# Proof: Let $C \subseteq [0,1]$ be uncountable, show that $C \cap [a,1]$ is uncountable for some $a \in (0,1)$

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 $a \notin S$.

• The trouble is that we don't know whether $\inf(S)\in S$.