求证$R^2$欧几里得拓扑基中开矩形、三角形的交并关系

Hey there! Let's break this down step by step—no hand-wavy geometric arguments, just solid inequalities and set theory to make everything rigorous.

Part 1: Intersections of Axis-Aligned Open Rectangles #

First, let's formalize what an axis-aligned open rectangle in ( \mathbb{R}^2 ) is: it's any set of the form ( R = (a, b) \times (c, d) ), where ( a < b ) and ( c < d ) are real numbers. This means every point ( (x, y) \in R ) satisfies ( a < x < b ) and ( c < y < d ).

Finite Intersections (the relevant case for topology bases) #

Suppose we have finite axis-aligned open rectangles ( R_1, R_2, ..., R_k ), where each ( R_i = (a_i, b_i) \times (c_i, d_i) ). Their intersection is:
[
\bigcap_{i=1}^k R_i = \left( \max{a_1, a_2, ..., a_k}, \min{b_1, b_2, ..., b_k} \right) \times \left( \max{c_1, c_2, ..., c_k}, \min{d_1, d_2, ..., d_k} \right)
]

We have two cases to check:

• If ( \max{a_i} < \min{b_i} ) and ( \max{c_i} < \min{d_i} ): The intersection is itself an axis-aligned open rectangle, so it's trivially a union of one such rectangle.
• If either ( \max{a_i} \geq \min{b_i} ) or ( \max{c_i} \geq \min{d_i} ): The intersection is the empty set, which can be represented as the union of zero rectangles (a standard edge case in set theory).

Note: Infinite intersections of open rectangles don't always work (e.g., ( \bigcap_{n=1}^\infty (-1/n, 1/n) \times (-1/n, 1/n) = {(0,0)} ), a single point that can't be written as a union of open rectangles). But since topology bases only require closure under finite intersections, this is the case we care about for your proof.

Part 2: Intersections of Open Triangles #

First, let's formalize an open triangle in ( \mathbb{R}^2 ): it's the set of all interior points of a triangle, which can be written as the intersection of three open half-planes. Each open half-plane is defined by a strict linear inequality: for a triangle ( T ), there exist real numbers ( A_j, B_j, C_j ) (not all zero for each ( j )) such that:
[
T = {(x, y) \in \mathbb{R}^2 \mid A_1x + B_1y + C_1 < 0, A_2x + B_2y + C_2 < 0, A_3x + B_3y + C_3 < 0}
]

Step 1: Any open triangle is a union of axis-aligned open rectangles #

Take any point ( (x_0, y_0) \in T ). Since ( (x_0, y_0) ) is in each open half-plane, we know:
[
A_1x_0 + B_1y_0 + C_1 = -\epsilon_1 < 0, \quad A_2x_0 + B_2y_0 + C_2 = -\epsilon_2 < 0, \quad A_3x_0 + B_3y_0 + C_3 = -\epsilon_3 < 0
]
where ( \epsilon_1, \epsilon_2, \epsilon_3 > 0 ).

We need to find a ( \delta > 0 ) such that the axis-aligned open rectangle ( R = (x_0 - \delta, x_0 + \delta) \times (y_0 - \delta, y_0 + \delta) ) is entirely contained in ( T ).

For the first half-plane inequality, consider any ( (x, y) \in R ):
[
A_1x + B_1y + C_1 = A_1(x - x_0) + B_1(y - y_0) + (A_1x_0 + B_1y_0 + C_1) = A_1(x - x_0) + B_1(y - y_0) - \epsilon_1
]
Using the triangle inequality:
[
|A_1(x - x_0) + B_1(y - y_0)| \leq |A_1||x - x_0| + |B_1||y - y_0| < (|A_1| + |B_1|)\delta
]
If we choose ( \delta < \frac{\epsilon_1}{|A_1| + |B_1|} ), this ensures ( A_1x + B_1y + C_1 < 0 ).

Repeat this logic for the other two inequalities, choosing ( \delta < \frac{\epsilon_2}{|A_2| + |B_2|} ) and ( \delta < \frac{\epsilon_3}{|A_3| + |B_3|} ). Finally, take ( \delta = \min\left( \frac{\epsilon_1}{|A_1| + |B_1|}, \frac{\epsilon_2}{|A_2| + |B_2|}, \frac{\epsilon_3}{|A_3| + |B_3|} \right) ). This ( \delta > 0 ) gives us a rectangle ( R ) where ( (x_0, y_0) \in R \subseteq T ).

Since every point in ( T ) is contained in such a rectangle, ( T ) is the union of all these axis-aligned open rectangles.

Step 2: Finite intersections of open triangles #

If we have finite open triangles ( T_1, T_2, ..., T_k ), their intersection ( T = T_1 \cap T_2 \cap ... \cap T_k ) is an open set (finite intersections of open sets are open in Euclidean topology).

For any ( (x_0, y_0) \in T ), we can find a rectangle ( R_i \subseteq T_i ) containing ( (x_0, y_0) ) for each ( i ). Take ( \delta = \min(\delta_1, \delta_2, ..., \delta_k) ) where ( \delta_i ) is the value from Step 1 for ( T_i ). The rectangle ( R = (x_0 - \delta, x_0 + \delta) \times (y_0 - \delta, y_0 + \delta) ) is contained in every ( T_i ), so ( R \subseteq T ).

Again, every point in ( T ) is in such a rectangle, so ( T ) is a union of axis-aligned open rectangles.

内容的提问来源于stack exchange，提问作者tobiasbriones