# [ilmath]n[/ilmath]-cell

(Redirected from Closed n-cell)
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Looking alright so far...

## Definitions

### Closed [ilmath]n[/ilmath]-cell

A topological space, [ilmath](X,\mathcal{ J })[/ilmath], is a "closed [ilmath]n[/ilmath]-cell" if it is homeomorphic to the closed unit ball[Note 1] in [ilmath]\mathbb{R}^n[/ilmath][1].

### Open [ilmath]n[/ilmath]-cell

A topological space, [ilmath](X,\mathcal{ J })[/ilmath], is a "open [ilmath]n[/ilmath]-cell" if it is homeomorphic to the open unit ball[Note 2] in [ilmath]\mathbb{R}^n[/ilmath][1].

## Characterisations

Caveat:There are (probably) other characterisations; like I should suspect any closed compact connected set with non-empty interior is a closed [ilmath]n[/ilmath]-cell

### If [ilmath]X\in\mathcal{P}(\mathbb{R}^n)[/ilmath] is compact and convex in [ilmath]\mathbb{R}^n[/ilmath] with non-empty interior then it is a closed [ilmath]n[/ilmath]-cell and its interior is an open [ilmath]n[/ilmath]-cell

Let [ilmath]X\in\mathcal{P}(\mathbb{R}^n)[/ilmath] be an arbitrary subset of [ilmath]\mathbb{R}^n[/ilmath][Note 3], 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 4] 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 5] 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]

## Notes

1. The closed unit ball, often denoted: [ilmath]\overline{\mathbb{B}^n} [/ilmath] or [ilmath]\overline{\mathbb{B} }^n[/ilmath], is a closed ball of radius [ilmath]1[/ilmath] based at the origin.
• [ilmath]\overline{\mathbb{B}^n}:\eq\{x\in\mathbb{R}^n\ \vert\ \Vert x\Vert\le 1\} [/ilmath], where the norm used is the standard Euclidean norm: [ilmath]\Vert x\Vert:\eq\sqrt{\sum^n_{i\eq 1} x_i^2} [/ilmath]
2. The open unit ball, denoted: [ilmath]\mathbb{B}^n [/ilmath], is an open ball of radius [ilmath]1[/ilmath] based at the origin.
• [ilmath]\mathbb{B}^n:\eq\{x\in\mathbb{R}^n\ \vert\ \Vert x\Vert < 1\} [/ilmath], where the norm is the same as it is in the note for the closed unit ball
3. Considered with its usual topology. Given by the Euclidean norm of course
4. 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]
5. 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.