# [ilmath]n[/ilmath]-cell

**Stub grade: A**

## 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]}.

- Often [ilmath]X[/ilmath] will be a subset of another topological space and the topology will be the subspace topology [ilmath]X[/ilmath] inherits from it.

### 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]}.

- Often [ilmath]X[/ilmath] will be a subset of another topological space and the topology will be the subspace topology [ilmath]X[/ilmath] inherits from it.

## 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]}:

- [ilmath]X[/ilmath] is a closed [ilmath]n[/ilmath]-cell and its interior is an open [ilmath]n[/ilmath]-cell

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:

- [ilmath]f(0)\eq p[/ilmath]
- [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 - [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

- ↑ 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]

- ↑ 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

- ↑ Considered with its usual topology. Given by the Euclidean norm of course
- ↑ 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]

- [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:
- ↑ 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

- ↑
^{1.0}^{1.1}^{1.2}Introduction to Topological Manifolds - John M. Lee