Excuse me , can you see this question, For each positive integer $n$ , let $S_n=\{n,n+1,\ldots\}$ . The collection of all subsets of natural number which contain some $S_n$ is a base for a topology on $\Bbb N$ , Discribe the closure operation in this space ...
I had proved that it is a base , but i can not Describe the closure ..
I think the topology which generated by this base is as same as indiscrete one is it true ?
It's not indiscrete, that only has one open set, namely $N$. While in your topology $S_2 = \{2,3,4,5,\ldots\} \neq N$ is also open.
A typical neighbourhood of $n \in N$ is $S_n$ and so $n \in \overline{A}$ iff every neighbourhood of $N$ intersects $A$ iff $S_n$ intersects $A$ iff $A$ contains a point $m \ge n$. So $$\overline{A}=A^{\downarrow}:= \{m: \exists n \in A: m \le n\}$$
so a "downward closure". So any infinite set is dense, while the closure of $\{100\}$ is just $\{0,1,2,\ldots,100\}$ etc.
In the indiscrete topology $\overline{A}=X$ whenever $A \neq \emptyset$, which is quite different, as you can see.