Let $\tau$ be a lower limit topology (also called the Sorgenfrey topology) on $\mathbb{R}$. If $a<b$, then for an interval $A=[a,b)$ on the real number line what is the boundary points w.r.t $\tau$?
I'm inclined to say that $a$ and $b$ are not boundary points. Explanations hereto are
for a: Let $a<\mu < b$. Since $[a, \mu) \in N(a)$, where $N(a)$ is the set of all neighborhoods of $a$, and $[a, \mu) \subset A$, $a$ can not be a boundary point of $A$.
for b: Let $b< \epsilon$. Since $[b, \epsilon) \in N(b)$ and $A \cap [b, \epsilon) = \emptyset$, $b$ can not be a boundary point of $A$.
Hence, to me it makes sense if the boundary points of $A$ would look something like $\verb!{!a-\varepsilon, b-\varepsilon \verb!}!$, where $\varepsilon>0$. However, $b-\varepsilon$ seems a little fishy.
Please don't hold back as any help is appreciated.
Best regards,
kasp9201