Theorem: Generators for Extended Real Sigma-Algebra

Let $\overline \BB$ be the extended real $\sigma$-algebra.
Then $\overline \BB$ is generated by each of the following collections of extended real intervals:
{{begin-eqn}}
{{eqn | n = 1
      | o = 
      | r = \set {\ \closedint a \to: a \in \R}
}}
{{eqn | n = 1'
      | o = 
      | r = \set {\ \closedint a \to: a \in \Q}
}}
{{eqn | n = 2
      | o = 
      | r = \set {\ \hointl b \to: b \in \R}
}}
{{eqn | n = 2'
      | o = 
      | r = \set {\ \hointl b \to: b \in \Q}
}}
{{eqn | n = 3
      | o = 
      | r = \set {\ \hointr \gets c: c \in \R}
}}
{{eqn | n = 3'
      | o = 
      | r = \set {\ \hointr \gets c: c \in \Q}
}}
{{eqn | n = 4
      | o = 
      | r = \set {\ \closedint \gets d: d \in \R}
}}
{{eqn | n = 4'
      | o = 
      | r = \set {\ \closedint \gets d: d \in \Q}
}}
{{end-eqn}}

Proof:
Let us first establish that $(1)$ up to $(4')$ all generate the same $\sigma$-algebra.
Denote $\GG_i$ for the collection at point $(i)$, and $\GG'_i$ for that at $(i')$, where $i = 1, 2, 3, 4$.
Furthermore, write $\Sigma_i$ for $\map \sigma {\GG_i}$ and $\Sigma'_i$ for $\map \sigma {\GG'_i}$.
Here $\sigma$ denotes generated $\sigma$-algebra.
By Generated Sigma-Algebra Preserves Subset, we have the following inclusions:
:$\Sigma'_i \subseteq \Sigma_i$
for $i = 1, 2, 3, 4$.
Since we have the following observations about complements in $\overline \R$ (for arbitrary $a \in \R$):
:$\relcomp {\overline \R} {\closedint a \to} = \hointr \gets a$
:$\relcomp {\overline \R} {\closedint \gets a} = \hointl a \to$
we deduce that:
:$\GG_3 \subseteq \Sigma_1, \GG'_3 \subseteq \Sigma'_1$
:$\GG_2 \subseteq \Sigma_4, \GG'_2 \subseteq \Sigma'_4$
and by definition of generated $\sigma$-algebra:
:$\Sigma_3 \subseteq \Sigma_1, \Sigma'_3 \subseteq \Sigma'_1$
:$\Sigma_2 \subseteq \Sigma_4, \Sigma'_2 \subseteq \Sigma'_4$
By Complement of Complement, the converse inclusions:
:$\Sigma_1 \subseteq \Sigma_3, \Sigma'_1 \subseteq \Sigma'_3$
:$\Sigma_4 \subseteq \Sigma_2, \Sigma'_4 \subseteq \Sigma'_2$
are derived.
Subsequently, remark that, for all $a \in \R$:
:$\closedint a \to = \ds \bigcap_{n \mathop \in \N_{>0}} \hointl {a - \frac 1 n} \to$
{{MissingLinks|I know, but please find the justification yourself}}
and by Sigma-Algebra Closed under Countable Intersection, it follows that:
:$\GG_1 \subseteq \Sigma_2, \GG'_1, \subseteq \Sigma'_2$
whence by definition of generated $\sigma$-algebra:
:$\Sigma_1 \subseteq \Sigma_2, \Sigma'_1 \subseteq \Sigma'_2$
For the converse inclusion, remark that:
:$\hointl a \to = \ds \bigcup_{n \mathop \in \N_{>0}} \closedint {a + \frac 1 n} \to$
{{MissingLinks|again, justification}}
and thus immediately establish:
:$\Sigma_2 \subseteq \Sigma_1, \Sigma'_2 \subseteq \Sigma'_1$
To summarize, the above arguments establish:
:$\Sigma'_1 = \Sigma'_2 = \Sigma'_3 = \Sigma'_4 \subseteq \Sigma_4 = \Sigma_3 = \Sigma_2 = \Sigma_1$
Finally, for all $a \in \R$, we have:
:$\hointl a \to = \ds \bigcup_{\substack {q \mathop \in \Q \\ q \mathop > a}} \hointl q \to$
{{explain|via Between two Real Numbers exists Rational Number}}
whence $\Sigma_2 \subseteq \Sigma'_2$, and all eight $\sigma$-algebras are equal; denote them by $\Sigma$ from now on.
It remains to establish that in fact they equal $\overline \BB$.
Since all elements of $\GG_2$ are open sets in the extended real number space, it follows that:
:$\Sigma_2 \subseteq \overline \BB$
By Intervals with Extended Rational Endpoints form Countable Basis for Extended Real Number Space, every open set in the extended real number space is of the form:
:$\ds \bigcup \QQ$
where $\QQ$ is a collection of extended real intervals with extended rational endpoints.
By Set of Intervals with Extended Rational Endpoints is Countable, $\QQ$ is a subset of a countable set.
By Subset of Countably Infinite Set is Countable, $\QQ$ is countable as well.
It follows by $\sigma$-algebra axiom $(3)$ that it suffices to show that:
:$\openint a b \in \Sigma$
:$\hointl a \to \in \Sigma$
:$\hointr \gets b \in \Sigma$
:$\closedint \to \gets \in \Sigma$
for all rational numbers $a, b \in \Q$.
The middle two are in $\Sigma'_2$ and $\Sigma'_4$, respectively, hence in $\Sigma$.
By Sigma-Algebra Closed under Intersection:
:$\openint a b = \hointl a \to \cap \hointr \gets b \in \Sigma$
and by Sigma-Algebra Closed under Union:
:$\closedint \to \gets = \closedint a \to \cup \closedint \gets a \in \Sigma$
Hence $\Sigma = \overline \BB$, as desired.
{{qed}}