How does one prove that an uncountable set has an uncountable subset whose complement is uncountable. I know it needs the axiom of choice but I've never worked with it, so I can't figure out how to use. Here is my attempt (which seems wrong from the start):
Let $X$ be an uncountable set, write $X$ as a disjoint uncountable union of the sets $\{x_{i_1},x_{i_2}\}$ i.e $X=\bigcup_{i\in I}\{x_{i_1},x_{i_2}\}$ where $I$ is an uncountable index (I'm pretty sure writing $X$ like this can't always be done), using the axiom of choice on the collection $\{x_{i_1},x_{i_2}\}$ we get an uncountable set which say is all the ${x_{i_1}}$ then the remaining ${x_{i_2}}$ are uncountable.
Anyway how is it done, properly?
I know the question has been asked in some form here but the answers are beyond my knowledge.
Best Answer
Your idea is generally correct.
Using the axiom of choice, $|X|=|X|+|X|$, so there is a bijection between $X$ and $X\times\{0,1\}$. Clearly the latter can be partitioned into two uncountable sets, $X\times\{0\}$ and $X\times\{1\}$.
Therefore $X$ can be partitioned to two uncountable disjoint sets.
Indeed you need the axiom of choice to even have that every infinite set can be written as a disjoint union of two infinite sets, let alone uncountable ones.