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 :


1. $p(A)$ is a subset of the power set of $A$.
2. The elements of $p(A)$ are pairwise disjoint.
3. 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 $A$ ?



Cardinalities of which there exists partitions of a set containing elements of the same size
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.


