# Proving a certain topology to be non-Hausdorff

I'm doing this exercise and I don't know how to attack the second part:

Let $$\tau$$ be a topology on $$\mathbb{N}$$, with $$\tau=\{\emptyset\}\cup\{\{0,1,2,...,n\}|n\in\mathbb{N}\}\cup\{\mathbb{N}\}$$.

$$i$$) Prove that $$(\tau,\mathbb{N})$$ is not Hausdorff.

$$ii$$) Prove that the sequence $$\{1,2,1,2,1,2,1,...\}$$ is not convergent to $$0$$ and $$1$$, but it does to any other natural number (it converges to $$2$$, to $$3$$, to $$4$$, ...)

I've tried to solve those 2 parts:

$$i$$) Let $$m,n \in \mathbb{N}$$, $$n\neq m$$. Suppose, wlog, that $$n. $$n\in I_n=\{0,1,2,3,...,n\}$$ and it's the smallest open set that contains $$n$$. $$m\in I_m=\{0,1,2,3,...,m\}$$ and it's the smallest open set that contains $$m$$. But $$I_n\subset I_m$$. So $$(\tau,\mathbb{N})$$ is not Hausdorff.

I don't know if that try is correct. And how we do attack the second part? I've been trying to find an effective methid to prove that but I'm not getting anywhere.

i) It is not Hausdorff because every open set that contains $$1$$ also contains $$0$$.
ii) The sequence does not converge to $$0$$ because $$\{0\}$$ is an open set that contains $$0$$ but no term of the sequence. It doesn't converge to $$1$$ because $$\{0,1\}$$ is an open set that contains $$1$$ for which it is not true that contains all but finitely many terms of the sequence. On the other hand, if $$n\in\mathbb{N}\setminus\{0,1\}$$ and if $$A$$ is an open set containing $$n$$, then $$A\supset\{0,1,\ldots,n\}$$ and therefore every term of the sequence belongs to $$A$$. So, the sequence converges to $$n$$.