Is multiplication of infinite cardinals defined in ZF without Choice?

  • 2
    $\begingroup$ This may sound silly, but what exactly do you mean by multiplication? Multiplication of (real/natural) numbers? Multiplication of cardinals? $\endgroup$
    – Srivatsan
    Jan 25, 2012 at 17:07
  • $\begingroup$ I'm sorry. I meant multiplication of infinite cardinals. $\endgroup$ Jan 25, 2012 at 17:20

2 Answers 2


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.

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.
  • $\begingroup$ So, without choice it is possible that A*0=C where C is nonzero? $\endgroup$ Jan 25, 2012 at 17:16
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    $\begingroup$ No. It is possible that there is an infinite family of sets and $\prod X_i=\varnothing$, so the cardinality $|\prod X_i|\neq\prod|X_i|$. $\endgroup$
    – Asaf Karagila
    Jan 25, 2012 at 17:21
  • $\begingroup$ Which cardinals would be "incomparable"? $\endgroup$ Jan 25, 2012 at 17:30
  • $\begingroup$ @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 $\alpha$. $\endgroup$
    – Asaf Karagila
    Jan 25, 2012 at 17:33
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Jan 25, 2012 at 19:15


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.

  • $\begingroup$ 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) $\endgroup$ Jan 25, 2012 at 17:13

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