Show $(0,1)$ is open but not closed in the Lower Limit Topology.

I know that $[a,b)$ is open and closed in the lower limit topology, but I am not sure how to prove this one.

Thanks for any help.

  • 1
    $\begingroup$ Do you know what the general open sets in the LL topology? $\endgroup$ – user99680 Dec 6 '13 at 3:54
  • 1
    $\begingroup$ The basis of the topology is all of the half-open intervals of the form [a,b). $\endgroup$ – kumhmb Dec 6 '13 at 3:59

To show $(0,1)$ is open in LL topology note the following $$ x \in [x,1) \subset (0,1) ~~\text{for all $x \in (0,1)$}. $$

Hence $(0,1)$ is open in LL topology.

To show $(0,1)$ not closed in LL topology, we shall show that closure of $(0,1)$ in LL topology is not $(0,1)$.

Take any neighborhood $N$ of $0$. There exists $a>0$ such that $0 \in [0,a) \subset N$

Hence $N \cap (0,1)$ is not empty.

Hence $0 \in cl(0,1)$. But $0 \notin (0,1)$.

Hence $(0,1)$ is not closed in LL topology.

Hope this helps.


  • $\begingroup$ Welcome to math.SE! You seem to know your way around TeX syntax. To have the site render it for you, you need to add $ or $$ around the expressions though. I took the liberty of doing it for you this time, and you can click edit to see how I did it. $\endgroup$ – Daniel R May 23 '14 at 9:13
  • $\begingroup$ Thanks a lot for editing my answer. $\endgroup$ – WhySee May 26 '14 at 8:13
  • $\begingroup$ Consider $ A = \cap_{n \in \Bbb N} [-1/n,1+1/n] $ then since any interval of type $[p,q]$ is closed in lower limit topology and arbitrary intersection of closed sets is closed hence (0,1) is also closed. Is this right? $\endgroup$ – Error 404 May 13 '15 at 11:45

To show that $(0,1)$ is open in the LL-topology, show that there is some basis element $[a,b) \subset (0,1)$ around each $x \in (0,1)$.

To show that $(0,1)$ is not closed, remember that the complement of any closed set must be open. So find the the complement of $(0,1)$, and show that it's not open in the LL-topology.

  • $\begingroup$ So on the closed part, the complement is just (-infinity,0]U[1,infinity). and then the first part is closed and the second part is open, so it is not open. Is that right? $\endgroup$ – kumhmb Dec 6 '13 at 4:19
  • $\begingroup$ Correct, but remember that "closed" in topology doesn't necessarily mean "not open". Can you use the negation of the definition of an open set to explain why $(-\infty, 0]$ isn't open? $\endgroup$ – Matt R. Dec 6 '13 at 4:34
  • $\begingroup$ I should probably be more specific when I say "the definition of an open set". The particular definition I mean is what I referred to in the first hint. $\endgroup$ – Matt R. Dec 6 '13 at 4:40
  • $\begingroup$ I understand that you need to use the negation, but is there any specific way of showing that there does not exist a basis element? Or it is just sort of obvious because of the closed part on b? $\endgroup$ – kumhmb Dec 9 '13 at 13:23
  • $\begingroup$ There are two points in $A = (-\infty, 0] \cup [1, \infty)$ such that no basis element containing either is a subset of $A$. Which points are they? $\endgroup$ – Matt R. Dec 9 '13 at 14:18

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.