# Show that $\overline{A}=\bigcap\limits_{n=1}^\infty U_n$

Let $(X,d)$ be a metric space and non-empty $A\subseteq X$. Show that there exists a sequence of open subsets $U_1, U_2,\dots,U_n,\dots$ in $(X,d)$ such that $\overline{A}=\bigcap\limits_{n=1}^\infty U_n$ (where $\overline{A}$ is the closure of $A$).

I was thinking about approaching this problem as follows:

• Take all $x\in X$ such that the distance between $x$ and $A$ is $0$. Where the distance from $x$ to $A$ is defined as $\mbox{dist}(x,A)=\inf\{d(x,a):a\in A\}$. Then $x\in \overline{A}$ (the closure of $A$).

• Form open balls of radius $1/k$ around these $x$.

• Mutually intersect these open balls for all $k$.

• Then $\bigcap\limits_k B(x_i,1/k)=\overline{A}$, since for each $x_i$, this intersection should converge to $\{x_i\}$.

However, if this approach is correct, then how do I formally define this intersection for all $i\in\mathbb{N}$?

UPDATE:

Let $f:X\to \mathbb{R}$, $f(x) := \mbox{dist}(x,A)$, then $f$ is continuous. Thus $f$ pulls back open sets to open sets. Then

$f^{-1}\left(\left(-\frac1n,\frac1n\right)\right)$ is open. Now define

$$\bigcap\limits_{n=1}^\infty f^{-1}\left(\left(-\frac1n,\frac1n\right)\right)=f^{-1}(\{0\})=\overline{A}$$

Do you think this is correct?

• No. Actually, $$\bigcap_kB(x_i,1/k)=\{x_i\}.$$ – bof Oct 27 '17 at 3:14
• @bof Please see my update. – sequence Oct 27 '17 at 3:38
• Your update is correct. – DanielWainfleet Oct 27 '17 at 3:49
• Yes, the update is fine. – bof Oct 27 '17 at 3:57

Consider $U_n=\bigcup _{x\in \overline A}B(x,{\frac{1}{n}})=\{y\in \overline A:d(x,y)<\frac{1}{n}\}$
Since each $U_n$ is an union of open balls with center at $x\in \overline A$ so $\overline A\subseteq \bigcap_{n=1}^\infty U_n$.
Conversely if $x\in U_n\forall n$ then $\exists x_n\in \overline A$ such that $x\in B(x_n,\frac{1}{n})\implies (x_n)_n$ converges to $x\implies x$ is a limit point of $\overline A$ .
But $\overline A$ is closed and hence $x\in \overline A\implies \bigcap U_n\subseteq \overline A$.