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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See also

TODO: todo

Notes

  1. ↑ Considered with its usual topology. Given by the Euclidean norm of course
  2. ↑ Recall the closed unit ball is:
    • [ilmath]\overline{\mathbb{B}^n}:\eq\{x\in\mathbb{R}^n\ \vert\ \Vert x\Vert\le 1\} [/ilmath] where the norm is the usual Euclidean norm:
      • [ilmath]\Vert x\Vert:\eq\sqrt{\sum^n_{i\eq 1}x_i^2} [/ilmath]
  3. ↑ As before:
    • [ilmath]\mathbb{B}^n:\eq\{x\in\mathbb{R}^n\ \vert\ \Vert x\Vert<1\} [/ilmath], with the Euclidean norm mentioned in the note for closed unit ball above.

References

  1. ↑ Introduction to Topological Manifolds - John M. Lee


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