I have a question regarding the following exercise from Stephen Abbott's Understanding Analysis - 2nd edition:
Let $C \subseteq [0,1]$ be uncountable. Show that exists $a \in (0,1)$ such that $C \cap [a,1] $ is uncountable
Actually, someone already asked this question here. However I came up with a slightly different proof and I would appreciate some comments on the correctness.
First we define $ S:= \{ a \in (0,1] \, | \, s.t. \ C \ \cap [a,1] \ \text{is countable} \}$. Clearly $S$ is bounded from below by 0 and non-empty as $1$ is always an element of S. Thus it follows from the Axiom of Completness that $ s := \inf(S)$ exists. It also holds that $ s > 0 $, as $s = 0$ would implicate that $C \cap (0,1]$ is countable, however then $ C = \{0 \} \ \cup \ (C \ \cap \ (0,1] ) $ would be countable as an union of two countable sets, which contradicts our initial assumption. It follows that for $a:= s/2 $ it holds that $1 > a > 0$ and also $C \ \cap [a,1]$ is uncountable as $s > a$ and thus $a \notin S $.
Thanks for your feedback, Max