A question from my homework is
Let $(X,<)$ be a totally ordered set. Let us examine $X$ with the order topology.
a. Prove that $\overline{(a,b)} \subseteq [a,b]$ (where the overline denotes closure of the set). Give an example where the inclusion is strict.
b. Give necessary and sufficient conditions such that $a\in\overline{(a,b)}$ and such that $b\in\overline{(a,b)}$
So I've managed a. easily enough. I think a proper example of strict inclusion would be a finite set - say $\{1, 2, 3, 4, 5\}$ with the obvious order. Then $(2,4)=\{3\}$ so indeed $ \overline{(2,4)} = \{3\}$ and not $[2,4]$.
I'm stumped about the conditions though. Can anyone give me some help? Thanks!
Indeed a) follows from the fact that $[a,b]$ is closed, which is clear as the complement is the union of $\{x: x < a \}$ and $\{x: x > b\}$, which are open by definition of the order topology. The example of strictness is fine.
For $a$ to be in $\overline{(a,b)}$, we need every neighbourhood of $a$ to contain points $> a$. This happens exactly when $a$ has no right neighbour, i.e $a^{+} = \min \{x : x > a \}$ does not exist. Equivalently, there is no $a^{+} \in X$ such that $(a, a^{+}) = \emptyset$.
Examples of ordered sets where these exist, besides the obvious discrete ones like $\mathbb{Z}$ include $[0,1] \times \{0,1\}$ in the lexicographic ordering, where $(x,0)$ has right neighbour $(x,1)$.
For the right hand side $b$ we want of course for no left neighbour to exist, the definition of which should be guessable...