# Is $X$ compact and connected

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

1. Compact and connected
2. Hausdorff and Connected
3. Hausdorff and compact
4. Neither Compact nor connected.

Since in $X,\tau$ we have no two disjoint open sets the set $X$ is not Hausdorff and also not connected. Also $X= A_1\cup A_2\cup..$. Then $X$ cant have any other cover otherwise if we delete any element from the cover then we are unable to cover $X$. So $X$ is connected and compact.

But the answer is given to be $4$. Am I wrong?

• The definition of "compact" varies according to authors: some define "compact" to imply "Hausdorff" (and use "quasi-compact" for "compact-but-not-necessarily-Hausdorff") whereas others do not. Any question of this sort should always clarify what definitions are being used. Jan 19, 2016 at 13:53
• Isn’t $\{A_n \mid n \in \mathbb{N}\} \cup \{\emptyset\}$ already a topology? If so, then this seems to be connected. Jan 19, 2016 at 13:56
• The correct answer is 1. Jan 19, 2016 at 13:56
• @Crostul At least for those who do not require Hausdorff for compact (which seems to be the case for the OP, given the formulation of 3) Jan 19, 2016 at 13:58
• I have edited the question @Crostul Jan 19, 2016 at 13:59

The only neighbourhood of $1$ is the whole space $X$ from where it follows that $X$ is not Hausdorff. This also implies that $X$ does not satisfy $T_1$ axiom. The space is connected since there does not exist two open disjoint sets as you claimed above. The space is compact since every open cover of $X$ contains an open set that contains $1$. The only open set that contains $1$ is $X$. Therefore $X$ is in every cover of $X$. Therefore every open cover of $X$ has a finite subcover.