Show that any uncountable set $A$ of $\mathbb{R}$ can be divided into two uncountably infinite disjoint subsets

Question

How would one prove using the Axiom of Choice that any uncountable set $A$ of $\mathbb{R}$ can be divided into two uncountably infinite disjoint subsets? I haven't seen this kind of proof exactly, because I'm not looking for a partition.

Formally, let $D$ be the set with the following property: $x$ is a member of $D$ just in case the set of all elements of $A$ strictly greater than $x$ is uncountable and the set of all elements of $A$ strictly less than $x$ is uncountable. How would one show that $D$ is nonempty, preferably without directly resorting to cardinal arithmetic?

Update: I appreciate all the feedback, including a solution. Although a nice proof has been given, I'm interested in seeing how it can be done by using the rationals in particular to split the set.

Answer 1

For every rational $q$ let $A_q^+=\{x\in A\mid x\geq q\}$ and $A_q^-=\{x\in A\mid x<q\}$. If for some $q$ both are uncountable, super. If not, what can you say about $A$ itself?

Now, you might wonder as to the use of the axiom of choice here, and it hides beneath the surface: in order to prove that every countable union of countable sets is countable, we have to use the axiom of choice.

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Edit: Let $r=\sup\{q\mid A^-_q\text{ is countable}\}$ and $s=\inf\{q\mid A^+_q\text{ is countable}\}$.

1. If either $r$ or $s$ is not a real number (i.e., $\pm\infty$) then we get that $A$ is countable.

2. If $r>s$, take a rational $q$ witnessing that, then $A^-_q$ and $A^+_q$ are both countable, what do we get?

3. If $r<s$, take a rational $q$ witnessing that, then $A^-_q$ and $A^+_q$ are both uncountable.

4. Finally, if $r=s=q$, take an increasing sequence of rationals $r_n$ and a decreasing sequence of rationals $s_n$, both converging to $q$. Then $A\cap\bigcup_{p<q}A_p^+=A\cap\bigcup_{n\in\Bbb N}A^+_{r_n}$ is countable; and $A\cap\bigcup_{p>r} A^-_p$ is countable. Again, we get that $A$ is countable.

So if $A$ is uncountable, there is a rational $q$ such that $r<q<s$ and $A_q^-$ and $A^+_q$ are both uncountable.

Answer 2

The question immediately reminded me of this. Here is an argument following the same basic idea at the beginning of that argument:

First, consider $B=\{x\in A\mid A\cap(-\infty,x]$ is countable$\}$, and note that $B$ itself is countable: The point is that if $x\in B$ then $A\cap(-\infty,x]\subseteq B$. Now, if $B\ne\emptyset$, let $t=\sup B$, fix an increasing sequence of elements of $B$ converging to $t$, $t_0\le t_1\le t_2\le\dots\to t$, and note that $$B\subseteq A\cap(-\infty,t]\subseteq \{t\}\cup\bigcup_n A\cap(-\infty,t_n],$$ and the expression on the right is a countable union of countable sets, and therefore countable (by the axiom of choice).

Now let $C=A\setminus B$. Note that $C$ is uncountable and that if $x\in C$ then $C\cap(-\infty,x]$ is itself uncountable. Otherwise, $x\in B$ (a contradiction), since $A\cap(-\infty,x]\subseteq B\cup(C\cap(-\infty,x])$. Forget about $A$ itself now and work with $C$. The same argument as above shows that $D=\{x\in C\mid C\cap[x,\infty)$ is countable$\}$ is countable as well.

Finally, let $E=C\setminus D$, and note that for any $x\in E$, both $E\cap(-\infty,x)$ and $E\cap(x,\infty)$ are uncountable.

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Let me present another argument, showing that we can actually find a rational $q$ such that both $A\cap(-\infty,q)$ and $A\cap(q,\infty)$ are uncountable:

Since $$A\subseteq\mathbb Z\cup\bigcup_{l\in\mathbb Z}A\cap(l,l+1),$$ at least one of the intervals $A\cap(l,l+1)$ is uncountable (by the axiom of choice). We may as well assume that, for such an $l$, $A\setminus(l,l+1)$ is countable, or we are done.

Let $I_0=(l,l+1)$. Suppose we have found $I_n=(a,b)$ of length at most $1/2^n$ with rational endpoints such that $A\cap I_n$ is uncountable but $A\setminus I_n$ is countable. Find rational points $q_m$ for $m\in\mathbb Z$ such that $a<q_k<q_m<b$ whenever $k<m$, $\lim_{k\to-\infty} q_k=a$, $\lim_{m\to\infty}q_m=b$, and $q_{m+1}-q_m\le 1/2^{n+1}$ for all $m\in\mathbb Z$. As before, $$A\cap I_n\subseteq\{q_m\}_{m\in\mathbb Z}\cup\bigcup_{m\in\mathbb Z}A\cap(q_m,q_{m+1}),$$ so for some $m$, $A\cap(q_m,q_{m+1})$ is uncountable. As before, we may assume that $A\setminus (q_m,q_{m+1})$ is countable, or else we are done. Now set $I_{n+1}=(q_m,q_{m+1})$.

The claim is that at some stage $n$ we must actually be done, that is, $A\setminus(q_m,q_{m+1})$ must also be uncountable and therefore at least one of $q_m,q_{m+1}$ works as the rational $q$ we were after.

The proof is by contradiction: Otherwise, the recursion produces an infinite, shrinking sequence of intervals $I_n=(a_n,b_n)$ with $A\cap I_n$ uncountable and $A\setminus I_n$ countable for all $n\in\mathbb N$. Notice that $\bigcap_n I_n$ is either empty or a singleton $\{t\}$, and set $t=0$ if $\bigcap_n I_n=\emptyset$. We have $$A\subseteq \{t\}\cup\bigcup_{n\in\mathbb N}\{a_n,b_n\}\cup\bigcup_{n\in\mathbb N}A\setminus I_n,$$ so $A$ is countable.