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

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## Statement

Let [ilmath]X\in\mathcal{P}(\mathbb{R}^n)[/ilmath] be an arbitrary subset of [ilmath]\mathbb{R}^n[/ilmath][Note 1], then, if [ilmath]X[/ilmath] is compact and convex, and has a non-empty interior then[1]:

Furthermore, given any point [ilmath]p\in\text{Int}(X)[/ilmath], there exists a homeomorphism, [ilmath]f:\overline{\mathbb{B}^n}\rightarrow X[/ilmath] (where [ilmath]\overline{\mathbb{B}^n} [/ilmath] is the closed unit ball[Note 2] in [ilmath]\mathbb{R}^n[/ilmath]) such that:

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

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

## Proof

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