $ \mathbb{R} $ requires some form of $ \mathsf{AC} $. – Berrick Caleb Fillmore Oct 10 '17 at 10:28

Yes.
Multiplication of finite (natural, real, complex, etc.) numbers or of ordinals does not depend on choice.
Cardinal multiplication can be given its usual definition $A\timesB=A\times B$, assuming that you have a suitable definition of what a cardinal is  the usual definition of cardinals as distinguished ordinals doesn't work without choice. But rules such as $A\timesB=\max(A,B)$ for infinite $A$ and $B$ may fail to hold without choice.
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I was concerned about the infinite cardinals. It says on Wikipedia that the result A*B=Max{A,B} assumes choice. (See the entry on Cardinal Numbers under Cardinal Multiplication) – mathNotebook Jan 25 '12 at 17:13