I am currently reading Apostol Mathematical Analysis. There was this question
Question : Prove that a closed set in $\mathbb{R}^1$ is the intersection of countable collection of open sets.
[Here $N_a(\varepsilon) =$ the open set $(a-\varepsilon, a+\varepsilon)$ ]
My Attempt : Let $A$ be a closed set. Let $A^c$ denote the complement of $A$. Take $G$ be a collection of neighborhoods where for each $N_a(\varepsilon)\in G $ where $a\in \mathbb{Q}\cap A^c=A'$(Let) and $\varepsilon$ is the minimum real number such that $A\subseteq N_a(\varepsilon) $
As the set of rational numbers are countable hence the set $G$ is also countable. Hence $$A=\bigcap_{S\in G}S$$
Can we write the last statement ?
Best Answer
Here's an approach that works more generally in any metric space. As in the statement of the question, we will let $N_r(x)$ be the set of points whose distance to $x$ is less than $r$. For all $n$, let $$U_n = \bigcup_{a \in A} N_{1/n}(a).$$ This is a union of open sets, and hence is open itself. I claim that $$\bigcap_{n \in \Bbb{N}} U_n = A.$$ It's clear that $A \subseteq U_n$ for each $n$, simply because each $a \in A$ belongs in $N_{1/n}(a) \subseteq U_n$.
On the other hand, suppose that $x \notin A$. Then some open neighbourhood of $x$ exists that fails to intersect $A$, i.e. there is some $\varepsilon > 0$ such that $N_{\varepsilon}(x) \cap A = \emptyset$. Fix some $n$ such that $1/n < \varepsilon$. If we had $x \in U_n$, then some $a \in A$ exists such that $x \in N_{1/n}(a)$. That is, the distance from $x$ to $a$ is no larger than $1/n$, which is less than $\varepsilon$. But then $a \in N_\varepsilon(x) \cap A = \emptyset$, a contradiction. Therefore, $x \notin U_n$, and hence $x \notin \bigcap_{n \in \Bbb{N}} U_n$, completing the proof.