what is the interior and the closure of $\{4\}$ and of $\{2,4,6,8.....\}$? let $n \in \mathbb{N}$ . Set $A_n = \{ n , n+1, n+2,....\} $ and $ T = \{ \emptyset , A_n\}_{ n\in \mathbb{N} }$
$1.$Is  $T $is separated ?
$2.$ what are dense  set in $T$ ?
$3.$ what is the interior and the  closure of $\{4\}$  and of $\{2,4,6,8.....\}$?
My attempt :  for  $ 1,$  I know  that $ T$ is not separated because  take $A_1 = \{ 1, 2,3,...\}$ and $A_2 = \{ 2, 3,4,,,,,,,\}$ here both $A_1 \cap A_2 \ne \emptyset$
For  $2)$ i thinks  $\mathbb{N} $
For $3)$ im confused
Any hints/solution will be appreciated
thanks u
 A: Your answers:
1) Incorrect. The fact that two particular open sets intersect does not imply at all that the topology is not separated. You should pick two natural numbers $m,n$ (call $m$ the lesser one, for example) and try to show that this points can not be separated.
2) Incorrect. Of course $\Bbb N$ is dense. But how are the closed sets? What would be the closure of the set of, say, even numbers?
For the third one, perhaps you have a clearer idea of how the topology is. Good try!
A: $1)$ Actually, $A_n\cap A_m\neq \emptyset$ for every $n,m\in\Bbb{N}$. That should help you prove that no pair of points can be separated. 
$2)$ A subspace $S$ is dense when its closure $\overline{S}$ is the whole space. A point $x\in\overline{S}$ is a point for which every neighbourhood contains a point of $S$. It is not hard to prove that $S$ is dense if and only if $S$ is infinite. If $S$ is finite, then it has a maximum element $s_0$, so any point $s>s_0$ has a neighbourhood which does not intersect $S$. Similarly, if for every $n\in \Bbb{N}$ there is a neighbourhood containing a point in $S$, then $S$ must be infinite.
$3)$ Their interior is empty because no open set is contained in $\{4\}$ nor in $\{2,4,6,\dots\}$.
