[proofplan] We show each family generates $\mathcal{B}(\mathbb{R})$ by establishing a cycle of inclusions: $\sigma(\text{open sets}) = \sigma(\text{open intervals}) = \sigma(\text{half-open intervals}) = \sigma(\text{open rays}) = \sigma(\text{closed rays}) = \sigma(\text{closed sets})$. The strategy for each step is: (i) express each generating set from one family as a countable union, countable intersection, or complement of sets from the other family, and (ii) invoke the closure of $\sigma$-algebras under these operations. Finally, the rational-endpoints claim follows from the fact that every open interval $(a,b)$ is a countable union of open intervals with rational endpoints. [/proofplan] [step:Establish $\sigma(\text{open sets}) = \sigma(\text{closed sets})$] Let $\mathcal{O}$ denote the open sets and $\mathcal{C}$ the closed sets of $\mathbb{R}$. Every closed set $F$ satisfies $F = \mathbb{R} \setminus U$ for some open set $U$. Since $U \in \sigma(\mathcal{O})$ and $\sigma$-algebras are closed under complements, $F \in \sigma(\mathcal{O})$. Therefore $\mathcal{C} \subset \sigma(\mathcal{O})$, and since $\sigma(\mathcal{C})$ is the smallest $\sigma$-algebra containing $\mathcal{C}$, $\sigma(\mathcal{C}) \subset \sigma(\mathcal{O})$. Conversely, every open set $U$ is the complement of a closed set, so $U \in \sigma(\mathcal{C})$, giving $\sigma(\mathcal{O}) \subset \sigma(\mathcal{C})$. Therefore $\sigma(\mathcal{O}) = \sigma(\mathcal{C})$. By definition, $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{O})$, so $\mathcal{B}(\mathbb{R})$ is also generated by the closed sets. [/step] [step:Show $\sigma(\text{open intervals}) = \mathcal{B}(\mathbb{R})$] Let $\mathcal{I} := \{(a,b) : a < b,\, a,b \in \mathbb{R}\}$. Since each $(a,b)$ is open, $\mathcal{I} \subset \mathcal{O}$, so $\sigma(\mathcal{I}) \subset \sigma(\mathcal{O}) = \mathcal{B}(\mathbb{R})$. For the reverse inclusion, every open set $U \subset \mathbb{R}$ is a countable union of open intervals: by the structure theorem for open subsets of $\mathbb{R}$, $U = \bigsqcup_{k=1}^\infty (a_k, b_k)$ for some (possibly finite) collection of pairwise disjoint open intervals. Each $(a_k, b_k) \in \sigma(\mathcal{I})$, and $\sigma$-algebras are closed under countable unions, so $U \in \sigma(\mathcal{I})$. Therefore $\mathcal{O} \subset \sigma(\mathcal{I})$, giving $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{O}) \subset \sigma(\mathcal{I})$. [guided] The structure theorem states that every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals (this follows from the second-countability of $\mathbb{R}$: each connected component of $U$ is an open interval, and there are at most countably many since each contains a rational). This is the key topological fact connecting open sets to open intervals. [/guided] [/step] [step:Show $\sigma(\text{half-open intervals}) = \mathcal{B}(\mathbb{R})$] Let $\mathcal{H} := \{(a,b] : a < b\}$. Since $(a,b] = (a, b+1/n] \cap (-\infty, b] \supset (a,b)$ — more directly, $(a,b) = \bigcup_{k=1}^\infty (a, b - 1/k]$ for $b - a > 0$ (choosing $k$ large enough so $b - 1/k > a$). Each $(a, b-1/k] \in \mathcal{H} \subset \sigma(\mathcal{H})$, and $\sigma$-algebras are closed under countable unions, so $(a,b) \in \sigma(\mathcal{H})$. Therefore $\mathcal{I} \subset \sigma(\mathcal{H})$, giving $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{I}) \subset \sigma(\mathcal{H})$. Conversely, $(a,b] = \bigcap_{k=1}^\infty (a, b + 1/k)$. Each $(a, b+1/k) \in \mathcal{I} \subset \sigma(\mathcal{I}) = \mathcal{B}(\mathbb{R})$, and $\sigma$-algebras are closed under countable intersections, so $(a,b] \in \mathcal{B}(\mathbb{R})$. Therefore $\sigma(\mathcal{H}) \subset \mathcal{B}(\mathbb{R})$. [/step] [step:Show $\sigma(\text{open rays}) = \mathcal{B}(\mathbb{R})$] Let $\mathcal{R}^+ := \{(a, \infty) : a \in \mathbb{R}\}$. Each $(a,\infty)$ is open, so $\sigma(\mathcal{R}^+) \subset \mathcal{B}(\mathbb{R})$. Conversely, $(a,b) = (a, \infty) \cap (-\infty, b) = (a,\infty) \cap (b, \infty)^c$. Since $(b, \infty) \in \sigma(\mathcal{R}^+)$ and $\sigma$-algebras are closed under complements, $(b, \infty)^c = (-\infty, b] \in \sigma(\mathcal{R}^+)$. Also $(a, \infty) \in \sigma(\mathcal{R}^+)$. We need $(-\infty, b)$, not $(-\infty, b]$. Write \begin{align*} (-\infty, b) = \bigcup_{k=1}^\infty (-\infty, b - 1/k] = \bigcup_{k=1}^\infty (b - 1/k, \infty)^c. \end{align*} Each $(b - 1/k, \infty)^c \in \sigma(\mathcal{R}^+)$, so $(-\infty, b) \in \sigma(\mathcal{R}^+)$. Therefore $(a,b) = (a, \infty) \cap (-\infty, b) \in \sigma(\mathcal{R}^+)$, giving $\mathcal{I} \subset \sigma(\mathcal{R}^+)$, and $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{I}) \subset \sigma(\mathcal{R}^+)$. By the same method, $\sigma(\{(-\infty, a) : a \in \mathbb{R}\}) = \mathcal{B}(\mathbb{R})$: since $(-\infty, a)$ is open, $\sigma$-inclusion in $\mathcal{B}(\mathbb{R})$ is immediate. For the reverse, $(a, \infty) = (-\infty, a)^c \in \sigma(\{(-\infty, a)\})$, so $\sigma(\mathcal{R}^+) \subset \sigma(\{(-\infty, a)\})$, and $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{R}^+) \subset \sigma(\{(-\infty, a)\})$. [guided] The strategy in each case is the same: express open intervals as countable set operations (unions, intersections, complements) on the generating family, thereby showing $\mathcal{I} \subset \sigma(\text{family})$, which gives $\mathcal{B}(\mathbb{R}) \subset \sigma(\text{family})$. The reverse inclusion $\sigma(\text{family}) \subset \mathcal{B}(\mathbb{R})$ is immediate since each generator is a Borel set. [/guided] [/step] [step:Show $\sigma(\text{closed rays}) = \mathcal{B}(\mathbb{R})$] Let $\mathcal{R}_c := \{[a, \infty) : a \in \mathbb{R}\}$. Since $[a, \infty)$ is closed, $\sigma(\mathcal{R}_c) \subset \mathcal{B}(\mathbb{R})$. Conversely, $(a, \infty) = \bigcup_{k=1}^\infty [a + 1/k, \infty)$. Each $[a+1/k, \infty) \in \sigma(\mathcal{R}_c)$, so $(a, \infty) \in \sigma(\mathcal{R}_c)$. Therefore $\mathcal{R}^+ \subset \sigma(\mathcal{R}_c)$, giving $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{R}^+) \subset \sigma(\mathcal{R}_c)$. [/step] [step:Rational endpoints suffice] Let $\mathcal{I}_\mathbb{Q} := \{(a,b) : a, b \in \mathbb{Q},\, a < b\}$. Since $\mathcal{I}_\mathbb{Q} \subset \mathcal{I}$, $\sigma(\mathcal{I}_\mathbb{Q}) \subset \sigma(\mathcal{I}) = \mathcal{B}(\mathbb{R})$. Conversely, for any $a, b \in \mathbb{R}$ with $a < b$, choose sequences $a_k \downarrow a$ and $b_k \uparrow b$ with $a_k, b_k \in \mathbb{Q}$ and $a_k < b_k$ (possible by density of $\mathbb{Q}$ in $\mathbb{R}$). Then \begin{align*} (a, b) = \bigcup_{k=1}^\infty (a_k, b_k). \end{align*} Each $(a_k, b_k) \in \sigma(\mathcal{I}_\mathbb{Q})$, so $(a,b) \in \sigma(\mathcal{I}_\mathbb{Q})$. Therefore $\mathcal{I} \subset \sigma(\mathcal{I}_\mathbb{Q})$, giving $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{I}) \subset \sigma(\mathcal{I}_\mathbb{Q})$. The same argument applies to each of the families (iv)--(vii) with rational endpoints: rational density in $\mathbb{R}$ allows every real-endpoint generator to be recovered as a countable union or intersection of rational-endpoint generators. [guided] The density of $\mathbb{Q}$ in $\mathbb{R}$ is the key: for any $a \in \mathbb{R}$, there exist rationals $q_k \downarrow a$ and $r_k \uparrow a$. This allows us to write $(a, b) = \bigcup_k (a_k, b_k)$ with $a_k, b_k \in \mathbb{Q}$, recovering every real-endpoint interval from countably many rational-endpoint intervals. Since $\sigma$-algebras are closed under countable unions, the rational-endpoint family generates the same $\sigma$-algebra. For rays: $(a, \infty) = \bigcup_{k=1}^\infty (a_k, \infty)$ with $a_k \in \mathbb{Q}$, $a_k \downarrow a$. For closed rays: $[a, \infty) = \bigcap_{k=1}^\infty (a_k, \infty)$ with $a_k \in \mathbb{Q}$, $a_k \uparrow a$ (and $a_k < a$). Each case uses the closure of $\sigma$-algebras under countable set operations and the density of $\mathbb{Q}$. [/guided] [/step]