Show that an uncountable set can be separated into two disjoint uncountable subsets

real-analysis

This is a question from Steven Abbott's Understanding Analysis and I am certainly aware of how many duplicates and duplicates there are on this website as well as the internet. However, of all the solutions, there is one essential part that still troubles me. For sake of completeness, I would show

  1. The full question
  2. Common solutions
  3. My question

Let A be an uncountable set and let B be the set of real number that divides A into two uncountable set; 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 B is nonempty and open.

The general approach to this question is

  1. Prove there exists a bounded uncountable subset
  2. Prove there exists a countable subset in this form $(-\infty,s)\cap A$
  3. Prove B is nonempty
  4. Prove B is open

Sorry for being so brief since I am stuck in point 1 and 2.Here's my confusion. Why does the following makes sense?

Since A is a bounded uncountable set, there must exists s such that $(-\infty,s)\cap A$ is uncountable.

Of course, if it is bounded and there doesn't exist a real number, including the supremum of this set, that make the intersection uncountable, then the entire set will be countable, contradicts A is uncountable.

My question, however, is, why bounded? What is the significance behind it?

Best Answer

It isn’t actually necessary to reduce the problem to the case in which $A$ is bounded. We can instead argue as follows.

Let

$$L=\{x\in\Bbb R:(\leftarrow,x)\cap A\text{ is countable}\}$$

and

$$R=\{x\in\Bbb R:(x,\to)\cap A\text{ is countable}\}\,.$$

  • If $L\ne\varnothing\ne R$, it is straightforward to show that $B=(\sup L,\inf R)\ne\varnothing$; this is the same argument as in the bounded case.
  • If $L=\varnothing=R$, then clearly $B=\Bbb R$.
  • If $L=\varnothing\ne R$, then $B=(\leftarrow,\inf R)$.
  • If $L\ne\varnothing=R$, then $B=(\sup L,\to)$.
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