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I am looking at the complement of the cantor set in $[0,1]$ as the union of open intervals of decreasing length $\displaystyle\bigcup_{i=1}^{\infty}A_i$ where $A_1 = (\frac{1}{3},\frac{2}{3}), A_2 = (\frac{1}{9},\frac{2}{9})\cup(\frac{7}{9},\frac{8}{9})$ and so on.

I am trying to prove that $\overline{\displaystyle\bigcup_{i=1}^{\infty}A_i} = \displaystyle\bigcup_{i=1}^{\infty}\overline{A_i}$.

I know that this is not true in general and is probably not true in this case, but can someone give a proof or a counterexample

Thank you very much

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1 Answer

up vote 2 down vote accepted

Here are some hints for one method of answering this:

  • The complement of the Cantor set is dense in $[0,1]$.
  • The closure of each individual $A_n$ only has finitely many extra points.
  • The Cantor set is uncountable.
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Jonas, as I recall if $X$ is dense in $Y$ then $Y\subseteq\operatorname{cl}(X)$, and if the axiom of countable choice for finite sets is assumed then a countable union of finite sets is countable. I also remember that uncountable-countable = uncountable. Which point am I missing here? –  Asaf Karagila Sep 9 '11 at 15:58
    
@Asaf, The conclusion Jonas is hinting at seems to be that the union of the closures is a proper subset of the closure of the union -- it misses the points in the Cantor set that are not interval endpoints (and such points must exist, because there are only countably many intervals, but the Cantor set itself is uncountable). –  Henning Makholm Sep 9 '11 at 16:21
    
Perhaps an explicit number would be good to work on. Show that $1/4$ belongs to the closure of the union, but not to the union of the closures. –  GEdgar Sep 9 '11 at 16:29
    
@Henning: Oh, right. For some reason it seemed to me that equality was hinted somehow.. :-) –  Asaf Karagila Sep 9 '11 at 16:34
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