5
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ Your proof looks good to me. Nice approach! $\endgroup$
    – Clayton
    Commented May 16, 2017 at 22:43
  • 4
    $\begingroup$ "It also holds that s>0, as s=0 would implicate that C∩(0,1] is countable" you might want to explain this a bit more $\endgroup$
    – user223391
    Commented May 16, 2017 at 22:43
  • 1
    $\begingroup$ @ZacharySelk Isnt this observation as hard as the original problem? $\endgroup$
    – clark
    Commented May 16, 2017 at 23:05
  • $\begingroup$ The trouble is that we don't know whether $\inf(S)\in S$. $\endgroup$
    – A.Γ.
    Commented May 16, 2017 at 23:31
  • $\begingroup$ @clark My point $\endgroup$
    – user223391
    Commented May 17, 2017 at 0:18

0

You must log in to answer this question.