A question about cardinal arithmetics without the Axiom of Choice

Question

Is multiplication of infinite cardinals defined in ZF without Choice?

Answer 1

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 non-empty. 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 $|A|\cdot|B|=\max\{|A|,|B|\}$. Without the axiom of choice there are cardinals which are not well-ordered 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:

$$\left|\prod X_i\right| = \prod|X_i| = \prod\lambda_i = |I|\cdot\sup\{\lambda_i\mid i\in I\}$$

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 well-defined, 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 well-defined without the axiom of choice.

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Further reading on this site:

1. There's non-Aleph transfinite cardinals without the axiom of choice?
2. About a paper of Zermelo
3. For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
4. Defining cardinality in the absence of choice
5. Somewhat similar question: Multiplying Infinite Cardinals (by Zero Specifically)
6. Also the comments here: The Axiom of Choice and the Cartesian Product.

Answer 2

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 $|A|\times|B|=|A\times B|$, 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 $|A|\times|B|=\max(|A|,|B|)$ for infinite $A$ and $B$ may fail to hold without choice.