Let $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \mathcal J^n_{ho}$. Then: :$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) = \left(({-\infty \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf a \,.\,.\, +\infty}\right))$ provides a way of writing this half-open $n$-rectangle as an intersection of an open and a closed set. By Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed, these are both in $\mathcal B \left({\R^n}\right)$, and so Sigma-Algebra Closed under Intersection yields: :$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \sigma \left({\mathcal O^n}\right)$ Hence, by definition of generated $\sigma$-algebra: :$\sigma \left({\mathcal J^n_{ho}}\right) \subseteq \sigma \left({\mathcal O^n}\right)$ Denote $\mathbf 1 = \left({1, \ldots, 1}\right) \in \R^n$. Define, for all $k \in \N$, $\mathcal S \left({k}\right)$ by: :$\mathcal S \left({k}\right) := \left\{{ \left[\left[{2^{-k}\mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) : \mathbf j \in \Z^n}\right\}$ It is immediate that $\displaystyle \bigcup \mathcal S \left({k}\right) = \R^n$ and $\mathcal S \left({k}\right) \subseteq \mathcal J^n_{ho}$. Also, $\mathcal S \left({k}\right)$ is countable from Cartesian Product of Countable Sets is Countable. Now define, again for all $k \in \N$, $U_k$ by: :$\displaystyle U_k := \bigcup \, \left\{{S \in \mathcal S \left({k}\right): S \subseteq U}\right\}$ From Set Union Preserves Subsets, $U_k \subseteq U$. Also, $U_k \in \sigma \left({\mathcal J^n_{ho}}\right)$ since the union is countable. It follows that also $\displaystyle \bigcup_{k \mathop \in \N} U_k \in \sigma \left({\mathcal J^n_{ho}}\right)$. Next, it is to be shown that $\displaystyle \bigcup_{k \mathop \in \N} U_k = U$. Note that Set Union Preserves Subsets ensures $\displaystyle \bigcup_{k \mathop \in \N} U_k \subseteq U$. For the converse, let $\mathbf x \in U$. As $U$ is open, there exists an $\epsilon > 0$ such that the open ball $B \left({\mathbf x; \epsilon}\right)$ is contained in $U$. Fix $k \in \N$ such that $\sqrt n \, 2^{-k} < \epsilon$, and find $\mathbf j \in \Z^n$ such that: :$\mathbf x \in \left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)$ Now it is to be shown that: :$\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) \subseteq B \left({\mathbf x; \epsilon}\right)$ To this end, observe that for any $\mathbf y \in \left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)$, it holds that: :$d \left({\mathbf x, \mathbf y}\right) \le \operatorname{diam} \left({\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)}\right)$ by definition of diameter. Now from Diameter of Rectangle, the right-hand side equals: :$\left\Vert{2^{-k} \left({\mathbf j + \mathbf 1}\right) - 2^{-k} \mathbf j}\right\Vert = \left\Vert{2^{-k} \mathbf 1}\right\Vert = \sqrt{n} \, 2^{-k}$ which is smaller than $\epsilon$ by the way $k$ was chosen. Hence: :$\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) \subseteq B \left({\mathbf x; \epsilon}\right)$ and so every $\mathbf x \in U$ is contained in some $U_k$. Thus it follows that $U \subseteq \displaystyle \bigcup_{k \mathop \in \N} U_k$. Thereby we have shown that: :$\sigma \left({\mathcal J^n_{ho}}\right) = \sigma \left({\mathcal O^n}\right)$ {{qed}}