2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Did you mean to say that $(X,\tau)$ is not disconnected, i.e. is connected? $\endgroup$ Mar 4, 2014 at 1:36
  • $\begingroup$ @Omnomnomnom:sorry I am edding , it is connectd. $\endgroup$
    – Struggler
    Mar 4, 2014 at 1:39
  • 1
    $\begingroup$ 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 $\mathcal U$. $\endgroup$
    – Dejan Govc
    Mar 4, 2014 at 1:51

1 Answer 1

2
$\begingroup$

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 $1$)?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.