Let $\kappa$ be an infinite cardinal number.

My question is whether there are $\lambda$ and $\mu$ such that both $<\kappa$ but $\lambda+\mu=\kappa$?

If AC holds, then the answer is definitely NO. However, can it become YES when refused AC?

Furthermore, what about some sets which are infinite but not Dedekind-infinite, especially amorphous sets?


1 Answer 1


Yes. It is certainly possible, if $\kappa$ is not well-orderable. For ordinals the laws of addition and multiplication are true without the axiom of choice because these are well-ordered sets. However for cardinals which are not ordinals this can certainly happen.

As you pointed out infinite Dedekind-finite sets are such example. In particular if $A$ is infinite Dedekind-finite and $B\subsetneqq A$ then $|B|<|A|$ and $|A\setminus B|<|A|$, but $|A|=|A\setminus B|+|B|$. Note, however, that if $A$ is amorphous then $B$ is either finite or $A\setminus B$ is finite. You cannot write an amorphous cardinal as the sum of two infinite cardinals.

Monro also proved that there are models of ZF in which every set which cannot be well-ordered can be written as the sum of two strictly smaller cardinals. One of these models is such that the real numbers cannot be well-ordered, and can therefore be written as the sum of two strictly smaller cardinals.

This is not provably true for sets which cannot be well-ordered. For example there are models in which every set of real numbers is either countable or have size continuum. In such model the real numbers cannot be written as the sum of two smaller cardinals.

Here is a short proof which you might like:

Theorem. The axiom of choice is equivalent to the fact that for every infinite set $A$, if $|A|=|X|+|Y|$ then $|X|=|A|$ or $|Y|=|A|$.

Proof. If the axiom of choice holds the implication is trivial. Suppose that the axiom of choice fails, then there exists a set which cannot be well-ordered. Denote such set $A$. Let $\kappa$ be $\aleph(A)=\min\{\alpha\in\mathsf{Ord}\mid \alpha\nleq A\}$. Consider the set $B=A\cup\kappa$, where we can assume that $A\cap\kappa=\varnothing$.

The cardinality of $B$ is clearly $|A|+\kappa$, and therefore $\kappa\leq|B|$ and $|A|\leq|B|$. It follows that both these inequalities have to be strict.

  • If $\kappa=|B|$ then $A$ can be well-ordered.
  • If $|A|=|B|$ then $\kappa\leq|A|$ which is a contradiction to the choice of $\kappa$.

In particular we have found an infinite set which can be written as the sum of two sets of strictly smaller cardinality. $\square$

  • $\begingroup$ The statement I gave and proved is probably known, but it is a simple example of how cardinal arithmetics are messed up without choice. $\endgroup$
    – Asaf Karagila
    Nov 10, 2012 at 11:24
  • $\begingroup$ Thank you. That's all right. $\endgroup$
    – Popopo
    Nov 10, 2012 at 12:32
  • $\begingroup$ You're welcome, again. Do you intend to ask more AC questions today? :-) $\endgroup$
    – Asaf Karagila
    Nov 10, 2012 at 12:53
  • $\begingroup$ Yes, one more of cardinal arithmetic,but first let me search it for avoiding duplicate. $\endgroup$
    – Popopo
    Nov 10, 2012 at 13:13
  • $\begingroup$ I have another question: Can a Dedkind-infinite set be such an example? $\endgroup$
    – Popopo
    Dec 1, 2012 at 13:24

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