Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle

Theorem
Let $\mathcal{O}^n$ be the collection of open subsets of the Euclidean space $\left({\R^n, \tau}\right)$.

Let $\mathcal{J}_{ho}^n$ be the collection of half-open rectangles in $\R^n$.

Then:


 * $\sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right)$

where $\sigma$ denotes generated $\sigma$-algebra.

Proof
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:


 * $\sigma \left({\mathcal{J}^n_{ho}}\right) = \sigma \left({\mathcal{O}^n}\right)$