# Does any uncountable set contain two disjoint uncountable sets?

Is it true that for any uncountable $S$, there exits two uncountable subsets $S_1,S_2 \subseteq S$ with $S_1 \cap S_2 = \emptyset$?

I can find no counter example, but no proof either.

I am aware of similar questions regarding the partition of uncountable subsets of the real line, and there are many examples of disjoints uncountable subsets of $\mathbb{R}$, say $R_1,R_2$.

It is not clear to me however whether this extends to any uncountable set. For uncountable sets $S$ with $S \succeq_{card} \mathbb{R}$, there exists an injection $f : \mathbb{R} \rightarrow S$ and choosing $S_1,S_2$ with $S_i = \{s\in S~|~ s = f(r_i)$ for some $r_i \in R_i\}$, works.

But from what I understood of the continuum hypothesis, there might exist an uncountable sets $H$ with $\mathbb{R} \succeq_{card} H$. So using the above argument, one cannot infer that the property holds for every uncountable sets from observing that it holds for $\mathbb{R}$. Is this right? Are there other ways to prove or disprove the assertion?

• Well, without axiom of choice, there exist infinite sets with no infinite subsets, so I'm guessing there should also exist an uncountable set with no uncountable subsets. Jul 1, 2014 at 10:53
• If you are willing to use choice, well-order your set and use the bijection between your set and its uncountable cardinal number. Put elements corresponding to ordinals of the form $\omega.\alpha+2n$ in one set and the rest into the other. Jul 1, 2014 at 11:04
• Also, you might want to take a look at this (Asaf's and Andres' answers) to see that you have to use axiom of choice in order to rule out the existence of pathological sets. Jul 1, 2014 at 11:28

So if $S$ is uncountable with cardinal $\alpha$, we have $\alpha+\alpha=\alpha$. So their exists two disjoint copies $S_1, S_2$ of $S$ such that their union has the same cardinality as $S$. So there is a bijection $f:S_1\cup S_2\to S$. So $f(S_1)$ and $f(S_2)$ are two disjoint uncountable subsets of $S$.
Here is an actual method to "construct" a bijection from $S$ to $S\times\{0,1\}$. Let $\mathcal{F}$ be the family of all bijections of some subset $T$ of $S$ to $T\times\{0,1\}$. Such a bijection always exists when $T$ is a countably infinite subset. Order these functions by set inclusion on their graph. Then the conditions of Zorn's Lemma are satisfied, and their exists a maximal such function of the form $f:T\to T\times\{0,1\}$. If $S\backslash T$ would be infinite, it would contain a countable subset and then we could extend $f$ to a larger such function in contradiction to it being maximal. So $S\backslash T$ must be finite. So there is a bijection $g:\{0,1,\ldots,n-1\}\to S\backslash T\times\{0,1\}$. Let $C=\{x_0,x_1,\ldots\}\subseteq S$ be countably infinite. You can construct now a new bijection $f':S\to S\times\{0,1\}$ by $$f'(x) = \begin{cases} g(m) &\mbox{if } x_m\in\{x_0,\ldots,x_{n-1}\}\subseteq C\\ f(x_{m-n}) & \mbox{if } x_m\in C\text{ and }m\geq n.\\ f(x) &\mbox{otherwise.}\end{cases}$$
• Nice argument. The axiom of choice is only required if you are dealing with arbitrary sets, you need the axiom to well order them. If you start with sets that are already well ordered, then you dont need it. So $\kappa+\kappa=\kappa$ does not require choice but $X\times \{0\} \cup X\times\{1\}\sim X$ for all $X$ does. Jul 1, 2014 at 11:53