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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – Gro-Tsen
    Jan 19, 2016 at 13:53
  • 1
    $\begingroup$ Isn’t $\{A_n \mid n \in \mathbb{N}\} \cup \{\emptyset\}$ already a topology? If so, then this seems to be connected. $\endgroup$ Jan 19, 2016 at 13:56
  • 1
    $\begingroup$ The correct answer is 1. $\endgroup$
    – Crostul
    Jan 19, 2016 at 13:56
  • $\begingroup$ @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) $\endgroup$ Jan 19, 2016 at 13:58
  • $\begingroup$ I have edited the question @Crostul $\endgroup$
    – Learnmore
    Jan 19, 2016 at 13:59

1 Answer 1


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.

I would say that 1. is the correct answer.


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