Assume I have a set M which is uncountable. Can I conclude that there exists a subset $A\subseteq M$ such that neither $A$ nor $M \backslash A$ is countable?
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With Axiom of Choice, yes. – André Nicolas Nov 13 '14 at 14:29
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one, two and three. – Asaf Karagila Nov 13 '14 at 15:40
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Hint: (Assuming the Axiom of Choice) $M\cong \{0,1\}\times M$ for any infinite set $M$.
Hagen von Eitzen
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Assuming the Axiom of Choice, every cardinal is comparable to $|\mathbb R|$, so your uncountable set $M$ has an uncountable subset $M_0$ such that $|M_0|\le|\mathbb R|$. Thus it suffices to show that an uncountable subset of $\mathbb R$ can be split into two disjoint uncountable sets.
In fact, for any uncountable set $M\subseteq\mathbb R$, there is a rational number $q$ such that both $\{x\in M:x\lt q\}$ and $\{x\in M:x\gt q\}$ are uncountable. (This uses the fact that a countable union of countable sets is countable, which also follows from the Axiom of Choice.)
bof
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