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.
 A: 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.
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A: 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.
A: 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.
