Prove the closure of $\{\frac{1}{n}, n \in \mathbb{N}\}$ is $\{ \frac{1}{n}, n \in \mathbb{N}\} \cup \{0\}$. Exercise 13.9 (a).

Notations:

 1. $A^C$ means the complement of $A$, i.e. $A \cup A^C = U$. In
    general, $U = \mathbb{R}$ unless there is a specific explanation.
 
 2. $(s_n)$ means a sequence. For example, $(\frac{1}{n})$ means the
        sequence $(1, 1/2,1/3, 1/4, 1/5, 1/6,...)$, i.e. $s_n = \frac{1}{n}$
        where $n \in \mathbb{N}$.

Proof:

Let $A = \{\frac{1}{n}: n \in \mathbb{N}\} \cup \{0\}$. For proving $A$ is a closed set, it equals to prove $A^C$ is an open set. Observing that the sequence $s_n = \frac{1}{n}$ ($n \in \mathbb{N}$) forming a strictly decreasing sequence, hence, we could construct a bunches of sets such that for set $S_n$: $\sup{S_n}=s_n=\frac{1}{n}$ and $\inf S_n = s_{n+1} = \frac{1}{n+1}$, $\sup S_n \notin S_n$ and $\inf S_n \notin S_n$, then no element from the sequence $s_n$ would be in the sets $S_n$.

Therefore, $A^C = \cup_{n=1}^{+\infty} (s_{n+1},s_n) \cup (-\infty,0) $.

Because the union of ***any*** collection of open sets is an open set, clearly, each $S_n$ is an open set and $(-\infty,0)$ is an open set. Therefore $A^C$ is open [1] (Contrastively, the *intersection* of ***finitely*** many open sets is an open set). Since $A = \mathbb{R} - A^C$ and $A^C$ is an open set, we proved that $A$ is a closed set by the definition of closed set.

Reference:

[1] Elementary Analysis: The Theory of Calculus, 2nd Edition. Kenneth A. Ross. Page 87, discussion 13.7, Springer, 2013.