The Invariance of Boundary Theorem

The Invariance of Boundary Theorem

Given, $M$, a manifold with a boundary, the interior and boundary are disjoint. (from Mathematical Foundations of Artificial Intelligence: Basics of Manifold Theory*)

Proof

Assume, for contradiction that $p$ is in Int $M$ and on $∂M$. Note that by definition, for a point $p$ to be in the interior of $M$ means there's a homeomorphism $\varphi$: Int $M \to \{(x_1, \dots, x_n) \in \mathbb{R}^n_{+} | x_n > 0 \}$. Recall that a point $p$ is on the boundary of $M$ means there's a homeomorphism $\psi: ∂M \to \{(x_1, \dots, x_n) \in \mathbb{R}^n_{+} | x_n = 0 \}$.

At $p$ there must be a transition map, $\psi(\varphi^{-1})$ that maps from $\{(x_1, \dots, x_n) \in \mathbb{R}^n_{+} | x_n > 0 \}$ to $\{(x_1, \dots, x_n) \in \mathbb{R}^n_{+} | x_n = 0 \}$. Since the transition map is a composition of bijections, it too, is a bijection.

Let $y,z \gt 0$ and $y \neq z$. Consider the points $p_1 = (x_1, \dots, x_{n-1}, y)$ and $p_2 = (x_1, \dots, x_{n-1}, z)$. Note that $p_1 \neq p_2$, but our transition map sends both $p1$ and $p_2$ to the same point $(x_1, \dots, x_{n-1}, 0)$. Thus we have our contradiction and reject the assumption that a point $p$ can be both in Int $M$ and on $∂M$. This concludes the proof. $\blacksquare$

*Xiong, M. (2026). Mathematical Foundations of Artificial Intelligence: Basics of Manifold Theory (1st ed.). page 29. Chapman and Hall/CRC. https://doi.org/10.1201/9781003641452