Show that if \(E \subset [ 0 , 1 ]\) is uncountable, then there is some \(t \in \mathbb { R }\) such that both \(E \cap ( - \infty , t )\) and \(E \cap ( t , \infty )\) are uncountable.
Solution
Set \[ I_0=[0,1]. \]
1. Starting from any closed interval\[ I_n=[a_n,b_n] \]for which \(E\cap I_n\) is uncountable, let\[ m_n=\frac{a_n+b_n}{2}. \]Either both\[ E\cap(-\infty,m_n) \qquad\text{and}\qquad E\cap(m_n,\infty) \]are uncountable, or there is a closed half \(I_{n+1}\) of \(I_n\) such that\(E\cap I_{n+1}\) is uncountable and\[ E\cap(I_n\sm I_{n+1}) \]is countable.
If both displayed sets are uncountable, the first alternative holds.
Otherwise at least one of them is countable. They cannot both be countable, because then \[ E = \bigl(E\cap(-\infty,m_n)\bigr) \cup \bigl(E\cap\{m_n\}\bigr) \cup \bigl(E\cap(m_n,\infty)\bigr) \] would be countable, contrary to the hypothesis.
Thus exactly one displayed side is countable. Choose \(I_{n+1}\) to be the closed half of \(I_n\) lying on the other side of \(m_n\). The discarded portion of \(E\) is contained in the countable side, while \(E\cap I_{n+1}\) must be uncountable because \(E\cap I_n\) is uncountable and only a countable set has been discarded.
2. If the first alternative in step 1 never occurs, then there is aunique point\[ s\in\bigcap_{n=0}^{\infty}I_n. \]
Under that assumption step 1 constructs a nested sequence \[ I_0\supseteq I_1\supseteq I_2\supseteq\cdots \] of nonempty closed intervals. At each stage the length is halved, so \[ \operatorname{length}(I_n)=2^{-n}\longrightarrow0. \] The nested interval theorem therefore gives exactly one point in their intersection; call it \(s\).
3. The first alternative in step 1 must occur at some finite stage.
Suppose it never occurs. For each \(n\), put \[ D_n=E\cap(I_n\sm I_{n+1}). \] Step 1 says every \(D_n\) is countable. By step 2, \[ \bigcap_{n=0}^{\infty}I_n=\{s\}. \] Every point of \(E\) other than possibly \(s\) must leave the nested intervals at some finite stage. Hence \[ E \subseteq \{s\} \cup \bigcup_{n=0}^{\infty}D_n. \] The right-hand side is a singleton together with a countable union of countable sets, and is therefore countable. This contradicts the assumption that \(E\) is uncountable.
4. There is a real number \(t\) such that both\[ E\cap(-\infty,t) \qquad\text{and}\qquad E\cap(t,\infty) \]are uncountable.
By step 3, the construction stops at some stage \(n\) because the first alternative in step 1 holds. Set \[ t=\boxed{m_n}. \] That alternative says exactly that both displayed subsets of \(E\) are uncountable.
5. Q.E.D.
Step 4 supplies the required splitting point.