Show that if \(E\subset [0, 1]\) is uncountable, then there is some \(t\in \RR\) such that \(E\intersect (-\infty ,t)\) and \(E\intersect (t, \infty)\) are also uncountable.
Solution
See 3.2.12 of Understanding analysis 2ed. of Abbott. Show something stronger, that the following set is nonempty and open: \[ S \da \ts{t\in \RR \st E \intersect (-\infty, t), E \intersect (t, \infty) \text{ are uncountable}} \subseteq \RR .\] Write \[ S_- &\da \ts{ t\in \RR \st E \intersect (- \infty, t) \text{ is countable}} \\ S_+ &\da \ts{ s\in \RR \st E \intersect (s, \infty) \text{ is countable}} .\]
Note that \(S_- \neq \RR\) since then we could write \(E = \Union_{n\in \ZZ} E \intersect (- \infty, n)\) as a countable union of countable sets.
Claim: \(S = (\sup S_-, \inf S_+)\).
We prove the two inclusions.
First, \(S \subseteq (\sup S_-, \inf S_+)\). Let \(t \in S\), so both \(E \cap (-\infty, t)\) and \(E \cap (t, \infty)\) are uncountable. Then \(t \notin S_-\) (since \(E \cap (-\infty, t)\) is uncountable, not countable), and \(t \notin S_+\). Now \(S_-\) is downward closed: if \(s \in S_-\) and \(s' < s\), then \(E \cap (-\infty, s') \subseteq E \cap (-\infty, s)\) is countable, so \(s' \in S_-\). Hence every element of \(S_-\) is \(< t\) (if some \(s \in S_-\) satisfied \(s \ge t\), then \(t \in S_-\) by downward closure, a contradiction), so \(t \ge \sup S_-\). Similarly \(S_+\) is upward closed, so every element of \(S_+\) is \(> t\), giving \(t \le \inf S_+\). Thus \(t \in [\sup S_-, \inf S_+]\). It remains to rule out the endpoints: if \(t = \sup S_-\), then for every \(s < t\) we have \(s \in S_-\), so \(E \cap (-\infty, s)\) is countable for all \(s < t\); writing \(E \cap (-\infty, t) = \bigcup_{n} E \cap (-\infty, t - 1/n)\) expresses \(E \cap (-\infty, t)\) as a countable union of countable sets, hence countable, contradicting \(t \in S\). So \(t > \sup S_-\), and symmetrically \(t < \inf S_+\). Therefore \(t \in (\sup S_-, \inf S_+)\).
Conversely, let \(t \in (\sup S_-, \inf S_+)\). Since \(t > \sup S_-\), we have \(t \notin S_-\) (if \(t \in S_-\) then \(t \le \sup S_-\)), so \(E \cap (-\infty, t)\) is uncountable. Since \(t < \inf S_+\), we have \(t \notin S_+\), so \(E \cap (t, \infty)\) is uncountable. Hence \(t \in S\).
Finally, \(S\) is nonempty and open. First note \(S_- \cap S_+ = \emptyset\): if \(t\) lay in both, then \(E \cap (-\infty, t)\) and \(E \cap (t, \infty)\) would both be countable, so \(E = (E \cap (-\infty, t)) \cup (E \cap (t, \infty)) \cup (E \cap \{t\})\) would be countable, contradicting that \(E\) is uncountable. Since \(S_-\) is downward closed and \(S_+\) is upward closed, their disjointness forces \(\sup S_- \le \inf S_+\) (if \(\sup S_- > \inf S_+\), some \(t\) with \(\inf S_+ < t < \sup S_-\) would lie in both). Moreover \(\sup S_- < \inf S_+\): if \(\sup S_- = \inf S_+ = c\), then \(E \cap (-\infty, c) = \bigcup_n E \cap (-\infty, c - 1/n)\) is a countable union of countable sets (each \(c - 1/n < c = \sup S_-\) lies in \(S_-\)), hence countable, so \(c \in S_-\); symmetrically \(E \cap (c, \infty) = \bigcup_n E \cap (c + 1/n, \infty)\) is countable, so \(c \in S_+\), contradicting \(S_- \cap S_+ = \emptyset\). Therefore \((\sup S_-, \inf S_+)\) is a nonempty open interval, and by the two inclusions above it equals \(S\), so \(S\) is nonempty and open.