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}\smallsetminus\cup 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

Answer 1

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

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

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

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

Answer 2

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 $n\in\mathbb{Z}$. 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 $C\subset\cap_k U_k$ because $C\subset U_k$ for all $k$.

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