Show that if \(E \subset [0, 1]\) is uncountable, then there exists some \(t \in (0, 1)\) such that both \(E \cap (-\infty, t)\) and \(E \cap (t, \infty)\) are uncountable.
Solution
For \(t\in[0,1]\), set \[ L_t=E\cap[0,t),\qquad R_t=E\cap(t,1]. \] Suppose no \(t\in(0,1)\) makes both sets uncountable.
Define \[ a=\sup\{t\in[0,1]:L_t\text{ is countable}\}. \] Because \(L_0=\varnothing\), the set is nonempty. Since \(E\) is uncountable, \(a<1\) cannot fail merely because all initial segments are countable; in any case the following argument handles the endpoint possibilities.
For every rational \(q<a\), \(L_q\) is countable. Hence \[ E\cap[0,a)=\bigcup_{q\in\mathbb Q,\,q<a}L_q \] is countable. For every rational \(q>a\), maximality of \(a\) implies \(L_q\) is uncountable; by the assumed failure of the conclusion, \(R_q\) must therefore be countable. Hence \[ E\cap(a,1]=\bigcup_{q\in\mathbb Q,\,q>a}R_q \] is countable. Thus \[ E=(E\cap[0,a))\cup(E\cap\{a\})\cup(E\cap(a,1] \] is countable, a contradiction.
Therefore some \(t\in(0,1)\) has both \[ E\cap(-\infty,t) \quad\text{and}\quad E\cap(t,\infty) \] uncountable.