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…)

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...]
