Locally Euclidean topological space of dimension [ilmath]n[/ilmath]

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See Locally euclidean, it's the same thing but with [ilmath]n[/ilmath] fixed before the [ilmath]\forall p\in x[/ilmath] part. Alec (talk) 17:07, 19 February 2017 (UTC)

• Fleshed out a little bit but more work is needed Alec (talk) 12:47, 20 February 2017 (UTC)

Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]n\in\mathbb{N}_0[/ilmath] be given. We say that [ilmath]X[/ilmath] is locally Euclidean of dimension [ilmath]n[/ilmath] if:

• [ilmath]\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big][/ilmath]
• Caveat:Or perhaps...
  • [ilmath]\exists n\in\mathbb{N}_0\forall p\in X\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists \varphi\in\mathcal{F}\big(U,B_\epsilon(0;\mathbb{R}^n)\big)\big[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)\big][/ilmath]
• Where the dimension, [ilmath]n[/ilmath], is the [ilmath]n[/ilmath] that must exist in the first quantifying clause.

TODO: Verdict needed after investigation

Equivalent definitions

We posit that there must be an open ball of radius [ilmath]\epsilon[/ilmath] about [ilmath]0\in\mathbb{R}^n[/ilmath], it actually works if:

1. We require there to be any open set containing [ilmath]p[/ilmath] to be homeomorphic to any open set of [ilmath]\mathbb{R}^n[/ilmath]
2. We require there be an open set containing [ilmath]p[/ilmath] homeomorphic to all of [ilmath]\mathbb{R}^n[/ilmath]
3. We require there be an open set containing [ilmath]p[/ilmath] homeomorphic to the open unit ball, [ilmath]\mathbb{B}^n[/ilmath]

See the Locally euclidean page for more information.