# Compactness and connectedness of the topological space?

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 .

Thank you.

• Did you mean to say that $(X,\tau)$ is not disconnected, i.e. is connected? Mar 4, 2014 at 1:36
• @Omnomnomnom:sorry I am edding , it is connectd. Mar 4, 2014 at 1:39
• Regarding compactness: suppose $\mathcal U$ is an arbitrary open cover. Choose an $U\in\mathcal U$. This $U$ contains all but finitely many natural numbers, so it only remains to make sure these missing numbers are covered using finitely many elements of $\mathcal U$. Mar 4, 2014 at 1:51

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