Let me quickly explain what I mean with my question.

Let $(\kappa_i)_{i\in I}$ be a collection of cardinal numbers, indexed by elements of some set $I$. We can try to define $\sum\limits_{i\in I}\kappa_i$ as follows: take collection $(A_i)_{i\in I}$ of sets such that $|A_i|=\kappa_i$, and let $A=\{(a,i):i\in I,a\in A_i\}$. Then we say $\sum\limits_{i\in I}\kappa_i=|A|$. We can do similar trick with multiplication, by setting $B=\{f:f\text{ is a function from }I\text{ such that }f(i)\in A_i\}$.

If we assume AC, then these notions are well-defined: First of all, we can choose such a sequence $(A_i)_{i\in I}$, and if we take two collections $(A_i)_{i\in I}$, $(A_i')_{i\in I}$ such that $|A_i|=|A_i'|=\kappa_i$, then we can use choice to get bijections between $A_i$ and $A_i'$ for every $i\in I$ and then use there to create bijection between $A=\{(a,i):i\in I,a\in A_i\}$ and $A'=\{(a',i):i\in I,a'\in A_i'\}$, so this procedure defines only one cardinal. Similar trick for multiplication.

However, if we do not assume choice, then 1. we don't know a priori that we can always choose any sequence $(A_i)_{i\in I}$, and even if we can, we have no guarantee that there is only one possible size of $A$ we can get in this way. I don't know of scenario where we cannot find any sequence of sets with given cardinalities, but I know that uniqueness of size can fail quite spectacularly, even if we add $2$ countably many times (see here).

I have three questions:

  1. Is it possible in abscence of choice that there is a collection of cardinals $(\kappa_i)_{i\in I}$, but there is no collection of sets $(A_i)_{i\in I}$ having respective cardinalities? (I suspect the answer is "yes")
  2. If the answer to 1. is yes, then is it known that existence of such a collection implies axiom of choice? (I suspect answer to this to be "doesn't imply" or "not known")
  3. Does existence of collections like above and "value of the sum and product of cardinals doesn't depend on choice of collection" imply axiom of choice? (no clue, but I'm hoping for "yes")

These are main questions I'm interested in, but we can ask more by asking which combinations of below three imply choice:

For any collection of cardinals, there is collection of sets of respective cardinalities. Sum of cardinals, if defined, is well-defined. (i.e. sum doesn't depend on choice of collection, but we allow possibility that the collection doesn't exist) Product of cardinals, if defined, is well-defined.

Thanks in advance for all feedback.


1 Answer 1


First let me answer the first question. Yes. It is possible. You can find the proof in Jech "The Axiom of Choice" in Theorems 11.2 and 11.3. Also you might be interested in reading Pincus' paper,

David Pincus, Cardinal representatives, Israel J. Math. 18 (1974), 321--344.

Where he shows that it is also consistent that there are representatives to all cardinals uniformly, but the axiom of choice fails.

To the third question, which seems to mix two separate ideas, I don't know the answer, and I don't know if it was asked in explicit details. But even without any assumptions on cardinal representatives, just talking about disjoint unions (and products), let me give you the following answer.

If you allow talking about products, then trivially you get the axiom of choice back.

Suppose that $\{Y_i\mid i\in I\}$ is a family of non-empty sets. Pick a large enough non-empty set $Y$ such that each $Y_i$ satisfies $|Y\times Y_i|=|Y|$.1 Now consider $\prod_{i\in I} Y$ and $\prod_{i\in I}(Y\times Y_i)$, since the former is non-empty, the latter is non-empty as well. Now pick some $f$ such that $f(i)\in Y\times Y_i$ and consider the projection onto $Y_i$, as a choice function.

For addition it seems to be open. The last progress I am aware of can be found in Higasikawa's paper,

Masasi Higasikawa, Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434.

Where you can find a proof that this assumption implies The Partition Principle (whose choice strength is quite open for quite some time now). You can find other proofs related to this principle, and that in the presence of some additional assumptions it will imply the axiom of choice.

Two remarks that might be useful, taken from Gregory Moore's "Zermelo's Axiom of Choice":

  1. The assumption is in fact equivalent to the statement that $\sum_{i\in I}|A|=|A|\cdot|I|$.

  2. If we also assume that "If $Y$ is a set and for every $y\in Y$, $|y|=|Y|$, then $|\bigcup Y|=|Y|$, then we can prove the axiom of choice.


  1. Such $Y$ always exists. For example $(\bigcup Y_i)^\omega$.
  • $\begingroup$ Thank you very much, I'll look into these references later. I wouldn't say that products "trivially" imply choice, since existence of such $Y$ is something that I honestly probably couldn't have figured out :P $\endgroup$
    – Wojowu
    Aug 2, 2015 at 12:32
  • 2
    $\begingroup$ You're welcome. The existence of such set is actually a trick I collected in my very first course in [axiomatic] set theory. If you want $A\times 2\approx A$, just replace $A$ by $(A\times\omega)$, and if you want $A\times A\approx A$, just replace it by $A^\omega$. This also allows you, often when needed, to assume that you are working with Dedekind-infinite sets. $\endgroup$
    – Asaf Karagila
    Aug 2, 2015 at 12:44
  • $\begingroup$ "non-empty either" $\: \mapsto \:$ "also non-empty" $\;\;\;\;$ $\endgroup$
    – user5810
    Aug 2, 2015 at 15:56
  • $\begingroup$ Just a clarifying question on that equivalent condition -- is it saying that if you sum an $I$-indexed collection of sets, all in bijection with a fixed set $A$, then the result has cardinality $|A||I|$? (I assume this is what you must mean as the other obvious interpretation would appear to be trivially true regardless of any assumption like this, but I figured I should ask.) $\endgroup$ Aug 3, 2015 at 2:50
  • $\begingroup$ @Harry: The question itself asks about cardinal arithmetic in presence and failure of "there is a choice of representatives from every set of cardinals". I don't think that answer appears in any paper or book. However questions about disjoint union of sets and the obtained cardinality do (which is morally the same question). It turns out that asking whether "the sum is well-defined" and "repeated sums is product" is the same thing, which is what I pointed out, and remarked both are little researched into. In general it's hard to work with "little bit" of cardinal structure. $\endgroup$
    – Asaf Karagila
    Aug 3, 2015 at 4:02

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