So I know that sets of the form $(0,b)$ for $b<1$ are open in the $(0,1)_l,$ where $l$ denotes the lower limit topology. But my question is, why is the set say $(0,0.5)$ open in $(0,1)_l?$ Because from how I understand open sets, given any $a\in (0,0.5)$, we can find some $B_{\epsilon}(a)\subset (0,0.5)$.
Indeed this seems to be true, but what I have trouble seeing is that for any $x\in (0,1)_l,$ say $x=0$, then $x \notin B_{\epsilon}(a)\subset (0,0.5)$.
My question is, if we choose some $x$ in the set $R$ where the set $U$ is open in, does it have to hold $\forall x \in R$, such that $x\in B_{\epsilon}(x)\subset U$ and $x \in U?$
$[x,y)$ is open in $(0,1)_l$ when $0<x<y<1.$ And $(0,0.5)=\bigcup_{(0<x<0.5)}[x,0.5).$