I have a conceptual confusion about the open sets in $\mathbb{R}_\mathrm{standard}$ and in the $\mathbb{R}_\mathrm{lower\, limit}$.
I know that open intervals, in the form of $(a,b)$, are basic open sets (i.e. basis elements) of $\mathbb{R}_\mathrm{standard}$. For $\mathbb{R}_\mathrm{lower\, limit}$, the basis elements are in the form of $[a,b)$. And I also understand that any open set is a union of basis elements.
But, is it correct to say that an open set of $\mathbb{R}_\mathrm{standard}$ is just an open interval $(a,b)$?
I've heard of this quite a few times. But I tend to disagree with this. For example, $(1,2)\cup(10,11)$ is open in $\mathbb{R}$, but it is not in the form of $(a,b)$. Similarly, $[1,2)\cup[10,11)$ is open in $\mathbb{R}_\mathrm{lower\, limit}$, but it's not in the form of $[a,b)$ either.
So rigorously, should we always talk about an arbitrary open set as a union of basis elements, or I have some misunderstanding about the concepts?
Thanks a lot!
You're right that the open sets in $\mathbb{R}$ are exactly the unions of open intervals.
Likewise for $\mathbb{R}_l$ and sets of the form $[a,b), a< b$.
Note that all usual (Euclidean) open sets are also lower-limit open, e.g. we can write $(a,b) = \bigcup\{[c,b): a < c < b\}$ so unions of open intervals are also unions of half-open ones.
The union fact is most often stated (and checked) as
$O$ is Euclidean open iff for all $x \in O$ there is some open interval $(a,b)$ such that $x \in (a,b) \subseteq O$.
The same holds mutatis mutandis for lower limit open sets and sets of the form $[a,b)$. It's what being a base for the topology means. We can often prove enough (e.g. about continuity, convergence etc.) by considering these basic open sets instead of general open sets (their unions).