Confusion with the proof that the Cantor set is closed

Question

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.

Answer 1

This is better interpreted via complementation: At each stage $n+1$ a set of open intervals is removed from stage $n$. The completion of this as $n$ 'goes infinite' is the union of all removed open intervals, which is an open set. The complement of an open set is a closed set.

Answer 2

For any large $n$, $F_n$ is the union of finitely many closed intervals so it is closed.

As you mentioned the intersection of an arbitrary family of closed sets is closed so $F=\cap_{n\to \infty} F_n$ is closed.

Note that for each $n$ we have finitely many closed sets, no matter how large $n$ is, so the union is closed.

Your example of the infinite union of closed sets ${ 1/n } not being closed does not contradict the counter set because you are dealing with an infinite union of closed sets not an infinite intersection of closed sets.

Answer 3

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.