# Prove that every closed set in $R^n$ is the intersection of countable collection of open sets(proof-verification)

I saw only the proof by Lindelof Convering Theorem. But I have different arguments:

My attempt: Suppose $V = (v_1,v_2,...)$ is a collection of open sets and let's define point $p \in \cap V_i$. By definition intersection means, that $p \in V_i$ for $i = 1...n$. Also we know, that every point in $R$ is a limit point. Because $p \in V_i$ we can find $N_\epsilon(p)$ s.t $N \subset \cap V_i$ (Actually here I have proved, that intersection of finite collection of open sets are open). Because $p$ is arbitrary and is a limit point => set $\cap V_i$ contains every limit point => set is closed

• The intersection of a finite collection of open sets is open. – Lord Shark the Unknown Jun 9 '17 at 16:52
• How can you "define" a point in the intersection of a given set of sets? If you prove the intersection is not empty then you can take a point there, but not "define it"... – DonAntonio Jun 9 '17 at 16:54
• finite --> countable makes it true, but this arument shows nothing. – Henno Brandsma Jun 9 '17 at 17:05
• The phrase "is a limit point" by itself has no meaning. You have to say "a limit point of a particular set." It's true that every point in R is a limit point of R. That doesn't help you prove that it's a limit point of A. – Nate Eldredge Jun 9 '17 at 17:30
• And now your argument is really wrong, because you when you write $V = \{V_1, \dots, V_n\}$ you are assuming that $V$ is finite. – Nate Eldredge Jun 9 '17 at 17:32

For every subset $A$, the function $f:X \to \mathbb{R},f(x) = d(x,A) = \inf\{d(x,a): a \in A\}$ is continuous. This holds in all metric spaces.
A set $A$ is closed iff $d(x,A) = 0 \leftrightarrow x \in A$.
So for a closed $A = f^{-1}[\{0\}] = \cap_n f^{-1}[(-\frac{1}{n}, \frac{1}{n})]$ is a $G_\delta$ in any metric space.