TITLE: Can every infinite set be divided into pairwise disjoint subsets of size $n\in\mathbb{N}$?
QUESTION [11 upvotes]: Let $S$ be an infinite set and $n$ be a natural number.  Does there exist partition of $S$ in which each subset has size $n$?


This is pretty easy to do for countable sets.  Is it true for uncountable sets?
If (1) is true, can it be proved without choice?

REPLY [4 votes]: There is a concept called amorphous sets, which becomes interesting when the axiom of choice fails. These sets are such that you cannot partition them into two distinct infinite sets.
One property of amorphous sets is that if we take an infinite partition of an amorphous set, then all but finitely many parts have the same size (and all these sizes are finite, of course). There are unbounded amorphous sets, which can be partitioned into arbitrarily large size, but there are also bounded amorphous sets.
So bounded amorphous sets form a counterexample to (1), they cannot be partitioned into sets of size $n$ for large enough $n$.
But this property of "non-divisibility" can be found in other sets, which are not amorphous. Although it is often harder to prove this property holds. For example in Cohen's first model there exists an infinite set of real numbers which does not have a countably infinite subset. This set is not amorphous because amorphous sets cannot be linearly ordered. And yet, it follows from the construction of the model that every partition of that set into finite parts is almost entirely singletons.