Splitting a set into two uncountable pieces

Question

Suppose I have an uncountable set $S$ that I obtained in a non-constructive way, like from the Axiom of Choice. In my particular example I'm thinking about a basis for $\mathbb R$ as a vector space over $\mathbb Q$.

I'd like to remove uncountably many elements into a set $T$ but in a way that leaves uncountably many elements behind. But because I don't know anything in particular about the specific elements of $S$ I don't know how to use the Axiom of Specification or some other method to be confident that I can correctly make a subset of the right size. For example, I could try $\{s \in S : s \geq 0\}$ but my basis doesn't have to have any negative elements, right? Or it could have all negative elements or a mix. If I just wanted $T$ to be countable or finite I could just draw a single element from $S$, say $s_1$, and then draw $s_2$ from $S \backslash \{s_1\}$, and so on (right?). But if I want $T$ to be uncountable this doesn't seem to help.

So can I simply say "Let $T$ be an uncountable subset of $S$ such that $S \backslash T$ is also uncountable"? Is that a thing I can just say that I can make? Or do I need to come up with a formula $\phi$ such that I can use Specification via $\{s\in S : \phi(s)\}$? For a set like $[0,2]$ I have no problem doing this because $\{s \in [0,2] : s \leq 1\}$ does the trick, and since I know it's possible in some way I'm not worried about not exhibiting a particular formula $\phi$. But with a set where I don't even know a single element of it, I'm not sure what to do.

Can I prove that $S$ must have such a subset $T$? If not, then I'd have an uncountable set where every subset is either countable or cocountable. Is there an issue with that?

Can you say that since $S$ is uncountable, there exists a surjection to ${1,2}\times\Bbb R$, call it $f$. Then consider $T=f^{-1}({1}\times\Bbb R$). — Clayton, Feb 04 '19 at 18:28
@Clayton In general, we cannot assume that an uncountable set is at least a large as $\Bbb R$ ... — Hagen von Eitzen, Feb 04 '19 at 18:32

score 5 · Accepted Answer · edited Feb 04 '19 at 18:47

5

If $A$ is an infinite set, then there exists a bijection $f\colon A\to A\times\{0,1\}$ (readily defined for example if we assume $A$ is well-ordered), hence $A$ can be written as disjoint union of two sets $f ^{-1}(A\times\{0\})$ and $f^{-1}(A\times\{1\})$ of equal cardinality.

edited Feb 04 '19 at 18:47

lisyarus

17,012

answered Feb 04 '19 at 18:31

Hagen von Eitzen

382,479

Good answer. Even though mine didn't work exactly, I'm glad to see the idea was in the right place. – Clayton Feb 04 '19 at 19:00
thank you so much, glad to see it's a simple process! – alfalfa Feb 04 '19 at 19:16

Splitting a set into two uncountable pieces

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