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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.

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    $\begingroup$ Do you know what the general open sets in the LL topology? $\endgroup$ – user99680 Dec 6 '13 at 3:54
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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