求证k维正则曲面上的任意点必为边界点或内点

Let’s break this down step by step, starting with formal definitions aligned with your problem statement, then moving to the proof itself.

Key Definitions #

First, restating the standard definition of a k-dimensional regular surface with boundary in ( \mathbb{R}^n ):

A set ( S \subset \mathbb{R}^n ) is a k-dimensional regular surface with boundary if for every ( p \in S ), there exists:

1. An open neighborhood ( V \subset \mathbb{R}^n ) containing ( p ),
2. An open set ( U \subset \mathbb{R}^k ),
3. An injective ( C^1 ) map ( \varphi: U \to \mathbb{R}^n ) with full rank (rank ( k )) at every point in ( U ),
   such that ( \varphi(U) = V \cap S ), and exactly one of the following holds for ( p ):

• Interior Point: ( p \in \varphi(\text{int}(\mathbb{R}^k_+)) ), where ( \mathbb{R}^k_+ = { (x_1,...,x_k) \mid x_k \geq 0 } ) and ( \text{int}(\mathbb{R}^k_+) ) is its interior (all points with ( x_k > 0 )).
• Boundary Point: ( p \in \varphi(\partial(\mathbb{R}^k_+)) ), where ( \partial(\mathbb{R}^k_+) = { (x_1,...,x_k) \mid x_k = 0 } ).

Proof #

Take any arbitrary point ( p \in S ). By the definition above, there exists a chart ( (\varphi, U) ) covering ( p )—meaning ( \varphi ) maps ( U ) to ( V \cap S ), and ( q = \varphi^{-1}(p) \in U ).

We have two mutually exclusive and exhaustive cases for ( q ):

Case 1: ( q \in \text{int}(\mathbb{R}^k_+) ) #

Since ( U ) is open in ( \mathbb{R}^k ), we can find an open ball ( B(q, \epsilon) \subset U ) where every point in the ball has ( x_k > 0 ). The image ( \varphi(B(q, \epsilon)) ) is an open subset of ( S ) (because ( \varphi ) is a homeomorphism onto its image) that contains ( p ). By definition, this means ( p ) has an open neighborhood entirely within ( S )—so ( p ) is an interior point.

Case 2: ( q \in \partial(\mathbb{R}^k_+) ) #

Here, ( q ) has coordinates ( (x_1,...,x_{k-1}, 0) ). Any open neighborhood of ( q ) in ( U ) will include points from ( \text{int}(\mathbb{R}^k_+) ) (with ( x_k > 0 )) but no points from the lower half-space (since ( \varphi ) only maps ( U \cap \mathbb{R}^k_+ ) to ( S )). The image of those ( x_k > 0 ) points are interior points of ( S ), but ( p ) itself cannot be contained in an open subset of ( S ) that’s entirely mapped from ( \text{int}(\mathbb{R}^k_+) ). This fits the definition of a boundary point of ( S ).

Since these two cases cover all possible positions of ( q ), and they can’t both be true for the same ( q ), every point ( p \in S ) must be either an interior point or a boundary point—there’s no third possibility.

内容的提问来源于stack exchange，提问作者rubikscube09