Uncountable Sets that can not be expressed as a disjoint union of two Uncountable sets

Question

"There does not exist an uncountable subset of the real numbers which can not be expressed as the union of two uncountable sets which are disjoint from one another" is obviously true assuming choice. However, I'm looking for a proof of this statement that does not involve choice. I believe I have come up with one myself, but it's a little complicated. Is there any easy way to go about this?

Answer 1

Let $x\in\mathbb{R}$. Define $$S_{\leq x}=S~\cap~(-\infty,x]~~~\mathrm{and}~~~S_{>x}=S~\cap~ (x,+\infty)$$ and define $S_{\leq}=\{$ the set of all $x$ such that $S_{\leq x}$ is uncountable $\}$, and $S_>=\{$ the set of all $x$ such that $S_{>x}$ is uncountable $\}$.

$S_{\leq}$ is a non empty$^{(*)}$ interval that extends to infinity on the right, and is therefore of the form $[a,+\infty)$ or $(a,+\infty)$ for some $a\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.

Similarly, $S_>$ is a non empty$^{(*)}$ interval that extends to infinity on the left, and is thus of the form $(-\infty,b]$ or $(-\infty,b)$ for some $b\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.

$^{(*)}$ : I have used some form of choice I believe when I say that $S_>$ and $S_{\leq}$ are non empty, because I use the fact that with some form of choice, the countable union of countable sets is still countable. Is it countable choice? Since $$S=\cup_{n\in\mathbb{N}}S_{\leq n}$$ is uncountable, at least one of the $S_{<n}$ must be uncountable, and so for some natural number $n$ you have $n\in S_{<}\neq\emptyset$.

Let's show that $S_>\cap S_{\leq}$ is non empty. If either one of $S_>$ or $S_{\leq}$ is the whole real line, there is no problem. So let's suppose none of them is the whole of $\mathbb{R}$.

Suppose $x\in S_{\leq}$ that is $S_{\leq x}=S~\cap~(-\infty,x]$ is uncountable. Since $$\bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}\subset S_{\leq x}=S~\cap~(-\infty,x]\subset \{x\}\cup \bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}$$ the already used fact that countable unions of countable sets are countable with some form of choice implies that at least one of the $S_{\leq x -\frac{1}{n}}$ is uncountable, which means that for some $n\in\mathbb{N}^*,~x -\frac{1}{n}\in L$ and thus $S_{\leq}$ is open, that is, $$\mathbf{S_{\leq}=(b,+\infty)}$$ for some real number $b$. Similarly, $$\mathbf{S_{>}=(-\infty,a)}$$ for some real number $a$.

Then $S_>\cap S_{\leq}$ is empty iff $a\leq b$. But this would entail that $$S=S_{\leq a}\cup S_{> a}$$ is countable as the union of two countable sets, since $a \notin (-\infty,a)=S_{\leq}$ means $S_{\leq a}$ is countable, and $a \notin (b,+\infty)=S_{>}$ means $S_{> a}$ countable.