This is the exercise 3.2.12 of Understanding analysis 2ed. of Abbott:
Let $A$ be an uncountable set and let $B$ be the set of real numbers that divides $A$ into two uncountable sets; that is, $s\in B$ if both $\{x: x\in A\text { and } x< s\}$ and $\{x: x\in A \text { and } x> s\}$ are uncountable. Show that $B$ is nonempty and open.
The problem that I have here is that as I understand the problem the statement is not true, I have a counterexample:
$$A=(0,1)\cup\{2\}\cup(3,4)$$
This set is uncountable and $2\in B$ cause the condition is hold but obviously $2$ is an isolated point, so $B$ is certainly nonempty but is not open.
Maybe I dont understand something in the exercise, if Im wrong, can someone clarify the exercise? Thank you in advance.
Best Answer
As pointed out in the comments, we are not looking for $B \subseteq A$, we are looking for $B \subseteq \mathbb R$. In particular, for your $A$, the associated $B$ is equal to $(0, 4)$. Let us also take a moment to see how to solve the exercise (since this answer has now been accepted).
Let $S^{\uparrow}$ be the set of $s$ with $(-\infty, s) \cap A$ countable, and $S^{\downarrow}$ be the set of $s$ with $(s, \infty) \cap A$ countable. Trivially $S^{\uparrow}$ is downward closed. It cannot be all of $\mathbb R$, because then every integer would be in it, and then $$ A = \bigcup_{n \in \mathbb Z} A \cap (-\infty, n) $$ would be a countable union of countable sets and thus countable. Finally, it is closed, because if $(s_i)_i \in \mathbb N$ is an increasing sequence converging to $s \in S^{\uparrow}$, then $$ A \cap (-\infty, s) = \bigcup_{i \in \mathbb N} A \cap (-\infty, s_i) $$ is a countable union of countable sets, and thus countable. Hence $S^{\uparrow} = (-\infty, r]$ for some $r$. By the same reasoning, $S^{\downarrow} = [r', \infty)$ for some $r'$. Now we must have $r < r'$, for if $r \geq r'$ then $$A = ((-\infty, r] \cap A) \cup ([r, \infty) \cap A)$$ would be countable.
But now note that $B$ is exactly equal to $(r, r')$ by definition, so we are done.