December 21, 2019

Proposition

Let $A \subset \mathbb{R}^n$ be an open set such that boundary $A$ is an $(n - 1)$-dimensional manifold. Show that $N = A \cup \text{boundary} A$ is an $n$-dimensional manifold-with-boundary.
Prove a similar assertion for an open subset of an $n$-dimensional manifold.

Solution

1

Let $x \in N$.

Case 1: $x$ is an interior point of $N$.
- Let $U$ be a neighborhood of $x$ contained in $N$. Then $\Id_U$ is a diffeomorphism from $U$ to $U$. Thus $x$ satisfies the condition (M) in [Spivak].
Case 2: $x$ is a boundary point of $N$.

Then $x$ is a boundary point of $A$.

For every open neighborhood $U$ of $x$, $U \cap A^c \supset U \cap N^c \ne \emptyset$.
Let $U$ be an open neighborhood of $x$. Then $U \cap N \ne \emptyset$. Let $y \in U \cap N$. If $y \in A$, then $U \cap A \ne \emptyset$. If $y \notin A$, then $y \in \bd(A)$, so $U \cap A \ne \emptyset$.

Therefore, $x$ is part of the $(n - 1)$-dimensional manifold.

TODO (Finish this proof)

2

TODO (Finish this)