Compactness and connectedness of the topological space?

Question

Let $X=\mathbb N$ be equipped with the topology generated by the basis consisting of sets $A_n = \{n,n+1,n+2,\ldots\} ,n \in \mathbb N $ . Then $X$ is

1. compact and connected

2. Hausdorff and connected

3. Hausdorff and compact

4. Neither compact nor connected

My attempt is

Let $ A_n \text{ and } A_m$ be two distinct basis elements . If $n > m$, then $A_n \cap A_m = A_n$. So intersection of two non empty open set is not empty. So $(X,\tau)$ is not Hausdorff but connected .

Please tell me about compactness.

Thank you.

Answer 1

First note that the topology generated by the basis you describe above consists of the following sets: $$\{ \varnothing \} \cup \{ A_n : n \in \mathbb{N} \},$$ i.e., the only open set you have not explicitly mentioned is the empty set. This follows from the fact that $A_m \cap A_n = A_{\max \{m,n\}}$, and for a nonempty $B \subseteq \mathbb{N}$, $\bigcup \{ A_n : n \in B \} = A_{\min B}$. From here you should be able to easily conclude that $X$ is not T₁, let alone Hausdorff. (I guess this only makes sense if you have been introduced to the T₁ property.)

For compactness I'll just leave the following

Hint: Which sets in this topology contain the least natural number (be that $0$ or $1$)?