Show that $\mathcal U$ forms a topology on $X= \mathbb N$, where $$\mathcal U := \{\emptyset,\mathbb N \} \cup \{O_n : n\geq 1\},$$ where $O_n = \{n,n+1,n+2,\dots \}$.
i) Clearly the emptyset and $\mathbb N$ are open by construction.
ii) Let $\{O_\alpha : \alpha \in I\}$ be an arbitrary family of sets indexed over some $I\subseteq \mathbb N$. Then $\bigcup_{\alpha \in I} O_\alpha = O_{\min I }$ because $O_j \subseteq O_k$ when $j \geq k$. Notably, if the family includes $O_1$ then the union is simply $\mathbb N$ itself.
iii) When $I$ is assumed to be a finite subset of $\mathbb N$, then the intersection $\bigcap_{\alpha \in I} O_\alpha = O_{\max I}$ since $O_a \cap O_b = O_b$ when $b\geq a$.
Is this correct?