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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.

Nebo Alex
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kumhmb
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3 Answers3

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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 $0 \in \overline{(0,1)}$. But $0 \notin (0,1)$.

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

Hope this helps.

$$\;\;\;$$

WhySee
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    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 09:13
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    Thanks a lot for editing my answer. – WhySee May 26 '14 at 08:13
  • 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
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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.

Matt R.
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  • 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 06 '13 at 04:19
  • 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 06 '13 at 04:34
  • 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 06 '13 at 04:40
  • 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 09 '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 09 '13 at 14:18
  • Are the points 0 and 1? – kumhmb Dec 09 '13 at 14:21
  • Indeed they are! So now you know that the complement of $(0, 1)$ isn't open, which means that $(0, 1)$ is not closed. – Matt R. Dec 09 '13 at 14:37
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I'm going to attempt an answer to this based upon the other comments, hints and answers as part of an exercise in undertsanding the lower limit topology - I'd appreciate any pointers and comments.

I think any base of $\Bbb R_l$ contains subsets only of $\Bbb R$ of the form: $\{x:a\leq x<b:a,b\in\Bbb R\}$

The complement of $(0,1)$ in $\Bbb R$ is $(-\infty,0]\cup[1,\infty)$

$0$ is in the complement but by definition of the topology, every open set containing $0$ must have as a subset one of the form $[0,x):x>0$. Since no such set is a subset of the complement, there is no open subset of the complement containing $0$ and therefore the complement is not open. Therefore $(0,1)$ is not the complement of an open set and is therefore not closed.

Now to show it is open:

$$\bigcup_{0<x<1}[x,1)=(0,1)$$

Since $(0,1)$ is a union of open sets, it's open.

it's a hire car baby
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