# Lower Limit Topology?

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.

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

$$\;\;\;$$

• 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. – Daniel R May 23 '14 at 9:13 • Thanks a lot for editing my answer. – WhySee May 26 '14 at 8:13 • Consider$ A = \cap_{n \in \Bbb N} [-1/n,1+1/n] $then since any interval of type$[p,q]$is closed in lower limit topology and arbitrary intersection of closed sets is closed hence (0,1) is also closed. Is this right? – Error 404 May 13 '15 at 11:45 To show that$(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)$, and show that it's not open in the LL-topology. • So on the closed part, the complement is just (-infinity,0]U[1,infinity). and then the first part is closed and the second part is open, so it is not open. Is that right? – kumhmb Dec 6 '13 at 4:19 • Correct, but remember that "closed" in topology doesn't necessarily mean "not open". Can you use the negation of the definition of an open set to explain why$(-\infty, 0]$isn't open? – Matt R. Dec 6 '13 at 4:34 • I should probably be more specific when I say "the definition of an open set". The particular definition I mean is what I referred to in the first hint. – Matt R. Dec 6 '13 at 4:40 • I understand that you need to use the negation, but is there any specific way of showing that there does not exist a basis element? Or it is just sort of obvious because of the closed part on b? – kumhmb Dec 9 '13 at 13:23 • There are two points in$A = (-\infty, 0] \cup [1, \infty)$such that no basis element containing either is a subset of$A\$. Which points are they? – Matt R. Dec 9 '13 at 14:18