Let us first establish that $(1)$ up to $(4')$ all generate the same $\sigma$-algebra. Denote $\mathcal G_i$ for the collection at point $(i)$, and $\mathcal G'_i$ for that at $(i')$, where $i = 1, 2, 3, 4$. Furthermore, write $\Sigma_i$ for $\sigma \left({\mathcal G_i}\right)$ and $\Sigma'_i$ for $\sigma \left({\mathcal G'_i}\right)$. 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 $a \in \R$): :$\complement_{\overline \R} \left({\left[{a \,.\,.\, +\infty}\right]}\right) = \left[{-\infty \,.\,.\, a}\right)$ :$\complement_{\overline \R} \left({\left[{-\infty \,.\,.\, a}\right]}\right) = \left({a \,.\,.\, +\infty}\right]$ we deduce that: :$\mathcal G_3 \subseteq \Sigma_1, \mathcal G'_3 \subseteq \Sigma'_1$ :$\mathcal G_2 \subseteq \Sigma_4, \mathcal G'_2 \subseteq \Sigma'_4$ 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 $a \in \R$: :$\left[{a \,.\,.\, +\infty}\right] = \displaystyle \bigcap_{n \mathop \in \N} \left({a - \frac 1 n \,.\,.\, +\infty}\right]$ {{MissingLinks|I know, but please find the justification yourself}} and by Sigma-Algebra Closed under Countable Intersection, it follows that: :$\mathcal G_1 \subseteq \Sigma_2, \mathcal G'_1, \subseteq \Sigma'_2$ whence by definition of generated $\sigma$-algebra: :$\Sigma_1 \subseteq \Sigma_2, \Sigma'_1 \subseteq \Sigma'_2$ For the converse inclusion, remark that: :$\left({a \,.\,.\, +\infty}\right] = \displaystyle \bigcup_{n \mathop \in \N} \left[{a + \frac 1 n \,.\,.\, +\infty}\right]$ {{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: :$\left({a \,.\,.\, +\infty}\right] = \displaystyle \bigcup_{\substack{q \mathop \in \Q \\ q \mathop > a}} \left({q \,.\,.\, +\infty}\right]$ {{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 $\overline{\mathcal B}$. Since all elements of $\mathcal G_2$ are open sets in the extended real number space, it follows that: :$\Sigma_2 \subseteq \overline{\mathcal B}$ 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: :$\displaystyle \bigcup \mathcal Q$ where $\mathcal Q$ is a collection of extended real intervals with extended rational endpoints. By Set of Intervals with Extended Rational Endpoints is Countable, $\mathcal Q$ is a subset of a countable set. By Subset of Countably Infinite Set is Countable, $\mathcal Q$ is countable as well. It follows by $\sigma$-algebra axiom $(3)$ that it suffices to show that: :$\left({a \,.\,.\, b}\right) \in \Sigma$ :$\left({a \,.\,.\, +\infty}\right] \in \Sigma$ :$\left[{-\infty \,.\,.\, b}\right) \in \Sigma$ :$\left[{-\infty \,.\,.\, +\infty}\right) \in \Sigma$ 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: :$\left({a \,.\,.\, b}\right) = \left({a \,.\,.\, +\infty}\right] \cap \left[{-\infty \,.\,.\, b}\right) \in \Sigma$ and by Sigma-Algebra Closed under Union: :$\left({-\infty \,.\,.\, +\infty}\right) = \left[{a \,.\,.\, +\infty}\right] \cup \left[{-\infty \,.\,.\, a}\right] \in \Sigma$ Hence $\Sigma = \overline{\mathcal B}$, as desired. {{qed}}