Every Closed Set In $R^1$ is intersection of countable collection of open set.

Question

This is question I tried to solve as follows 
Consider $A$ be closed set in $Bbb R$ Therefore $Bbb{R}smallsetminus A$ is Open set .Now By Representation theorem of open set ,Every open set in $Bbb R$ can be written as union of countable collection of disjoint open interval.$Bbb{R}smallsetminus A=bigcup I_n$ Where $I_n$ is open interval where n is form countable index set.$Bbb{R}smallsetminus(Bbb{R}smallsetminus A)=A$ $=Bbb{R}smallsetminuscup I_n $ $=bigcap (Bbb{R}smallsetminus I_n)$ which implies $A$ is countable intersection of Closed set . I had to prove that it is intersection of countable intersection of open set but I got other answer Where is my mistake in argument ? Any Help will be appreciated

Berci · Accepted Answer

This works in every metric space $(X, d) $:

Hints:
Let $A$ be closed in $X$, then

$A={x:d(x, A) =0} $
Consider $A_n:={x:d(x, A) <frac1n} $.

where $d(x, A) $ denotes the distance of point $x$ to set $A$, i.e. 
$$d(x, A) =inf_{ain A} d(x, a) $$
And specifically for $Bbb R$,  the distance function is given by $d(x, y) :=vert y-xvert$.

user553213 · Answer

For $k=1,2,...$, cover the real line with the intervals $(n/2^k-1/2^{k+1},(n+1)/2^k+1/2^{k+1})$, for $ninmathbb{Z}$. Define $U_k$ to be the union of those intervals in this collection that intersect your closed set $C$.

The claim is that $cap_k U_k=C$. Clearly $Csubsetcap_k U_k$ because $Csubset U_k$ for all $k$.

Assume that $xincap_k U_k$. Then for each $k$ there is an $n_kinmathbb{Z}$ such that $xin(n_k/2^k-1/2^{k+1},(n_k+1)/2^k+1/2^{k+1})subset U_k$. But this interval contains some point $c_kin C$ by definition of $U_k$. Therefore, $c_kto x$ because the sizes of those intervals $1/2^k-1/2^{k+1}to0$. Therefore $xin C$ because $C$ is closed.

Every Closed Set In $R^1$ is intersection of countable collection of open set.

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