Open Rectangles Closed under Intersection

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Theorem

Let $\left(({\mathbf a \,.\,.\, \mathbf b}\right))$ and $\left(({\mathbf c \,.\,.\, \mathbf d}\right))$ be open $n$-rectangles.


Then $\left(({\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left(({\mathbf c \,.\,.\, \mathbf d}\right))$ is also an open $n$-rectangle.


Proof

From Cartesian Product of Intersections: General Case, we have:

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


Therefore, it suffices to show that the intersection of two open intervals is again an open interval.

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

Then $x$ is subject to:

  • $x > a_i$ and $x > c_i$, i.e., $x > \max \left\{{a_i, c_i}\right\}$
  • $x < b_i$ 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 an 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 an open $n$-rectangle.

$\blacksquare$