A compact and convex subset of Euclidean n-space with non-empty interior is a closed n-cell and its interior is an open n-cell/Statement

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Sort out interior and boundary stuff!

Statement

Let $X\in\mathcal{P}(\mathbb{R}^n)$ be an arbitrary subset of $\mathbb{R}^n$, then, if $X$ is compact and convex, and has a non-empty interior then:

• $X$ is a closed $n$-cell and its interior is an open $n$-cell

Furthermore, given any point $p\in\text{Int}(X)$, there exists a homeomorphism, $f:\overline{\mathbb{B}^n}\rightarrow X$ (where $\overline{\mathbb{B}^n}$ is the closed unit ball in $\mathbb{R}^n$) such that:

1. $f(0)\eq p$
2. $f\left(\mathbb{B}^n\right)\eq\text{Int}(X)$ (where $\mathbb{B}^n$ is the open unit ball in $\mathbb{R}^n$), and
3. $f(\mathbb{S}^{n-1})\eq\partial X$ (where $\mathbb{S}^{n-1}\subset\mathbb{R}^n$ is the $(n-1)$-sphere, and $\partial X$ denotes the boundary of $X$)

Caveat:When we speak of interior and boundary here, we mean considered as a subset of $\mathbb{R}^n$, not as $X$ itself against the subspace topology on $X$