$A=\{1,\frac{1}{2},\frac{1}{3},...\}=\{\frac{1}{n},n\in \mathbb{N}\}$ is closed or open.? my friend say that $A$ is closed.
$$A=\{1,\frac{1}{2},\frac{1}{3},...\}=\{\frac{1}{n},n\in \mathbb{N}\}$$
I cant prove it.. 
Although I think it is not closed.
if it set is closed,, how prove it? 
 A: $0$ is a limit point of $A$ by Archimedean Principle but $0 \notin A$, so $A$ isn't closed.
A: $A$ is not open(with repsect to the usual topology) because it does not contain an open  interval around every point of it.
In general $A$ does not contain any interval.It has empty interior.
Also if $A$ was open then $A=A^o=\emptyset$ which is a contradiction.

Now if you know the sequential characterization of a closed set in a metric space then taking $x_n \in A$ where $x_n= \frac{1}{n}$ then $x_n \to 0$ but $0 \notin A$.
Thus we found a sequence in $A$ which has a limit that does not belong to $A$ thus it is not closed.

Also for another proof we will prove  that $A$ is not closed because its complement is not open.

Assume that $A^c$ is open.We have that $0 \in A^c$ thus exists $\epsilon>0$ such that $0 \in (-\epsilon,\epsilon) \subseteq A^c.$
Now we can use the Archimedeian property.
So for this $\epsilon>0$  exists $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0}< \epsilon$
Thus $\frac{1}{n_0} \in (-\epsilon,\epsilon) \subseteq A^c$ and also $\frac{1}{n_0} \in A$ .
This is a contradiction.So $A^c$ is not open.

