Exercise 3.2.12

Question

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 $\{x$ : $x \in A$ and $x<s\}$ and $\{x: x \in A$ and $x>s\}$ are uncountable. Show $B$ is nonempty and open.

Ulisse Mini · Accepted Answer

Our primary tool will be that countably infinite unions preserve countability (see Exercise \ref{ex:countable_union}).

Consider $B_1 = \{x \in \mathbf{R} : (-\infty, x) \cap A \text{ is uncountable}\}$. $B_1$ must be nonempty; otherwise, $A = \bigcup^\infty_{n=1} (-\infty, n) \cap A$ is a union of countable or finite sets, which by \ref{ex:countable_union} means that $A$ is countable (which it isn't). Note that if $x \in B_1$ and $y>x$, then $y \in B_1$. Moreover, $\exists \epsilon > 0$ so that $x - \epsilon \in B_1$; we can prove this by contradiction. If there is no such $\epsilon$, then $(-\infty, x - 1/n) \cap A$  must be countable for all $n \in \mathbf{N}$; by \ref{ex:countable_union},
  $$
  \bigcup_{n=1}^\infty (-\infty, x - 1/n) \cap A = (-\infty, x) \cap A
  $$
  is also countable, a contradiction. Therefore, $B_1$ is open. Now note that $B_1$ must be of the form $(-\infty, b_1)$, where $b_1 = \inf B_1$ (or $-\infty$ if $\inf B_1$ is undefined). Similarly, $B_2 = \{x \in \mathbf{R} : (x, \infty) \cap A \text{ is uncountable}\}$ is of the form $(b_2, \infty)$.

Note that $B_1 \cup B_2 = \mathbf{R}$; therefore $b_1 > b_2$ and so $B = B_1 \cap B_2 \neq \emptyset$. Moreover since $B_1$ and $B_2$ are both open, so is $B$.

Exercise 3.2.12

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Exercise 3.2.12

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