How do I derive the closure of $\{1/n: n \in \mathbb{N}\}$? So, I have to prove that
$$\mathrm{cl}\left(\left\{\frac{1}{n}: n \in \mathbb N\right\}\right) = \left\{\frac{1}{n}: n \in \mathbb N\right\}\cup\{0\}$$
Here $\mathrm{cl}(A)$ is the closure of set $A$.
Can anyone assist me with starting this proof? To prove that these two sets are equal, do I show that A ⊆ B and B ⊆ A? 
(Sorry, I'm still trying to get the hang of proving…)
 A: Sure, showing one is a subset of the other is essentially what you need to do.

Version 1. Definition of closure of a set $A$ is $\operatorname{cl}(A):= A \cup A'$ where $A'$ is the set of limit points of $A$.
To show your statement, you just need to show that $0$ is the only limit point of $A$ that lies outside of $A$. This boils down to two things:


*

*showing that $0$ is a limit point of $A$

*showing that any other point outside of $A$ is not a limit point



Version 2. Definition of closure of a set $A$ is the intersection of all closed sets containing $A$.
To show your statement, first show that the right-hand side is indeed a closed set. Then by definition, it must contain the closure of the left-hand side. So, the closure is either $\{1/n: n\in \mathbb{N}\}$ or $\{1/n : n \in \mathbb{N}\} \cup \{0\}$. If you show that the former is not closed, then you are finished. [Note that doing this proof will require you to do the same things as in Version 1...]
A: Stating the question:
Let $A = \{\,\tfrac{1}{n}:n = 1\,,2\,, \ldots \,\}$, and let $\overline{A}$ denote the closure of $A$.
We want to show that $\overline{A} = A \cup \{0\}$.
Defition of the closure $\overline{A}$ of $A$:
$\overline{A} = A \cup A'$, where $A'$ is the set of limit point of $A$ in $\mathbb{R}$.
Assuming that we work in $\mathbb{R}$ with the usual $|\, \cdot \,|$-metric (i.e. the distance function $(x,y) \mapsto |\,x-y\,|$). If so, a point $p \in \mathbb{R}$ is a limit point of $A$ iff for every $\epsilon > 0$ $$B(\epsilon , p) \cap (A \setminus \{p\}) \ne \varnothing,$$
where $B(\epsilon,x) = \{\,x \in \mathbb{R} : |\,x-p\,| < \epsilon \,\}$ (called the open ball of radius $\epsilon$ and center $x$.)
Something like this (or how is it done in your book?).
Solution
From this we see that our task is to show that the only limit point to $A$ is $0$.
So a good begining might be to show that $0$ is a limit point and from there go on to show that any other point $p \ne 0$ is not a limit point.
