TITLE: Confusion with the proof that the Cantor set is closed
QUESTION [1 upvotes]: I have encountered the definition of Cantor set and its property.
Cantor set is constructed by removing the middle third open set of each interval .So each time we get some union of closed sets.
As $F_0=[0,1]$
$F_1=[0,1/3]\cup [2/3,1]$
$F_2=[0,1/9]\cup [2/9,1/3]\cup [2/3,7/9]\cup [8/9,1]$
So on ,
That for $F_n$ we get the Union of $2^n$ closed interval .
I know that finite union of closed interval is closed, but this is not true for arbitarly union.
As I had a counterexample:  $\cup$ {{1/n}|for $n$ $\in N$} as this is not closed as $0$ is not in  that union.
So what is the argument here to say that $F_n$ is closed for any $ n$ .
As we are using this fact to prove the Cantor set to be closed as $F=\cap_{n\to \infty} F_n$ i.e arbitarly intersection of closed set is closed .
Where am I misunderstanding?
Any help will be appreciated.

REPLY [2 votes]: Since the  $F_n^{\prime s}$  are closed their complements are open sets. So,the complement of it's intersection is given by
$$
\left(\bigcap_nF_n\right)^c=\bigcup_nF_n^c
$$
which is a union of open sets and therefore an open set.
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