Half-Open Rectangles Closed under Intersection

Theorem
Let $\lefthalf-open $n$-rectangles.

Then $\lefthalf-open $n$-rectangle.

Proof
As $\varnothing$ is trivially a half-open $n$-rectangle, let us assume that:


 * $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf c \,.\,.\, \mathbf d}\right)) \ne \varnothing$

By Cartesian Product of Intersections: General Case, it follows that:


 * $\displaystyle \left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf c \,.\,.\, \mathbf d}\right)) = \prod_{i \mathop = 1}^n \left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right)$

which leaves only to prove that the $\left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right)$ are half-open intervals.

Now let $x \in \left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right)$.

Then $x$ is subject to:


 * $x \ge a_i$ and $x \ge c_i$, i.e., $x \ge \max \left\{{a_i, c_i}\right\}$
 * $x < b_i$ and $x < d_i$, i.e., $x < \min \left\{{b_i, d_i}\right\}$

and we see that these conditions are satisfied precisely when:


 * $x \in \left[{\max \left\{{a_i, c_i}\right\} \,.\,.\, \min \left\{{b_i, d_i}\right\}}\right)$

Thus, we conclude:


 * $\left[{a_i \,.\,.\, b_i}\right) \cap \left[{c_i \,.\,.\, d_i}\right) = \left[{\max \left\{{a_i, c_i}\right\} \,.\,.\, \min \left\{{b_i, d_i}\right\}}\right)$

showing that indeed the intersection is a half-open interval.

Combining this with the above reasoning, it follows that indeed:


 * $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf c \,.\,.\, \mathbf d}\right))$

is again a half-open $n$-rectangle.