I was working with the book Extensions and Absolutes of Hausdorff Spaces book by Porter and I got stuck with an exercise.
Let $\{(r_j,t_j) \mid j\in J \}$ be a pairwise disjoint collection of nonempty open intervals of the ordered space $L$. Then:
(1) Let $b\in\text{cl}_{L}\left(\bigcup \{(r_j,t_j) \mid j\in J \} \right)\setminus \bigcup\left\{ \text{cl}_{L}(r_j,t_j)\mid j\in J \right\}$. If $b\in (c,d)$ then there exist $j_{0}\in J$ such that $(r_{j_0},t_{j_0})\subseteq (c,d)$.
(2) Let $x\in L$. Then $\left|\left\{j\in J\mid x\in \text{cl}_{L}(r_j,t_j) \right\} \right|\leq 2$
For (1), the contradiction is the first tool that I tried, but doesn't works. I don't know how use the fact that $(r_{j},t_j)\not\subseteq(c,d)$ for all $j\in J$. Then, I tried to prove that $c<\sup\{r_j\mid j\in J \}$ and $\inf \left\{t_j\mid j\in J \right\}<d$. Then, use the two inequalities to take $r_j$ and $t_k$ such that $c<r_j<t_k<d$ but, there is a problem because $(r_j,t_k)$ is not an element of the original collection. How can I refine the argument to obtain the desired result? Or am I missing something?
For (2) is clearly that there exist a unique $j\in J$ such that $x\in (r_j, t_j)$ or $x\in L\setminus\bigcup\{(r_j,t_j)\mid j\in J \}$. In the second case, we have the desired result. With the first case, suposse that there exist $k,\ell\in J$ such that $x\in\text{cl}_{L}(r_k,t_k)\cap\text{cl}_{L}(r_\ell,t_\ell)$ and by hypothesis $x\in\text{cl}_{L}(r_j,t_j)$. Thus $x\in \text{cl}_{L}(r_k,t_k)\cap\text{cl}_{L}(r_\ell,t_\ell)\cap \text{cl}_{L}(r_j,t_j)$ but $(r_j,t_j)$ is an open neighborhood of $x$ such that $(r_j,t_j)\cap(r_k,t_k)=\emptyset$ and therefore $x\notin\text{cl}_{L}(r_k,t_k)$ and similarly, $x\notin\text{cl}_L(r_\ell,t_\ell)$. With this, if $x$ is in the closure of at least 3 elements of the family, we obtain a contradiction. Finally, the cardinality can be 2 because we could have the case where two elements of the family have the form $(r_k,t_k)$ and $(t_k,t_\ell)$. Clearly, $t_k$ is in the closure of both sets.
I don't know how to solve (1). Any hint? I think that my proof of (2) is correct, but, it is?
Here is an example for which b actually exists.
Perhaps it will give us some insight.
Let the intervals be for each n in N, (1/(n+1), 1/n).
b = 0. The desired statement is true.
There must be an infinite number of intervals for b to exist.