TITLE: Cardinalities of which there exists partitions of a set containing elements of the same size
QUESTION [8 upvotes]: Let $A$ be a nonempty set. Then we call a nonempty set $p(A)$ to be a partition set of $A$ if and only if all the following are true :

$p(A)$ is a subset of the power set of $A$.
The elements of $p(A)$ are pairwise disjoint.
Every element of $A$ is present in some element of $p(A)$.

Now, for a set $A$ and a set $D$ both nonempty we say that $D$ divides $A$ if and only if there is a partition set $p_D(A)$ of $A$ such that every element of $p_D(A)$ is bijective to $D$. 
Question
Suppose given two infinite sets $A,B$. Is it true that : either $A$ divides $B$ or $B$ divides $A$ ?

REPLY [16 votes]: This is equivalent to the axiom of choice.
If the axiom of choice holds, then given $A$ and $B$ which are infinite, then $|A\times B|=\max\{|A|,|B|\}$. So let's say $|A|$ is the maximal one, then this means there is a bijection between $A$ and $A\times B$, so we can partition $A$ to sets of size $|B|$ by considering $\{a\}\times B$ as a partition of $A\times B$.
In the other direction, assume that any two sets one divides the other. Take $A$ to be some infinite set, and let $B$ be the least ordinal such that there is no injection from that ordinal into $A$. This ordinal exists, as per Hartogs' theorem. Now since $B$ does not inject into $A$, this means that you cannot partition $A$ into sets of size $B$. But now we can partition $B$ into sets of size $A$, so pick one of these sets, and this gives us an injection from $A$ into $B$ which well-orders $A$. So every set can be well-ordered, and therefore the axiom of choice holds.
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