Proof. It is sufficient to show that the $\sigma$-algebra generated by any of the above two families coincide, and that the resulting $\sigma$-algebra is the Borel $\sigma$-algebra on $\real$.

(1) $\to$ (2): For any $a \in \real$, $(a, \infty) = (-\infty, a)^{c}$.

(2) $\to$ (3): For any $a \in \real$, $[a, \infty) = \bigcap_{n \in \natp}(a - 1/n, \infty)$.

(3) $\to$ (4): For any $a \in \real$, $(-\infty, a) = [a, \infty)^{c}$.

(4) $\to$ (5): For any $a, b \in \real$, $[a, b) = (-\infty, b) \cap (-\infty, a)^{c}$.

(5) $\to$ (6): For any $a, b \in \real$, $[a, b]= \bigcap_{n \in \natp}(a - 1/n, b]$.

(6) $\to$ (7): For any $a, b \in \real$, $(a, b] = \bigcup_{n \in \natp}[a + 1/n, b]$.

(7) $\to$ (8): For any $a, b \in \real$, $(a, b) = \bigcup_{n \in \natp}(a, b - 1/n]$.

(8) $\to$ (1): For any $a \in \real$, $(-\infty, a] = \bigcup_{n \in \natp}\bigcap_{k \in \natp}(-n, a + 1/k]$.

For any $U \subset X$ open and $q \in U \cap \rational$, there exists $r_{q} > 0$ such that $(q - r_{q}, q + r_{q}) \subset U$. In which case,

so (8) generates all open sets in $\real$. Conversely, every element of (8) is open, so the $\sigma$-algebra generated by (8) is the Borel $\sigma$-algebra on $\real$.$\square$