Excuse me , can you see this question , the collection of all open intervals (a,b) together with the one-point sets {n} for all positive and negative integers n is a base for a topology on a real line , i have already prove that it is a base , it remains to describe the interior operation in the space ?
Wh know that x belong to interior of A iff there exist nhood of x contained in A ,,, can we take the nhood as the interval (a,infinity) ?
You probably already know the usual (metric) topology $\tau_e$ on $\Bbb R$, generated by all open intervals. There you have $$\operatorname{Int}_{\tau_e}(A)= \bigcup \{(a,b): a < b; (a,b) \subseteq A\}$$
(The interior of $A$ is just the union of all sets contained in $A$, in general).
This new topology $\tau$ adds $\{n\}$ as new open sets where $n \in \Bbb Z$, and so whenever $n \in A$ fr an integer $n$, this $n$ is also in the interior. So the interior consists of two "parts": the old-fashioned interior, plus all points of $\Bbb Z$:
$$\operatorname{Int}_{\tau}(A)= \operatorname{Int}_{\tau_e}(A) \cup \left(A \cap \Bbb Z\right)$$
We can also (bonus challenge?) find a metric $d$ on $\Bbb R$ such that $\tau$ is exactly the metric topology w.r.t. $d$.