Can every compact subset of $\mathbb R^n$ be realized as a manifold with boundary?

Question

I think it should be true since in Lee's introduction to smooth manifolds, he referred to a differential form on a compact subset $D \subset \mathbb R^n$. It is easy to provide charts for $\operatorname{Int} D$, but for $\partial D$, how should I set up the charts for it?

Answer 1

No. Just look at the Cantor set in $\mathbb R$ as an example. It has no neighborhoods homeomorphic to any Euclidean space (or half space), but it is compact.

For a less pathological example, consider $$\left\{\frac1n:n\in \mathbb N\right\}\cup\{0\}$$ This is compact, but no neighborhood of $0$ is homeomorphic to Euclidean space or half space.

Answer 2

I'm guessing that you're referring to Chapter 16 ("Integration on Manifolds") in ISM, where I talk about the integral of a differential form over a domain of integration in $\mathbb R^n$. This is defined as follows (see page 402): Suppose $D\subseteq\mathbb R^n$ is a domain of integration (meaning a bounded subset whose boundary has measure zero). An $n$-form on $D$ is just a continuous section of the bundle $\Lambda^n(\mathbb R^n)$ defined only at points of $D$. Such a form can be written $\omega = f\,dx^1\wedge\dots\wedge dx^n$ for some continuous function $f\colon D\to \mathbb R$. Given such a form $\omega$, we define the integral of $\boldsymbol \omega$ over $\boldsymbol D$ as follows: $$ \int_D\omega = \int_D f(x^1,\dots,x^n)\, dx^1\dotsm dx^n, $$ where the right-hand integral is to be interpreted in the ordinary sense of multivariable calculus.

There is a smooth manifold structure here, but it's the structure of $\mathbb R^n$, not $D$ itself.