This is a proof-verification (or more like counterexample-verification) request.
Let $\Omega$ be a non-empty set and $\geq$ a linear order (a complete, transitive, antisymmetric order) on it. Define \begin{align*} \mathscr L\equiv&\;\big\{\{\psi\in\Omega\,|\,\psi<\omega\}\,\big|\,\omega\in\Omega\big\},\\ \mathscr U\equiv&\;\big\{\{\psi\in\Omega\,|\,\psi>\omega\}\,\big|\,\omega\in\Omega\big\} \end{align*} to be the family of lower sets and that of upper sets, respectively.
The order topology $\tau$ on $\Omega$ is the topology which has $\mathscr L\cup\mathscr U$ as its topological subbasis; it is the coarsest topology according to which all lower and upper sets are open. From this topology, one can generate the Borel $\sigma$-algebra $\mathscr B\equiv\sigma(\tau)$, which is the smallest $\sigma$-algebra according to which sets open according to the order topology are measurable.
My goal is to show that, in general, \begin{align*} \mathscr B\neq\sigma(\mathscr L\cup\mathscr U);\tag{$\clubsuit$} \end{align*} that is, the $\sigma$-algebra generated by the lower and upper sets is not large enough to give the Borel $\sigma$-algebra.
Remark: In some special cases, $\mathscr B$ and $\sigma(\mathscr L\cup \mathscr U)$ do coincide—for example, when $\Omega=\mathbb R$ with the usual order, in which case the order topology is the standard Euclidean topology, whose Borel $\sigma$-algebra is well-known to be generated by $\mathscr L\cup\mathscr U$ (and even by $\mathscr L$ alone or by $\mathscr U$ alone). More generally, if the order topology $\tau$ is second-countable, then $\mathscr L\cup\mathscr U$ does generate $\mathscr B$; the details can be found here. However, as a rule, the two $\sigma$-algebras $\mathscr B$ and $\sigma(\mathscr L\cup\mathscr U)$ may be different.
In particular, I have the following supposed counterexample in mind. Let $(\Omega,\geq)$ be the set of countable ordinals. This is a linearly ordered set such that
- $\Omega$ is uncountable; but
- each lower set in $\mathscr L$ is countable; furthermore,
- every non-empty subset of $\Omega$ has a least element with respect to the order $\geq$.
The first step is to show that \begin{align*} \sigma(\mathscr L\cup\mathscr U)=\mathscr C, \end{align*} where $\mathscr C$ is the $\sigma$-algebra of countable and co-countable sets: \begin{align*} \mathscr C\equiv\{A\subseteq\Omega\,|\,\text{$A$ is countable or $A^{\mathsf c}$ is countable}\}. \end{align*} Clearly, each set in $\mathscr L$ is countable and each set in $\mathscr U$ is co-countable, so that $\sigma(\mathscr L\cup\mathscr U)\subseteq\mathscr C$. Conversely, note that for each $\omega\in\Omega$, the singleton $\{\omega\}$ is the intersection of the complement of a set from $\mathscr L$ and the complement of another set from $\mathscr U$. Therefore, the singletons are in $\sigma(\mathscr L\cup\mathscr U)$, which implies that $\mathscr C$ is included in $\sigma(\mathscr L\cup\mathscr U)$.
Therefore, in order to prove ($\clubsuit$), it is sufficient to exhibit a set that is open in the order topology (and hence a fortiori Borel measurable) but is neither countable nor co-countable. To this end, define the immediate-successor function $S:\Omega\to\Omega$ as follows: for each $\omega\in\Omega$, $S(\omega)\in\Omega$ is the (unique) element of $\Omega$ such that
- $S(\omega)>\omega$; and
- there exists no $\psi\in\Omega$ such that $S(\omega)>\psi>\omega$.
More formally, $S(\omega)$ is the least element of the upper set $\{\psi\in\Omega\,|\,\psi>\omega\}$.
It is easily seen that for each $\omega\in\Omega$, one has \begin{align*} \{S(\omega)\}=\{\psi\in\Omega\,|\,\psi<S(S(\omega))\}\cap\{\psi\in\Omega\,|\,\psi>\omega\}, \end{align*} so that the set \begin{align*} U\equiv S(\Omega)=\bigcup_{\omega\in\Omega}\{S(\omega)\} \end{align*} is open in the order topology. What is more, since $U$ is a union of open singletons, any subset of $U$ is open, too. In addition, the function $S$ is injective, so that $U$ is uncountable. Therefore, $U$ can be partitioned into two disjoint uncountable subsets, say $V$ and $W$. Hence, $V$ is an uncountable open set and since $W\subseteq V^{\mathsf c}$, the complement of $V$ is also uncountable. It follows that $V\notin\sigma(\mathscr L\cup\mathscr U)$. This observation concludes the construction of the desired counterexample.
Any comments and suggestions are greatly appreciated.
