Principle of Real Induction

Problem 1. Fix $K \subseteq [a, b]$ . Suppose the following three properties are required:

$a \in K$ ,
if $[a, x] \subseteq K$ , then there exists $\delta > 0$ such that $[x, x + \delta] \subseteq K$ ,
if $[a, x) \subseteq K$ , then $x \in K$ .

Then $K = [a, b]$ . You could think of this result as the principle of “real induction”.

(Click for Solution)

Solution. Define $L := [a, b] \backslash K$ . We claim that $L = \emptyset$ . Suppose otherwise that $L \neq \emptyset$ . Since $L$ is bounded below, it has a greatest lower bound (i.e. infimum), which we will denote $m$ . By the first hypothesis, $a \in K$ , so that $m \neq a$ , and therefore, $m > a$ . By the second hypothesis, since $[a,a] = \{a\} \subseteq K$ , there exists $\delta > 0$ such that $[a, a+\delta] \subseteq K$ . In particular, if $m \leq a + \delta$ , then $m \in K$ , a contradiction. Therefore, $m > a + \delta$ . Finally, for any $n < m$ , if $n \notin K$ , then $m \leq n < m$ , a contradiction. Therefore, $n \in K$ , which implies $[a, m) \subseteq K$ . By the third hypothesis, $m \in K$ , a contradiction. Therefore, $L = \emptyset$ .